Answer to Question #246368 in Linear Algebra for Trimmy

Question

Show that if be the Eigenvalues of the matrix, then has the Eigenvalues

.

λ1, λ2, λ3, . . . λn A An

λn

1 , λ n

2 , λn

3 . . . λn

n

Answer 1

Answer:-

We use the fact that characteristics polynomial of matrix A given by

$P(X)=det(A-\lambda 1)=(\lambda 1-X)(\lambda 2-X).....(\lambda n-X)$

We know that if A is n×n matrix and $\lambda i$iλiiλi is an Eigenvalue of A. It is a root of characteristics polynomial of A that is given by $P(X)=det(A-Xl)=0$

In =n×n identity matrix

$P(\lambda 1)=0$

If $\lambda 1, \lambda 2.......\lambda n$ are Eigenvalue of matrix A then

$P(\lambda 1)=0$ which is a monic polynomial

$P(X)=(\lambda 1 -X)(\lambda 2-X)........(\lambda n-X)\\ det(A-XIn)=(\lambda 1 -X)(\lambda 2-X)........(\lambda n-X)$

Put X=0

We get

$det(A)=(\lambda 1 -0)(\lambda 2-0).......(\lambda n-0)$

$=\lambda 1,\lambda 2......\lambda n$

$det(A)=(\lambda 1,\lambda 2)......(\lambda n)$

Hence proved.