1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Chapter 11 Additional Topics andApplications
11.1 Quadratic Forms
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Definition
DefinitionThe quadratic form is a function Q : Rn → R that has theform
Q(x) = xTAx
where A is a symmetric n × n matrix called thematrix of the quadratic form.
Facts About Symmetric Matrices
I The eigenvalues of a symmetric matrix are real
I There is a set of orthonormal eigenvectors of A
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Definition
DefinitionThe quadratic form is a function Q : Rn → R that has theform
Q(x) = xTAx
where A is a symmetric n × n matrix called thematrix of the quadratic form.
Facts About Symmetric Matrices
I The eigenvalues of a symmetric matrix are real
I There is a set of orthonormal eigenvectors of A
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Definition
DefinitionThe quadratic form is a function Q : Rn → R that has theform
Q(x) = xTAx
where A is a symmetric n × n matrix called thematrix of the quadratic form.
Facts About Symmetric Matrices
I The eigenvalues of a symmetric matrix are real
I There is a set of orthonormal eigenvectors of A
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 1
Example
Evaluate the quadratic form Q : R3 → R at the vector
x0 =
21−3
for
Q(x) = 3x21 + 2x2
2 − x23 − 4x1x2 + 2x2x3
Q(x0) = Q
21−3
= 3(2)2 + 2(1)2 − (−3)2 − 4(2)(1) + 2(1)(−3)
= −9
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 1
Example
Evaluate the quadratic form Q : R3 → R at the vector
x0 =
21−3
for
Q(x) = 3x21 + 2x2
2 − x23 − 4x1x2 + 2x2x3
Q(x0) = Q
21−3
= 3(2)2 + 2(1)2 − (−3)2 − 4(2)(1) + 2(1)(−3)
= −9
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 1
Example
Evaluate the quadratic form Q : R3 → R at the vector
x0 =
21−3
for
Q(x) = 3x21 + 2x2
2 − x23 − 4x1x2 + 2x2x3
Q(x0) = Q
21−3
= 3(2)2 + 2(1)2 − (−3)2 − 4(2)(1) + 2(1)(−3)
= −9
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 2
Example
Evaluate the quadratic form Q : R3 → R at the vector
x0 =
1−12
for
Q(x) = 6x21 + x2
2 + 3x23 + 6x1x3 − 2x2x3
Q(x0) = Q
1−12
= 6(1)2 + (−1)2 + 3(2)2 + 6(1)(2)− 2(−1)(2)
= 35
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 2
Example
Evaluate the quadratic form Q : R3 → R at the vector
x0 =
1−12
for
Q(x) = 6x21 + x2
2 + 3x23 + 6x1x3 − 2x2x3
Q(x0) = Q
1−12
= 6(1)2 + (−1)2 + 3(2)2 + 6(1)(2)− 2(−1)(2)
= 35
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 2
Example
Evaluate the quadratic form Q : R3 → R at the vector
x0 =
1−12
for
Q(x) = 6x21 + x2
2 + 3x23 + 6x1x3 − 2x2x3
Q(x0) = Q
1−12
= 6(1)2 + (−1)2 + 3(2)2 + 6(1)(2)− 2(−1)(2)
= 35
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 3
Example
Find the matrix A associated with the quadratic form
Q(x) = 3x21 + 2x2
2 − x23 − 4x1x2 + 2x2x3
Diagonal Matrices
If A is diagonal
A =
a1 . . . 00 a2 . . ....
......
0 . . . an
then xTAx = c is equivalent to a1x
21 + . . .+ anx
2n = c
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 3
Example
Find the matrix A associated with the quadratic form
Q(x) = 3x21 + 2x2
2 − x23 − 4x1x2 + 2x2x3
Diagonal Matrices
If A is diagonal
A =
a1 . . . 00 a2 . . ....
......
0 . . . an
then xTAx = c is equivalent to a1x
21 + . . .+ anx
2n = c
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 3
So, we know that the matrix we seek has the form
A =
32−1
Now, the off diagonal entries ... we need to look at theindices of the non-squared terms, i.e. −4x1x2 + 2x2x3
Remember, the matrix A is symmetric, so we need to usewhat we have to determine values for the (1, 2) and (2, 3)positions, which will in turn give us the (2, 1) and (3, 2)positions.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 3
So, we know that the matrix we seek has the form
A =
32−1
Now, the off diagonal entries ... we need to look at theindices of the non-squared terms, i.e. −4x1x2 + 2x2x3
Remember, the matrix A is symmetric, so we need to usewhat we have to determine values for the (1, 2) and (2, 3)positions, which will in turn give us the (2, 1) and (3, 2)positions.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 3
So, we know that the matrix we seek has the form
A =
32−1
Now, the off diagonal entries ... we need to look at theindices of the non-squared terms, i.e. −4x1x2 + 2x2x3
Remember, the matrix A is symmetric, so we need to usewhat we have to determine values for the (1, 2) and (2, 3)positions, which will in turn give us the (2, 1) and (3, 2)positions.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 3
So, at this point, we have
A =
3 02
0 −1
For the (1, 2) and (2, 1) positions, we need to evenly splitthe coefficient of the x1x2 term.
A =
3 −2 0−2 20 −1
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 3
So, at this point, we have
A =
3 02
0 −1
For the (1, 2) and (2, 1) positions, we need to evenly splitthe coefficient of the x1x2 term.
A =
3 −2 0−2 20 −1
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 3
So, at this point, we have
A =
3 02
0 −1
For the (1, 2) and (2, 1) positions, we need to evenly splitthe coefficient of the x1x2 term.
A =
3 −2 0−2 20 −1
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 3
Similarly, we find the (2, 3) and (3, 2) entries in the samemanner.
A =
3 −2 0−2 2 10 1 −1
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 3
Similarly, we find the (2, 3) and (3, 2) entries in the samemanner.
A =
3 −2 0−2 2 10 1 −1
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Why Does This Work?
xTAx =[x1 x2 x3
] a11 a12 a13
a21 a22 a23
a31 a32 a33
x1
x2
x3
=[x1 x2 x3
] a11x1 + a12x2 + a13x3
a21x1 + a22x2 + a23x3
a31x1 + a32x2 + a33x3
= a11x
21 + a12x2x1 + a13x3x1
+a21x1x2 + a22x22 + a23x3x2
+a31x1x3 + a32x2x3 + a33x23
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Why Does This Work?
xTAx =[x1 x2 x3
] a11 a12 a13
a21 a22 a23
a31 a32 a33
x1
x2
x3
=[x1 x2 x3
] a11x1 + a12x2 + a13x3
a21x1 + a22x2 + a23x3
a31x1 + a32x2 + a33x3
= a11x21 + a12x2x1 + a13x3x1
+a21x1x2 + a22x22 + a23x3x2
+a31x1x3 + a32x2x3 + a33x23
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Why Does This Work?
xTAx =[x1 x2 x3
] a11 a12 a13
a21 a22 a23
a31 a32 a33
x1
x2
x3
=[x1 x2 x3
] a11x1 + a12x2 + a13x3
a21x1 + a22x2 + a23x3
a31x1 + a32x2 + a33x3
= a11x
21 + a12x2x1 + a13x3x1
+a21x1x2 + a22x22 + a23x3x2
+a31x1x3 + a32x2x3 + a33x23
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 4
Example
Find the matrix A associated with the quadratic form
Q(x) = 6x21 + x2
2 + 3x23 + 6x1x3 − 2x2x3
A =
6 0 30 1 −13 −1 3
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 4
Example
Find the matrix A associated with the quadratic form
Q(x) = 6x21 + x2
2 + 3x23 + 6x1x3 − 2x2x3
A =
6 0 30 1 −13 −1 3
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 5
Example
Find a formula for the quadratic form with the matrix
A =
2 3 83 8 18 1 −2
Q(x) = 2x21 + 8x2
2 − 2x23 + 6x1x2 + 16x1x3 + 2x2x3
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 5
Example
Find a formula for the quadratic form with the matrix
A =
2 3 83 8 18 1 −2
Q(x) = 2x21 + 8x2
2 − 2x23
+ 6x1x2 + 16x1x3 + 2x2x3
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 5
Example
Find a formula for the quadratic form with the matrix
A =
2 3 83 8 18 1 −2
Q(x) = 2x21 + 8x2
2 − 2x23 + 6x1x2 + 16x1x3 + 2x2x3
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Classifying Quadratic Form
DefinitionSuppose Q : Rn → R is a quadratic form.
(a) A is positive definite if Q(x) > 0 for all nonzero vectorsx ∈ Rn and positive semidefinite if Q(x) ≥ 0 for allnonzero vectors x ∈ Rn
(b) A is negative definite if Q(x) < 0 for all nonzero vectorsx ∈ Rn and negative semidefinite if Q(x) ≤ 0 for allnonzero vectors x ∈ Rn
(c) A is indefinite if Q(x) is positive for some x ∈ Rn andnegative for others.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Classifying Quadratic Form
This would be tough to verify if we didn’t have a better waythan checking all vectors ...
TheoremSuppose Q : Rn → R is a quadratic form and A is the matrixof Q. Then
(a) A is positive definite iff all of the eigenvalues for A arepositive.
(b) A is negative definite iff all of the eigenvalues for A arenegative.
(c) A is indefinite iff A has both positive and negativeeigenvalues.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Classifying Quadratic Form
This would be tough to verify if we didn’t have a better waythan checking all vectors ...
TheoremSuppose Q : Rn → R is a quadratic form and A is the matrixof Q. Then
(a) A is positive definite iff all of the eigenvalues for A arepositive.
(b) A is negative definite iff all of the eigenvalues for A arenegative.
(c) A is indefinite iff A has both positive and negativeeigenvalues.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Classifying Quadratic Form
This would be tough to verify if we didn’t have a better waythan checking all vectors ...
TheoremSuppose Q : Rn → R is a quadratic form and A is the matrixof Q. Then
(a) A is positive definite iff all of the eigenvalues for A arepositive.
(b) A is negative definite iff all of the eigenvalues for A arenegative.
(c) A is indefinite iff A has both positive and negativeeigenvalues.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Classifying Quadratic Form
This would be tough to verify if we didn’t have a better waythan checking all vectors ...
TheoremSuppose Q : Rn → R is a quadratic form and A is the matrixof Q. Then
(a) A is positive definite iff all of the eigenvalues for A arepositive.
(b) A is negative definite iff all of the eigenvalues for A arenegative.
(c) A is indefinite iff A has both positive and negativeeigenvalues.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 6
Example
Classify the quadratic form of Q(x) = 5x21 + 3x2
2 + 4x1x2.
What do we need to do first?
A =
[5 22 3
]|A− λI | =
∣∣∣∣ 5− λ 22 3− λ
∣∣∣∣= (5− λ)(3− λ)− 4 = 0
λ2 − 8λ+ 11 = 0
λ =8±√
64− 44
2
= 4±√
5 > 0
So, A is positive definite.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 6
Example
Classify the quadratic form of Q(x) = 5x21 + 3x2
2 + 4x1x2.
What do we need to do first?
A =
[5 22 3
]|A− λI | =
∣∣∣∣ 5− λ 22 3− λ
∣∣∣∣= (5− λ)(3− λ)− 4 = 0
λ2 − 8λ+ 11 = 0
λ =8±√
64− 44
2
= 4±√
5 > 0
So, A is positive definite.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 6
Example
Classify the quadratic form of Q(x) = 5x21 + 3x2
2 + 4x1x2.
What do we need to do first?
A =
[5 22 3
]
|A− λI | =
∣∣∣∣ 5− λ 22 3− λ
∣∣∣∣= (5− λ)(3− λ)− 4 = 0
λ2 − 8λ+ 11 = 0
λ =8±√
64− 44
2
= 4±√
5 > 0
So, A is positive definite.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 6
Example
Classify the quadratic form of Q(x) = 5x21 + 3x2
2 + 4x1x2.
What do we need to do first?
A =
[5 22 3
]|A− λI | =
∣∣∣∣ 5− λ 22 3− λ
∣∣∣∣
= (5− λ)(3− λ)− 4 = 0
λ2 − 8λ+ 11 = 0
λ =8±√
64− 44
2
= 4±√
5 > 0
So, A is positive definite.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 6
Example
Classify the quadratic form of Q(x) = 5x21 + 3x2
2 + 4x1x2.
What do we need to do first?
A =
[5 22 3
]|A− λI | =
∣∣∣∣ 5− λ 22 3− λ
∣∣∣∣= (5− λ)(3− λ)− 4 = 0
λ2 − 8λ+ 11 = 0
λ =8±√
64− 44
2
= 4±√
5 > 0
So, A is positive definite.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 6
Example
Classify the quadratic form of Q(x) = 5x21 + 3x2
2 + 4x1x2.
What do we need to do first?
A =
[5 22 3
]|A− λI | =
∣∣∣∣ 5− λ 22 3− λ
∣∣∣∣= (5− λ)(3− λ)− 4 = 0
λ2 − 8λ+ 11 = 0
λ =8±√
64− 44
2
= 4±√
5 > 0
So, A is positive definite.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 6
Example
Classify the quadratic form of Q(x) = 5x21 + 3x2
2 + 4x1x2.
What do we need to do first?
A =
[5 22 3
]|A− λI | =
∣∣∣∣ 5− λ 22 3− λ
∣∣∣∣= (5− λ)(3− λ)− 4 = 0
λ2 − 8λ+ 11 = 0
λ =8±√
64− 44
2
= 4±√
5 > 0
So, A is positive definite.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 6
Example
Classify the quadratic form of Q(x) = 5x21 + 3x2
2 + 4x1x2.
What do we need to do first?
A =
[5 22 3
]|A− λI | =
∣∣∣∣ 5− λ 22 3− λ
∣∣∣∣= (5− λ)(3− λ)− 4 = 0
λ2 − 8λ+ 11 = 0
λ =8±√
64− 44
2
= 4±√
5
> 0
So, A is positive definite.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 6
Example
Classify the quadratic form of Q(x) = 5x21 + 3x2
2 + 4x1x2.
What do we need to do first?
A =
[5 22 3
]|A− λI | =
∣∣∣∣ 5− λ 22 3− λ
∣∣∣∣= (5− λ)(3− λ)− 4 = 0
λ2 − 8λ+ 11 = 0
λ =8±√
64− 44
2
= 4±√
5 > 0
So, A is positive definite.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 6
Example
Classify the quadratic form of Q(x) = 5x21 + 3x2
2 + 4x1x2.
What do we need to do first?
A =
[5 22 3
]|A− λI | =
∣∣∣∣ 5− λ 22 3− λ
∣∣∣∣= (5− λ)(3− λ)− 4 = 0
λ2 − 8λ+ 11 = 0
λ =8±√
64− 44
2
= 4±√
5 > 0
So, A is positive definite.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Principle Axes Theorem
TheoremIf A is a symmetric matrix then there is an orthogonal matrixP such that the transform y = PTx changes the quadraticform xTAx into the quadratic form yTDy (where D isdiagonal) that has no cross-product terms.
Do we remember this form? (chapter 6 ...)
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Principle Axes Theorem
TheoremIf A is a symmetric matrix then there is an orthogonal matrixP such that the transform y = PTx changes the quadraticform xTAx into the quadratic form yTDy (where D isdiagonal) that has no cross-product terms.
Do we remember this form? (chapter 6 ...)
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
Example
Find the change of coordinates necessary to express thequadratic form with matrix
A =
[3 11 3
]as a quadratic form with no cross-product terms.
Who remembers what to do?
|A− λI | =
∣∣∣∣ 3− λ 11 3− λ
∣∣∣∣(3− λ)2 − 1 = 0
λ2 − 6λ+ 8 = 0
λ = 2, 4
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
Example
Find the change of coordinates necessary to express thequadratic form with matrix
A =
[3 11 3
]as a quadratic form with no cross-product terms.
Who remembers what to do?
|A− λI | =
∣∣∣∣ 3− λ 11 3− λ
∣∣∣∣(3− λ)2 − 1 = 0
λ2 − 6λ+ 8 = 0
λ = 2, 4
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
Example
Find the change of coordinates necessary to express thequadratic form with matrix
A =
[3 11 3
]as a quadratic form with no cross-product terms.
Who remembers what to do?
|A− λI | =
∣∣∣∣ 3− λ 11 3− λ
∣∣∣∣
(3− λ)2 − 1 = 0
λ2 − 6λ+ 8 = 0
λ = 2, 4
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
Example
Find the change of coordinates necessary to express thequadratic form with matrix
A =
[3 11 3
]as a quadratic form with no cross-product terms.
Who remembers what to do?
|A− λI | =
∣∣∣∣ 3− λ 11 3− λ
∣∣∣∣(3− λ)2 − 1 = 0
λ2 − 6λ+ 8 = 0
λ = 2, 4
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
Example
Find the change of coordinates necessary to express thequadratic form with matrix
A =
[3 11 3
]as a quadratic form with no cross-product terms.
Who remembers what to do?
|A− λI | =
∣∣∣∣ 3− λ 11 3− λ
∣∣∣∣(3− λ)2 − 1 = 0
λ2 − 6λ+ 8 = 0
λ = 2, 4
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
Example
Find the change of coordinates necessary to express thequadratic form with matrix
A =
[3 11 3
]as a quadratic form with no cross-product terms.
Who remembers what to do?
|A− λI | =
∣∣∣∣ 3− λ 11 3− λ
∣∣∣∣(3− λ)2 − 1 = 0
λ2 − 6λ+ 8 = 0
λ = 2, 4
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
For λ = 4:
A− 4I =
[−1 11 −1
]∼[
1 −10 0
]So, we get the eigenvector of A associated with λ = 4 to be[
11
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
For λ = 4:
A− 4I =
[−1 11 −1
]
∼[
1 −10 0
]So, we get the eigenvector of A associated with λ = 4 to be[
11
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
For λ = 4:
A− 4I =
[−1 11 −1
]∼[
1 −10 0
]
So, we get the eigenvector of A associated with λ = 4 to be[11
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
For λ = 4:
A− 4I =
[−1 11 −1
]∼[
1 −10 0
]So, we get the eigenvector of A associated with λ = 4 to be[
11
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
For λ = 2:
A− 2I =
[1 11 1
]∼[
1 10 0
]So, we get the eigenvector of A associated with λ = 2 to be[
1−1
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
For λ = 2:
A− 2I =
[1 11 1
]
∼[
1 10 0
]So, we get the eigenvector of A associated with λ = 2 to be[
1−1
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
For λ = 2:
A− 2I =
[1 11 1
]∼[
1 10 0
]
So, we get the eigenvector of A associated with λ = 2 to be[1−1
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
For λ = 2:
A− 2I =
[1 11 1
]∼[
1 10 0
]So, we get the eigenvector of A associated with λ = 2 to be[
1−1
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
So, what is P?
P =
[1 11 −1
]What is D?
D =
[4 00 2
]So we are good?
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
So, what is P?
P =
[1 11 −1
]
What is D?
D =
[4 00 2
]So we are good?
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
So, what is P?
P =
[1 11 −1
]What is D?
D =
[4 00 2
]So we are good?
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
So, what is P?
P =
[1 11 −1
]What is D?
D =
[4 00 2
]
So we are good?
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
So, what is P?
P =
[1 11 −1
]What is D?
D =
[4 00 2
]So we are good?
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
We need to normalize now ...
How do we do that?
Both vectors have length√
2, so ...
P =1√2
[1 11 −1
]How do we know the vectors are orthogonal? They arebecause they are the eigenvectors associated with distincteigenvalues.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
We need to normalize now ...
How do we do that?
Both vectors have length√
2, so ...
P =1√2
[1 11 −1
]How do we know the vectors are orthogonal? They arebecause they are the eigenvectors associated with distincteigenvalues.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
We need to normalize now ...
How do we do that?
Both vectors have length√
2, so ...
P =1√2
[1 11 −1
]How do we know the vectors are orthogonal? They arebecause they are the eigenvectors associated with distincteigenvalues.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
We need to normalize now ...
How do we do that?
Both vectors have length√
2, so ...
P =1√2
[1 11 −1
]
How do we know the vectors are orthogonal? They arebecause they are the eigenvectors associated with distincteigenvalues.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
We need to normalize now ...
How do we do that?
Both vectors have length√
2, so ...
P =1√2
[1 11 −1
]How do we know the vectors are orthogonal?
They arebecause they are the eigenvectors associated with distincteigenvalues.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
We need to normalize now ...
How do we do that?
Both vectors have length√
2, so ...
P =1√2
[1 11 −1
]How do we know the vectors are orthogonal? They arebecause they are the eigenvectors associated with distincteigenvalues.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
Now, we have that A = PDPT . So, for
y =
[y1
y2
]= PTx
we have
Q(y) = 4y21 + 2y2
2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 7
Now, we have that A = PDPT . So, for
y =
[y1
y2
]= PTx
we have
Q(y) = 4y21 + 2y2
2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
Example
Let Q(x) = x21 − 8x1x2 − 5x2
2 . Find the change ofcoordinates necessary to express the quadratic form as aquadratic form with no cross-product term.
What do we need to do here?
I Write A, the matrix of the quadratic form
I Find the eigenvalues and associated eigenvectors
I If distinct, the eigenvalues are orthogonal; if not thenwe need to apply Gram-Schmidt
I Normalize the vectors and form the matrix P
I Form the matrix D
I Find Q(y) for y = PTx
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
Example
Let Q(x) = x21 − 8x1x2 − 5x2
2 . Find the change ofcoordinates necessary to express the quadratic form as aquadratic form with no cross-product term.
What do we need to do here?
I Write A, the matrix of the quadratic form
I Find the eigenvalues and associated eigenvectors
I If distinct, the eigenvalues are orthogonal; if not thenwe need to apply Gram-Schmidt
I Normalize the vectors and form the matrix P
I Form the matrix D
I Find Q(y) for y = PTx
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
Example
Let Q(x) = x21 − 8x1x2 − 5x2
2 . Find the change ofcoordinates necessary to express the quadratic form as aquadratic form with no cross-product term.
What do we need to do here?
I Write A, the matrix of the quadratic form
I Find the eigenvalues and associated eigenvectors
I If distinct, the eigenvalues are orthogonal; if not thenwe need to apply Gram-Schmidt
I Normalize the vectors and form the matrix P
I Form the matrix D
I Find Q(y) for y = PTx
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
Example
Let Q(x) = x21 − 8x1x2 − 5x2
2 . Find the change ofcoordinates necessary to express the quadratic form as aquadratic form with no cross-product term.
What do we need to do here?
I Write A, the matrix of the quadratic form
I Find the eigenvalues and associated eigenvectors
I If distinct, the eigenvalues are orthogonal; if not thenwe need to apply Gram-Schmidt
I Normalize the vectors and form the matrix P
I Form the matrix D
I Find Q(y) for y = PTx
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
Example
Let Q(x) = x21 − 8x1x2 − 5x2
2 . Find the change ofcoordinates necessary to express the quadratic form as aquadratic form with no cross-product term.
What do we need to do here?
I Write A, the matrix of the quadratic form
I Find the eigenvalues and associated eigenvectors
I If distinct, the eigenvalues are orthogonal; if not thenwe need to
apply Gram-Schmidt
I Normalize the vectors and form the matrix P
I Form the matrix D
I Find Q(y) for y = PTx
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
Example
Let Q(x) = x21 − 8x1x2 − 5x2
2 . Find the change ofcoordinates necessary to express the quadratic form as aquadratic form with no cross-product term.
What do we need to do here?
I Write A, the matrix of the quadratic form
I Find the eigenvalues and associated eigenvectors
I If distinct, the eigenvalues are orthogonal; if not thenwe need to apply Gram-Schmidt
I Normalize the vectors and form the matrix P
I Form the matrix D
I Find Q(y) for y = PTx
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
Example
Let Q(x) = x21 − 8x1x2 − 5x2
2 . Find the change ofcoordinates necessary to express the quadratic form as aquadratic form with no cross-product term.
What do we need to do here?
I Write A, the matrix of the quadratic form
I Find the eigenvalues and associated eigenvectors
I If distinct, the eigenvalues are orthogonal; if not thenwe need to apply Gram-Schmidt
I Normalize the vectors and form the matrix P
I Form the matrix D
I Find Q(y) for y = PTx
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
Example
Let Q(x) = x21 − 8x1x2 − 5x2
2 . Find the change ofcoordinates necessary to express the quadratic form as aquadratic form with no cross-product term.
What do we need to do here?
I Write A, the matrix of the quadratic form
I Find the eigenvalues and associated eigenvectors
I If distinct, the eigenvalues are orthogonal; if not thenwe need to apply Gram-Schmidt
I Normalize the vectors and form the matrix P
I Form the matrix D
I Find Q(y) for y = PTx
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
Example
Let Q(x) = x21 − 8x1x2 − 5x2
2 . Find the change ofcoordinates necessary to express the quadratic form as aquadratic form with no cross-product term.
What do we need to do here?
I Write A, the matrix of the quadratic form
I Find the eigenvalues and associated eigenvectors
I If distinct, the eigenvalues are orthogonal; if not thenwe need to apply Gram-Schmidt
I Normalize the vectors and form the matrix P
I Form the matrix D
I Find Q(y) for y = PTx
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
A =
[1 −4−4 −5
]
|A− λI | =
∣∣∣∣ 1− λ -4-4 −5− λ
∣∣∣∣ = 0
(1− λ)(−5− λ)− 16 = 0
λ2 + 4λ− 21 = 0
(λ+ 7)(λ− 3) = 0
λ = 3,−7
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
A =
[1 −4−4 −5
]|A− λI | =
∣∣∣∣ 1− λ -4-4 −5− λ
∣∣∣∣ = 0
(1− λ)(−5− λ)− 16 = 0
λ2 + 4λ− 21 = 0
(λ+ 7)(λ− 3) = 0
λ = 3,−7
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
A =
[1 −4−4 −5
]|A− λI | =
∣∣∣∣ 1− λ -4-4 −5− λ
∣∣∣∣ = 0
(1− λ)(−5− λ)− 16 = 0
λ2 + 4λ− 21 = 0
(λ+ 7)(λ− 3) = 0
λ = 3,−7
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = 3, we have
A− 3I =
[−2 −4−4 −8
]∼[
1 20 0
]So, the eigenvector of A associated with λ = 3 is
v1 =
[2−1
]and ‖v1‖ =
√5, so we have
p1 =
[2√5
− 1√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = 3, we have
A− 3I =
[−2 −4−4 −8
]∼[
1 20 0
]So, the eigenvector of A associated with λ = 3 is
v1 =
[2−1
]and ‖v1‖ =
√5, so we have
p1 =
[2√5
− 1√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = 3, we have
A− 3I =
[−2 −4−4 −8
]∼
[1 20 0
]So, the eigenvector of A associated with λ = 3 is
v1 =
[2−1
]and ‖v1‖ =
√5, so we have
p1 =
[2√5
− 1√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = 3, we have
A− 3I =
[−2 −4−4 −8
]∼[
1 20 0
]
So, the eigenvector of A associated with λ = 3 is
v1 =
[2−1
]and ‖v1‖ =
√5, so we have
p1 =
[2√5
− 1√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = 3, we have
A− 3I =
[−2 −4−4 −8
]∼[
1 20 0
]So, the eigenvector of A associated with λ = 3 is
v1 =
[2−1
]and ‖v1‖ =
√5, so we have
p1 =
[2√5
− 1√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = 3, we have
A− 3I =
[−2 −4−4 −8
]∼[
1 20 0
]So, the eigenvector of A associated with λ = 3 is
v1 =
[2−1
]
and ‖v1‖ =√
5, so we have
p1 =
[2√5
− 1√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = 3, we have
A− 3I =
[−2 −4−4 −8
]∼[
1 20 0
]So, the eigenvector of A associated with λ = 3 is
v1 =
[2−1
]and ‖v1‖ =
√5, so we have
p1 =
[2√5
− 1√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = 3, we have
A− 3I =
[−2 −4−4 −8
]∼[
1 20 0
]So, the eigenvector of A associated with λ = 3 is
v1 =
[2−1
]and ‖v1‖ =
√5, so we have
p1 =
[2√5
− 1√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = −7, we have
A + 7I =
[8 −4−4 2
]∼[
2 −10 0
]So, the eigenvector of A associated with λ = −7 is
v2 =
[12
]and ‖v2‖ =
√5, so we have
p2 =
[1√5
2√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = −7, we have
A + 7I =
[8 −4−4 2
]∼[
2 −10 0
]So, the eigenvector of A associated with λ = −7 is
v2 =
[12
]and ‖v2‖ =
√5, so we have
p2 =
[1√5
2√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = −7, we have
A + 7I =
[8 −4−4 2
]∼
[2 −10 0
]So, the eigenvector of A associated with λ = −7 is
v2 =
[12
]and ‖v2‖ =
√5, so we have
p2 =
[1√5
2√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = −7, we have
A + 7I =
[8 −4−4 2
]∼[
2 −10 0
]
So, the eigenvector of A associated with λ = −7 is
v2 =
[12
]and ‖v2‖ =
√5, so we have
p2 =
[1√5
2√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = −7, we have
A + 7I =
[8 −4−4 2
]∼[
2 −10 0
]So, the eigenvector of A associated with λ = −7 is
v2 =
[12
]and ‖v2‖ =
√5, so we have
p2 =
[1√5
2√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = −7, we have
A + 7I =
[8 −4−4 2
]∼[
2 −10 0
]So, the eigenvector of A associated with λ = −7 is
v2 =
[12
]
and ‖v2‖ =√
5, so we have
p2 =
[1√5
2√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = −7, we have
A + 7I =
[8 −4−4 2
]∼[
2 −10 0
]So, the eigenvector of A associated with λ = −7 is
v2 =
[12
]and ‖v2‖ =
√5, so we have
p2 =
[1√5
2√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
For λ = −7, we have
A + 7I =
[8 −4−4 2
]∼[
2 −10 0
]So, the eigenvector of A associated with λ = −7 is
v2 =
[12
]and ‖v2‖ =
√5, so we have
p2 =
[1√5
2√5
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
This gives
P =
[2√5
1√5
− 1√5
2√5
]
and
D =
[3 00 −7
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
This gives
P =
[2√5
1√5
− 1√5
2√5
]
and
D =
[3 00 −7
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
This gives
P =
[2√5
1√5
− 1√5
2√5
]
and
D =
[3 00 −7
]
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
So, for x = Py, we have
x21 − 8x1x2 − 5x2
2 = xTAx
= (Py)TA(Py)
= yTPTAPy
= yTDy
= 3y21 − 7y2
2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
So, for x = Py, we have
x21 − 8x1x2 − 5x2
2 = xTAx
= (Py)TA(Py)
= yTPTAPy
= yTDy
= 3y21 − 7y2
2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
So, for x = Py, we have
x21 − 8x1x2 − 5x2
2 = xTAx
= (Py)TA(Py)
= yTPTAPy
= yTDy
= 3y21 − 7y2
2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
So, for x = Py, we have
x21 − 8x1x2 − 5x2
2 = xTAx
= (Py)TA(Py)
= yTPTAPy
= yTDy
= 3y21 − 7y2
2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
Example 8
So, for x = Py, we have
x21 − 8x1x2 − 5x2
2 = xTAx
= (Py)TA(Py)
= yTPTAPy
= yTDy
= 3y21 − 7y2
2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
The level curves to a quadratic form Q : R2 → R are conicsections. If D is the diagonal matrix from the Principle AxesTheorem, that is, such that A = PDPT , then the levelcurves are:
(a) ellipses if the diagonal entries of D are nonzero and havethe same sign
(b) hyperbolae if the diagonal entries of D are nonzero andhave opposite sign
(c) parabolae if exactly one of the diagonal entries is zero
The axes of the conic sections are the x and y axesrespectively rotated by P.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
The level curves to a quadratic form Q : R2 → R are conicsections. If D is the diagonal matrix from the Principle AxesTheorem, that is, such that A = PDPT , then the levelcurves are:
(a) ellipses if the diagonal entries of D are nonzero and havethe same sign
(b) hyperbolae if the diagonal entries of D are nonzero andhave opposite sign
(c) parabolae if exactly one of the diagonal entries is zero
The axes of the conic sections are the x and y axesrespectively rotated by P.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
The level curves to a quadratic form Q : R2 → R are conicsections. If D is the diagonal matrix from the Principle AxesTheorem, that is, such that A = PDPT , then the levelcurves are:
(a) ellipses if the diagonal entries of D are nonzero and havethe same sign
(b) hyperbolae if the diagonal entries of D are nonzero andhave opposite sign
(c) parabolae if exactly one of the diagonal entries is zero
The axes of the conic sections are the x and y axesrespectively rotated by P.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
The level curves to a quadratic form Q : R2 → R are conicsections. If D is the diagonal matrix from the Principle AxesTheorem, that is, such that A = PDPT , then the levelcurves are:
(a) ellipses if the diagonal entries of D are nonzero and havethe same sign
(b) hyperbolae if the diagonal entries of D are nonzero andhave opposite sign
(c) parabolae if exactly one of the diagonal entries is zero
The axes of the conic sections are the x and y axesrespectively rotated by P.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
What is the equation of an ellipse?
x21a2 +
x22b2 = 1 with a, b > 1
x1
x2
a
b
So, we have here that a =√
ca11
and b =√
ca22
where
Q(x) = xTAx = c .
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
What is the equation of an ellipse?x2
1a2 +
x22b2 = 1 with a, b > 1
x1
x2
a
b
So, we have here that a =√
ca11
and b =√
ca22
where
Q(x) = xTAx = c .
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
What is the equation of an ellipse?x2
1a2 +
x22b2 = 1 with a, b > 1
x1
x2
a
b
So, we have here that a =√
ca11
and b =√
ca22
where
Q(x) = xTAx = c .
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
What is the equation of an ellipse?x2
1a2 +
x22b2 = 1 with a, b > 1
x1
x2
a
b
So, we have here that a =√
ca11
and b =√
ca22
where
Q(x) = xTAx = c .
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
What if we do not have a diagonal matrix?
The ellipse isrotated ...
x1
x2y1y2
This is the ellipse given by Q(x) = 5x21 − 4x1x2 + 5x2
2 = 48and by Q(y) = 3y2
1 + 7y22 if we perform the change of
variable here.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
What if we do not have a diagonal matrix? The ellipse isrotated ...
x1
x2y1y2
This is the ellipse given by Q(x) = 5x21 − 4x1x2 + 5x2
2 = 48and by Q(y) = 3y2
1 + 7y22 if we perform the change of
variable here.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
What if we do not have a diagonal matrix? The ellipse isrotated ...
x1
x2y1y2
This is the ellipse given by Q(x) = 5x21 − 4x1x2 + 5x2
2 = 48and by Q(y) = 3y2
1 + 7y22 if we perform the change of
variable here.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
What if we do not have a diagonal matrix? The ellipse isrotated ...
x1
x2y1y2
This is the ellipse given by Q(x) = 5x21 − 4x1x2 + 5x2
2 = 48and by Q(y) = 3y2
1 + 7y22 if we perform the change of
variable here.
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
Who remembers the equation of a hyperbola?
x21a2 −
x22b2 = 1
x1
x2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
Who remembers the equation of a hyperbola?x2
1a2 −
x22b2 = 1
x1
x2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
Who remembers the equation of a hyperbola?x2
1a2 −
x22b2 = 1
x1
x2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
And if not a diagonal matrix, we rotate here as well.
x1
x2y1y2
1 Definition andFacts
2 Examples
3 Justification
4 Back toExamples
5 Classification
6 Principle AxesTheorem
7 The Geometry ofQuadratic Form
The Geometry of Quadratic Form
And if not a diagonal matrix, we rotate here as well.
x1
x2y1y2