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Lecture8, Eigenvalues and Eigenvectors

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Lecture 8
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Eigenvalues and Eigenvectors 1. Eigenvalues and Eigenvectors 2. Diagonalization 3.Symmetric Matrices and Orthogonal Diagonalization 4.Application of Eigenvalues and Eigenvectors
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Eigenvalues and Eigenvectors 1. Eigenvalues and Eigenvectors 2. Diagonalization 3.Symmetric Matrices and Orthogonal Diagonalization 4.Application of Eigenvalues and Eigenvectors 1. Eigenvalues and Eigenvectors Eigenvalue problem (one of the most important problems in the linear algebra): If A is an nn matrix, do there exist nonzero vectors x in Rn such that Ax is a scalar multiple of x Eigenvalue and Eigenvector: A: an nn matrix : a scalar (could be zero) x: a nonzero vector in Rn A = x xEigenvalue Eigenvector Geometrical Interpretation (The term eigenvalue is from the German word Eigenwert, meaning proper value) Ex 1:Verifying eigenvalues and eigenvectors ((

=1 00 2A110 (=( x1 12 0 1 2 12 20 1 0 0 0A ((((= = = = (((( x xEigenvalue 2 22 0 0 0 01 ( 1)0 1 1 1 1A ((((= = = = (((( x xEigenvalue Eigenvector Eigenvector 201 (=( x In fact, for each eigenvalue, it has infinitely many eigenvectors. For = 2, [3 0]T or [5 0]T are both corresponding eigenvectors. Moreover, ([3 0] + [5 0])Tis still an eigenvector. Thm.1: The eigenspace corresponding to of matrix A If A is an nn matrix with an eigenvalue , then the set of all eigenvectors of together with the zero vector is a subspace of Rn. This subspace is called the eigenspace of Pf: x1 and x2 are eigenvectors corresponding to 1 1 2 2(i.e., ,) A A = = x x x x1 2 1 2 1 2 1 21 2(2)( ) ( ) (i.e.,is also an eigenvector corresponding to)A A A + = + = + = ++x x x x x x x xx x1 1 1 11(1)( ) ( ) ( ) ( ) (i.e.,is also an eigenvector corresponding to)A c c A c cc = = = x x x xxSince this space is closed under vector addition and scalar multiplication, this space is a subspace of Rn Ex 3: Examples of eigenspaces on the xy-plane For the matrix A as follows, the corresponding eigenvalues are 1 = 1 and 2 = 1: ((

=1 00 1ASol: 0 1 0 0 0 010 1Ay y y y (((((= = = ((((( For the eigenvalue 1 = 1, corresponding vectors are any vectors on the x-axis 1 010 0 1 0 0 0x x x xA (((((= = = ((((( For the eigenvalue 2 = 1, corresponding vectors are any vectors on the y-axis Thus, the eigenspace corresponding to = 1 is the x-axis, which is a subspace of R2 Thus, the eigenspace corresponding to = 1 is the y-axis, which is a subspace of R2 1 0 0 1x xAy y (((= = ((( vIf( , ), x y = v Geometrically, multiplying a vector (x, y) in R2by the matrix A corresponds to a reflection to the y-axis The above reflection result also can be derived using the eigenvalues and eigenvectors as follows 0 00 001 10x x xA A A A Ay y yx xy y| | (((((= = + = + | ((((( \ . (((= + = ((( v(1) An eigenvalue of A is a scalar such that Thm.2: Finding eigenvalues and eigenvectors of a matrix AeMnn det( ) 0 I A =(2) The eigenvectors of A corresponding to are the nonzero solutions of Characteristic polynomial of AeMnn: 11 1 0det( ) ( )n nnI A I A c c c = = + + + + Characteristic equation of A: det( ) 0 I A =( ) I A = x 0Let A is an nn matrix. has nonzero solutions for x iff ( ) I A = x 0 det( ) 0 I A = Note: following the definition of the eigenvalue problem (homogeneous system) ( ) A A I I A = = = x x x x x 0 Ex 4: Finding eigenvalues and eigenvectors ((

=5 112 2A Sol: Characteristic equation: 22 12det( )1 53 2 ( 1)( 2) 0I A = += + + = + + =Eigenvalue:2 , 12 1 = = 2 , 1 = 2(2)2 = 122G.-J. E.124 12 0( )1 3 04 12 1 31 3 0 03 3,01xI Axx ss sx s ((( = = ((( (( (( ((( = = = ((( x1(1)1 = 112G.-J. E.123 12 0( )1 4 03 12 1 41 4 0 04 4,01xI Axx tt tx t ((( = = ((( (( (( ((( = = = ((( x(((

=2 0 00 2 00 1 2A Sol: Characteristic equation: 32 1 00 2 0 ( 2) 00 0 2I A = = =Eigenvalue: 2 = Ex 5: Finding eigenvalues and eigenvectors Find the eigenvalues and corresponding eigenvectors for the matrix A. What is the dimension of the eigenspace of each eigenvalue? The eigenspace of = 2: 1230 1 0 0( ) 0 0 0 00 0 0 0xI A xx ((( ((( = = ((( ((( x0 , ,1000010321=(((

+(((

=(((

=(((

t s t stsxxx1 00 0 , : the eigenspace of corresponding to20 1s t s t R A (( ((+ e = ` (( (( )Thus, the dimension of its eigenspace is 2. Notes: (1)If an eigenvalue 1 occurs as a multiple root (k times) for the characteristic polynominal, then 1 has multiplicity k. (2)The multiplicity of an eigenvalue is greater than or equal to the dimension of its eigenspace. (In Ex. 5, k is 3 and the dimension of its eigenspace is 2) Ex 6Find the eigenvalues of the matrix A and find a basis for each of the corresponding eigenspaces (((((

=3 0 0 10 2 0 110 5 1 00 0 0 1A Sol: Characteristic equation: 21 0 0 00 1 5 101 0 2 01 0 0 3( 1) ( 2)( 3) 0I A = = =Eigenvalues:3 , 2 , 13 2 1= = = According to the note on the previous slide, the dimension of the eigenspace of 1 = 1 is at most to be 2 For 2 = 2 and 3 = 3, the dimensions of their eigenspaces are at most to be 1 1(1)1 =121340 0 0 0 00 0 5 10 0( )1 0 1 0 01 0 0 2 0xxI Axx ((( ((( ((( = = ((( ((( x12342 0 21 0,, 02 0 20 1x tx ss t s tx tx t (((( (((( (((( = = + = (((( (((( 1202,0010)`(((((

(((((

is a basis for the eigenspace corresponding to11 =The dimension of the eigenspace of 1 = 1 is 2 2(2)2 =122341 0 0 0 00 1 5 10 0( )1 0 0 0 01 0 0 1 0xxI Axx ((( ((( ((( = = ((( ((( x12340 05 5,010 0xx tt tx tx ((( ((( ((( = = = ((( ((( 0150)`(((((

is a basis for the eigenspace corresponding to 22 =The dimension of the eigenspace of 2 = 2 is 1 3(3)3 =123342 0 0 0 00 2 5 10 0( )1 0 1 0 01 0 0 0 0xxI Axx ((( ((( ((( = = ((( ((( x12340 05 5,00 01xx tt txx t ((( ((( ((( = = = ((( ((( 1050)`(((((

is a basis for the eigenspace corresponding to 33 =The dimension of the eigenspace of 3 = 3 is 1 Thm.3: Eigenvalues for triangular matrices If A is an nn triangular matrix, then its eigenvalues are the entries on its main diagonal Ex 7: Finding eigenvalues for triangular and diagonal matrices 2 0 0(a)1 1 05 3 3A ( (= ( ( 1 0 0 0 00 2 0 0 0(b)0 0 0 0 00 0 0 4 00 0 0 0 3A ( ( ( ( = ( ( ( Sol: 2 0 0(a)1 1 0 ( 2)( 1)( 3) 05 3 3I A = = + = +1 2 32, 1, 3 = = = 1 2 3 4 5(b)1, 2, 0, 4, 3 = = = = =2. Diagonalization Diagonalization problem : For a square matrix A, does there exist an invertible matrix P such that P1AP is diagonal? Diagonalizable matrix : Definition 1: A square matrix A is called diagonalizable if there exists an invertible matrix P such that P1AP is a diagonal matrix (i.e., P diagonalizes A) Definition 2: A square matrix A is called diagonalizable if A is similar to a diagonal matrix Thm. 4: Similar matrices have the same eigenvalues If A and B are similar nn matrices, then they have the same eigenvalues Pf: AP P B B A1similar are and= 1 1 1 11 1 1( ) I B I P AP P IP P AP P I A PP I A P P P I A P P I AI A = = = = = = = Since A and B have the same characteristic equation, they are with the same eigenvalues For any diagonal matrix in the form of D = I, P1DP = D Considering the characteristic equation of B: Ex 1: Eigenvalue problems and diagonalization programs (((

=2 0 00 1 30 3 1A Sol:Characteristic equation: 21 3 03 1 0 ( 4)( 2) 00 0 2I A = = + =+1 2 3The eigenvalues : 4, 2, 2 = = = (1)4the eigenvector = 1110 ( (=( ( p(2)2the eigenvector = 2 31 01 ,00 1 (( ((= = (( (( p p11 2 31 1 0 4 0 0[ ] 1 1 0 ,and0 2 00 0 1 0 0 2P P AP (( ((= = = (( (( p p p2 1 31[ ]1 1 0 2 0 01 1 0 0 4 00 0 1 0 0 2PP AP= (( ((= = (( (( p p p Note:If Thm.5: Condition for diagonalization An nn matrix A is diagonalizable if and only if it has n linearly independent eigenvectors Notethatiftherearenlinearlyindependent eigenvectors, it does not imply that there are n distinct eigenvalues. It is possible to have only one eigenvalue withmultiplicityn,andtherearenlinearly independent eigenvectors for this eigenvalue However,iftherearendistincteigenvalues,then there are n linearly independent eivenvectors, and thus A must be diagonalizable Ex 4: A matrix that is not diagonalizable Show that the following matrix is not diagonalizable1 2

0 1A (=( Sol: Characteristic equation: 21 2( 1) 00 1I A = = =1 1The eigenvalue1,and then solve ( )for eigenvectors I A = = x 01 10 2 1eigenvector 0 0 0I A I A p (( = = = (( Since A does not have two linearly independent eigenvectors, A is not diagonalizable. Steps for diagonalizing an nn square matrix: Step 2: Let 1 2[ ]nP =pp pStep 1: Find n linearly independent eigenvectorsfor A with corresponding eigenvalues1 2, ,np p pStep 3: (((((

= =nD AP P 0 00 00 0211where,1,2, ,i i iA i n = = p p1 2, , ,n Ex 5: Diagonalizing a matrix diagonal. is such thatmatrixa Find1 1 31 3 11 1 1 1AP P PA((((

=Sol:Characteristic equation: 1 1 11 3 1 ( 2)( 2)( 3) 03 1 1I A = = + = +1 2 3The eigenvalues : 2, 2, 3 = = =21 = G.-J. E.11 1 1 1 0 11 1 1 0 1 03 1 3 0 0 0I A (( (( = (( (( 12 1310eigenvector01x txx t ((( (((= = ((( ((( p22 = 14G.-J. E.12 43 1 1 1 01 5 1 0 13 1 1 0 0 0I A (( (( = (( (( 11 412 2 431 eigenvector14x tx tx t ((( (((= = ((( ((( p33 = G.-J. E.32 1 1 1 0 11 0 1 0 1 13 1 4 0 0 0I A (( (( = (( (( 12 331 eigenvector11x tx tx t ((( (((= = ((( ((( p1 2 311 1 1[ ] 0 1 1and it follows that1 4 12 0 00 2 00 0 3PP AP ( (= = ( ( ( (= ( ( p p p Note: a quick way to calculate Ak based on the diagonalization technique 11220 00 00 00 0(1) 0 00 0kkkknnD D ( ( ( ( ( (= =( ( ( ( ( 1 1 1 1 1repeat times11 2(2) 0 00 0,where 0 0k kkkkk k kknD P AP D P AP P AP P AP P A PA PD P D = = = ( ( (= =( ( ( Thm. 6: Sufficient conditions for diagonalization IfannnmatrixAhasndistinct eigenvalues,thenthecorresponding eigenvectorsarelinearlyindependentand thus A is diagonalizable. Ex 7: Determining whether a matrix is diagonalizable ((((

=3 0 01 0 01 2 1A Sol: Because A is a triangular matrix, its eigenvalues are 1 2 31, 0, 3. = = = According to Thm. 6, because these three values are distinct, A is diagonalizable. 7.3 Symmetric Matrices and Orthogonal Diagonalization Symmetric matrix : A square matrix A is symmetric if it is equal to its transpose: TA A= Ex 1: Symmetric matrices and nonsymetric matrices (((

=5 0 20 3 12 1 0A((

=1 33 4B(((

=5 0 10 4 11 2 3C(symmetric) (symmetric) (nonsymmetric) Thm.7: Eigenvalues of symmetric matrices If A is an nn symmetric matrix, then the following properties are true.(1) A is diagonalizable (symmetric matrices are guaranteed tohas n linearly independent eigenvectors and thus be diagonalizable). (2) All eigenvalues of A are real numbers. (3) If is an eigenvalue of A with multiplicity k, then has k linearly independent eigenvectors. That is, the eigenspace of has dimension k. The above theorem is called the Real Spectral Theorem, and the set of eigenvalues of A is called the spectrum of A. Ex 2: Prove that a 2 2 symmetric matrix is diagonalizable. ((

=b cc aA Pf:Characteristic equation: 0 ) (2 2= + + = = c ab b ab cc aA I 2 22 2 22 2 2 2 24 ) (4 24 4 2 ) ( 4 ) (c b ac b ab ac ab b ab a c ab b a+ =+ + =+ + + = +0 >As a function in , this quadratic polynomial function has a nonnegative discriminant as follows 0 4 ) ( (1)2 2= + c b a0 , = = c b a0itself is a diagonal matrix.0a c aAc b a ((= = (( 0 4 ) ( ) 2 (2 2> + c b aThe characteristic polynomial of A has two distinct real roots, which implies that A has two distinct real eigenvalues. According to Thm. 6, A is diagonalizable. A square matrix P is called orthogonal if it is invertible and Orthogonal matrix : 1(or)T T TP P PP P P I= = = Thm. 8: Properties of orthogonal matrices An nn matrix P is orthogonal if and only if its column vectors form an orthonormal set. 1 1 1 2 11 1 1 2 12 1 2 2 2 12 1 2 2 2 11 21 2T T TnnT T TTnT T Tn n n nn n n nP P I ( ( ( ( ( (= = = ( ( ( ( ( p p p p p pp p p p p pp p p p p pp p p p p pp p p p p pp p p p p pPf: Suppose the column vectors of P form an orthonormal set, i.e., | |1 2,where0 for and1.n i j i iP i j = = = = p p p p p p pIt implies that P1 = PT and thus P is orthogonal. Ex 5: Show that P is an orthogonal matrix. ((((

= 5 355 345 3251523232310 P Sol: If P is a orthogonal matrix, then 1 T TP P PP I= =1 2 21 2 235 3 53 3 32 1 2 1 435 5 5 3 55 5 2 4 233 5 3 5 3 5 3 51 0 00 0 1 00 0 10TPP I (( ( (( (= = = (( ( (( ( (( 1 2 1 3 2 3 1 12 2 3 3we can produce0 and 1. = = = = = =p p p p p p p pp p p p1 2 23 3 32 11 2 35 552 43 53 5 3 5Moreover, let, ,and0 , (( ( (( (= = = (( ( (( ( (( p p pset. l orthonorma anis } p , p , {p So,3 2 1 Thm.9: Properties of symmetric matrices Let A be an nn symmetric matrix. If 1 and 2 are distinct eigenvalues of A, then their corresponding eigenvectors x1 and x2are orthogonal. (Thm. 6 only states that eigenvectors corresponding to distinct eigenvalues are linearly independent)1 1 2 1 1 2 1 2 1 2 1 2( ) ( ) ( ) ( ) ( )T T TA A A = = = = x x x x x x x x x xbecause is symmetric1 2 1 2 1 2 2 2 2 2 2( ) ( ) ( ) ( ) )AT T TA A = = = = = 1 1x x x x x x x x (x x1 2 1 21 2 1 2 1 2The above equation implies ( )( ) 0,and because ,it follows that0. So, and are orthogonal. == =x xx x x xPf: For distinct eigenvalues of a symmetric matrix, their corresponding eigenvectors are orthogonal and thus linearly independent to each other Note that there may be multiple x1 and x2 corresponding to 1 and 2

Thm. 10: Fundamental theorem of symmetric matrices Let A be an nn matrix. Then A is orthogonally diagonalizable and has real eigenvalues if and only if A is symmetric.A matrix A is orthogonally diagonalizable if there exists an orthogonal matrix P such that P1AP = D is diagonal. Orthogonal diagonalization : Ex 7: Determining whether a matrix is orthogonally diagonalizable (((

=1 1 11 0 11 1 11A(((

=0 8 18 1 21 2 52A((

=1 0 20 2 33A((

=2 00 04AOrthogonallydiagonalizable Symmetric matrix Ex 9: Orthogonal diagonalization Find an orthogonal matrix that diagonalizes.2 2 2 2 1 42 4 1P AA ( (= ( ( Sol: 0 ) 6 ( ) 3 ( ) 1 (2= + = A I1 26, 3 (has a multiplicity of 2) = =1 1 2 21 1 1 3 3 31(2) 6, (1, 2,2)( , , ) = = = =vv uv2 2 3(3) 3, (2,1,0), ( 2,0,1) = = = v vLinearly independent but not orthogonal Verify Thm. 7.9 that v1 v2 = v1 v3 = 0 4. Applications of Eigenvalues and Eigenvectors The rotation for quadratic equation: ax2+bxy+cy2+dx+ey+f = 0 Ex 5: Identify the graphs of the following quadratic equations 2 2(a) 4 9 36 0 x y + =Sol: 2 22 2(a) In standard form, we can obtain1.3 2x y+ =2 2(b) 13 10 13 72 0 x xy y + = Since there is no xy-term, it is easy to derive the standard form and it is apparent that this equation represents an ellipse. 2 2(b) 13 10 13 72 0 x xy y + = Since there is a xy-term, it is difficult to identify the graph of this equation. In fact, it is also an ellipse, which is oblique on the xy-plane. There is a easy way to identify the graph of quadratic equation. The basic idea is to rotate the x- and y-axes to x- and y-axes such that there is no more xy-term in the new quadratic equation. In the above example, if we rotate the x- and y-axes by 45 degree counterclockwise, the new quadratic equation can be derived, which represents an ellipse apparently. 2 22 2( ') ( ')13 2x y+ = In Section 4.8, the rotation of conics is achieved by changing basis, but here the diagonalizating technique based on eigenvalues and eignvectors is applied to solving the rotation problem Matrix of the quadratic form: If we define X = , then XTAX= ax2 + bxy + cy2 . In fact, the quadratic equation can be expressed in terms of X as follows. Quadratic form :ax2 + bxy + cy2 is the quadratic form associated with the quadratic equation ax2 + bxy + cy2 + dx + ey + f = 0. / 2/ 2a bAb c (=( xy ( ( | |.TX AX d e X f + + Principal Axes Theorem For a conic whose equation is ax2 + bxy + cy2 + dx + ey + f = 0, the rotation to eliminate the xy-term is achieved by X = PX, where P is an orthogonal matrix that diagonalizes A. That is, where 1 and 2 are eigenvalues of A. The equation for the rotated conic is given by 1 120,0TP AP P AP D (= = =( | |2 21 2( ') ( ') 0. x y d e PX f '+ + + =Pf: According to Thm. 10, since A is symmetric, we can conclude that there exists an orthogonal matrix P such that P1AP = PTAP = D is diagonal. Replacing X with PX, the quadratic form becomes 2 21 2( ) ( ) ( )( ) ( ) ( ) .T T T TTX AX PX A PX X P APXX DX x y ' ' ' '= =' ' ' '= = + It is obvious that the new quadratic form in terms of X has no xy-term, and the coefficients for (x)2 and (y)2 are the two eigenvalues of the matrix A. | |1 2 1 21 2 Since and are the orignal and new coodinates, the roles of and are like the basis vectors (or the axis vectors ) in the new coordinax x x xX PX x yy y y y' ' ((((' ' '= = = + ((((' ' v v v vv vte system. Sol: Ex 6: Rotation of a conic Perform a rotation of axes to eliminate the xy-term in the following quadratic equation 2 213 10 13 72 0 x xy y + =13 5.5 13A (=( The matrix of the quadratic form associated with this equation isThe eigenvalues are 1 = 8 and 2 = 18, and the corresponding eigenvectors are1 21 1 and.1 1 ((= = (( x xAfter normalizing each eigenvector, we can obtain the orthogonal matrix P as follows. 28( ) 18( ) 72 0, x y' '+ =1 1cos 45 sin 452 21 1sin 45 cos 452 2P ( ( ( (= =( ( ( Then by replacing X with PX, the equation of the rotated conic iswhich can be written in the standard form 2 22 2( ) ( )1.3 2x y' '+ = The above equation represents an ellipse on the xy-plane. According to the results in p. 268 in Ch4, X=PXis equivalent to rotate the xy-coordinates by 45 degree to form the new xy-coordinates, which is also illustrated in the figure on Slide 7.62. Matrix of the quadratic form: If we define X = [x y z]T, then XTAX= ax2 + by2 + cz2 + dxy + exz + fyz, and the quadratic surface equation can be expressed as In three-dimensional version: ax2 + by2 + cz2 + dxy + exz + fyz is the quadratic form associated with the equation of quadric surface: ax2 + by2 + cz2 + dxy + exz + fyz + gx + hy + iz + j = 0 / 2 / 2/ 2 / 2/ 2 / 2a d eA d b fe f c ( (=( ( | | .TX AX g h i X j + +


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