Eigenvalues and Eigenvectors Raibatak Sen Gupta 2019 Eigenvalues - PowerPoint PPT Presentation

Eigenvalues and Eigenvectors Raibatak Sen Gupta Eigenvalues and Eigenvectors Raibatak Sen Gupta 2019

Eigenvalues Characteristic Equation and Eigenvectors Raibatak Sen Gupta Let A be an n × n matrix. Then det ( A − xI n ) gives a polynomial in x of degree n . This polynomial is called the Characteristic Polynomial of the matrix A .

Eigenvalues Characteristic Equation and Eigenvectors Raibatak Sen Gupta Let A be an n × n matrix. Then det ( A − xI n ) gives a polynomial in x of degree n . This polynomial is called the Characteristic Polynomial of the matrix A . Correspondingly, the equation det ( A − xI n ) = 0 is called the Characteristic equation of A .

Eigenvalues An example and Eigenvectors Raibatak Sen   1 − 1 0 Gupta Consider the matrix A = 1 2 − 1   3 2 − 2

Eigenvalues An example and Eigenvectors Raibatak Sen   1 − 1 0 Gupta Consider the matrix A = 1 2 − 1   3 2 − 2 Then the characteristic polynomial of A is given by the determinant of the matrix ( A − xI 3 )   1 − x − 1 0 = det 1 2 − x − 1   3 2 − 2 − x = (1 − x )( x 2 − 2)+(1 − x ) = (1 − x )( x 2 − 1) = − x 3 + x 2 + x − 1.

Eigenvalues An example and Eigenvectors Raibatak Sen   1 − 1 0 Gupta Consider the matrix A = 1 2 − 1   3 2 − 2 Then the characteristic polynomial of A is given by the determinant of the matrix ( A − xI 3 )   1 − x − 1 0 = det 1 2 − x − 1   3 2 − 2 − x = (1 − x )( x 2 − 2)+(1 − x ) = (1 − x )( x 2 − 1) = − x 3 + x 2 + x − 1. Thus the characteristic equation of A is (1 − x )( x 2 − 1) = 0

Eigenvalues Characteristic Polynomial: Properties and Eigenvectors Raibatak Sen Gupta So the characteristic polynomial of an n × n matrix A , given by det ( A − xI n ), is an n -degree polynomial of the form c 0 + c 1 x + c 2 x 2 + . . . + c n x n , where the c i ’s belong to the field from which the entries of A are coming. It is easy to see, that c n = ( − 1) n . Also, c 0 is obtained by putting x = 0 in the polynomial, so c 0 = det ( A ). And also, c n − 1 = ( − 1) n − 1 trace ( A ).

Eigenvalues Cayley-Hamilton theorem and Eigenvectors Raibatak Sen Gupta Theorem Every square matrix satisfies its own characteristic equation. So if characteristic equation of a matrix A is c 0 + c 1 x + c 2 x 2 + . . . + c n x n = 0, then we will have that c 0 I n + c 1 A + c 2 A 2 + . . . + c n A n = O n For example, we can check that in the previous example, we will have − A 3 + A 2 + A − I 3 = O 3 .

Eigenvalues Finding inverse by Cayley-Hamilton theroem and Eigenvectors Raibatak Sen Gupta If A is invertible, then Cayley-Hamilton theorem gives an easy way of finding A − 1 .

Eigenvalues Finding inverse by Cayley-Hamilton theroem and Eigenvectors Raibatak Sen Gupta If A is invertible, then Cayley-Hamilton theorem gives an easy way of finding A − 1 . Let c 0 + c 1 x + c 2 x 2 + . . . + c n x n = 0 be the characteristic equation of A , then by Cayley-Hamilton theorem we have c 0 I n + c 1 A + c 2 A 2 + . . . + c n A n = O n ⇒ − ( c 1 A + c 2 A 2 + . . . + c n A n ) = c 0 I n = ⇒ − c − 1 0 ( c 1 I n + c 2 A + . . . + c n A n − 1 ) A = I n = (note that c 0 = det ( A ) � = 0). Thus A − 1 = − c − 1 0 ( c 1 I n + c 2 A + . . . + c n A n − 1 )

Eigenvalues Example and Eigenvectors Raibatak Sen Gupta In the previous example, characteristic equation was − x 3 + x 2 + x − 1 = 0, so we have that − A 3 + A 2 + A − I 3 = O 3 . Thus ( − A 2 + A + I 3 ) A = I 3 , which gives that A − 1 = − A 2 + A + I 3

Eigenvalues Eigenvalue and Eigenvectors Raibatak Sen Gupta Let p ( x ) = 0 be the characteristic equation of a square matrix A . Then the roots of this equation are called the Eigenvalues of A .

Eigenvalues Eigenvalue and Eigenvectors Raibatak Sen Gupta Let p ( x ) = 0 be the characteristic equation of a square matrix A . Then the roots of this equation are called the Eigenvalues of A .   1 − 1 0  are For example, the eigenvalues of A = 1 2 − 1  3 2 − 2 1 , − 1 , 1, since the characteristic equation is (1 − x )( x 2 − 1) = 0.

Eigenvalues Eigenvalue and Eigenvectors Raibatak Sen Gupta If A is a n × n matrix, then its characteristic polynomial p ( x ) is an n -degree polynomial, and hence it has n roots. Thus an n × n matrix has exactly n eigenvalues. However, the eigenvalues may not be all distinct, as seen in the previous example.

Eigenvalues Eigenvalue and Eigenvectors Raibatak Sen Gupta If A is a n × n matrix, then its characteristic polynomial p ( x ) is an n -degree polynomial, and hence it has n roots. Thus an n × n matrix has exactly n eigenvalues. However, the eigenvalues may not be all distinct, as seen in the previous example. A real matrix may have complex eigenvalues, e.g., the matrix � 0 � − 1 has the characteristic polynomial x 2 + 1. So its B = 1 0 eigenvalues are i , − i .

Eigenvalues Eigenvalues: Properties and Eigenvectors Raibatak Sen Theorem Gupta The product of all the eigenvalues of a matrix is equal to its determinant. Proof. Let c 0 + c 1 x + c 2 x 2 + . . . c n x n = 0 be the char. eqn. of a matrix A . Then the product of the roots of this equation is c n = ( − 1) n det ( A ) ( − 1) n c 0 ( − 1) n = det ( A ). Corollary A matrix is non-invertible if and only if 0 is an eigenvalue of that matrix. Proof. This happens because a matrix is non-invertible iff det ( A ) = 0.

Eigenvalues Eigenvalues: Properties and Eigenvectors Raibatak Sen Gupta Theorem The eigenvalues of a diagonal matrix is its diagonal elements. Theorem If c is an eigenvalue of an invertible matrix A, then 1 c is an eigenvalue of A − 1 . Proof. Let A be an n × n matrix. As c is an eigenvalue, we have det ( A − cI n ) = 0. Now det ( A − 1 − 1 det ( A ) det ( AA − 1 − 1 1 c I n ) = c AI n ) = det ( A ) det ( I n − 1 1 1 1 c A ) = c n det ( cI n − A ) = 0. det ( A ) c is a root of det ( A − 1 − xI n ) = 0, and thus it is an Hence 1 eigenvalue of A − 1 .

Eigenvalues Eigenvectors and Eigenvectors Raibatak Sen Gupta Definition Let A be an n × n matrix with entries from a field F . A non-null vector v ∈ F n is called an eigenvector of A if there exists c ∈ F such that Av = cv . In fact, in this case, c is seen to be an eigenvalue of A and v is called an eigenvector of A corresponding to c . Av = cv = ⇒ ( A − cI n ) v = θ , which shows that the system of equation ( A − cI n ) X = θ has a non-null solution, which happens if and only if det ( A − cI n ) = 0, i.e., c is an eigenvalue of A .

Eigenvalues Eigenvectors: Properties and Eigenvectors Raibatak Sen Gupta Theorem Let A ∈ M n × n ( F ) . To an eigenvector of A corresponds a unique eigenvalue of A. Proof. If possible let an eigenvector v correspond to two distinct eigenvalues c 1 , c 2 of A . Then Av = c 1 v = c 2 v , which gives ( c 1 − c 2 ) v = θ . However this is not possible as c 1 − c 2 � = 0 , v � = θ .

Eigenvalues and Eigenvectors Raibatak Sen Gupta Theorem Let A ∈ M n × n ( F ) . To each eigenvalue of A, there corresponds at least one eigenvector. Proof. If c is an eigenvalue, then det ( A − cI n ) = 0, so ( A − cI n ) X = θ for some non-null X ∈ F n . So Ax = cX . Thus X is an eigenvector of A corresponding to c . Clearly, if v is an eigenvector corresponding to c , any scalar multiple kv is also an eigenvector corresponding to c (as A ( kv ) = k ( Av ) = k ( cv ) = c ( kv )).

Eigenvalues Finding eigenvectors and Eigenvectors Raibatak Sen Gupta � 1 � 3 Let A = . The characteristic equation is 4 5 ( x + 1)( x − 7) = 0, so the eigenvalues are − 1 , 7. Let X 1 = ( x , y ) be an eigenvector corresponding to − 1, then we have ( A + I 2 ) X 1 = θ , which gives 2 x + 3 y = 0 , 4 x + 6 y = 0. So the eigenvectors are given by the set { k ( − 3 2 , 1) | k ∈ R } . Similarly let X 2 = ( x , y ) be an eigenvector corresponding to 7. Then ( A − 7 I 2 ) X 2 = θ , which gives − 6 x + 3 y = 0 , 4 x − 2 y = 0. So the eigenvectors corresponding to 7 are given by the set { k (1 , 2) | k ∈ R } .

Eigenvalues Eigenvectors: Properties and Eigenvectors Raibatak Sen Gupta Theorem Two eigenvectors corresponding to two distinct eigenvalues of a matrix are linearly independent. Proof. Let v 1 , v 2 be two eigenvectors corresponding to two distinct eigenvalues d 1 , d 2 of a matrix A . Let c 1 v 1 + c 2 v 2 = θ . Multiplying by A , we get c 1 Av 1 + c 2 Av 2 = θ , i.e., c 1 d 1 v 1 + c 2 d 2 v 2 = θ . This gives that d 1 ( − c 2 v 2 ) + d 2 c 2 v 2 = θ or c 2 ( d 1 − d 2 ) v 2 = θ . As v 2 is non-null and d 1 � = d 2 , we must have c 2 = 0. Thus c 1 v 1 = θ , which gives c 1 = 0. This shows that v 1 , v 2 are linearly independent.

Eigenvalues Eigenvectors: Properties and Eigenvectors Raibatak Sen Theorem Gupta Let E c be the set of all eigenvectors corresponding to any particular eigenvalue c of a matrix A ∈ M n ( F ) . Then S c = E c ∪ { θ } forms a subspace F n , which is called the eigenspace or characteristic subspace corresponding to c. Proof. First of all, θ ∈ S c . Now let v 1 , v 2 ∈ E c . Then for f , g ∈ F , A ( fv 1 + gv 2 ) = f ( Av 1 )+ g ( Av 2 ) = f ( cv 1 )+ g ( cv 2 ) = c ( fv 1 + gv 2 ). So fv 1 + gv 2 ∈ S c . This shows that S c is a subspace of F n .