Answer to Question #246981 in Linear Algebra for Anuj

Question #246981

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


1
Expert's answer
2021-10-07T14:20:21-0400

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)\\\\ \n\ndet(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.


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