# Suppose that A is diagonalized by the matrix P and that the eigen-values of A are `lambda1` .........`lambdan` Show that the eigenvalues of (A-`lambda1I` ) are...

Suppose that * A *is diagonalized by the matrix

*and that the eigen-values of*

*P**are `lambda1` .........`lambdan`*

*A*

Show that the eigenvalues of (A-`lambda1I` ) are

0,`lambda2-lambda1,lambda3-lambda1,.....................lambdan-lambda1`

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`P^(-1)AP=[lambda_i]_(1<=i<=n)`

and is [lambda_i] is diagonal matrix. Now consider

P^(-1)(A-lambda_1 **I**)P=(P^(-1)AP-Lambda_1P^(-1)**I**P)

matrix multplication is distributive.

=[lambda_i]-lambda_1[1]

**[lambda_i] **is diagonal matrix

**I **is an unit identity matrix whose diagonal elements are 1

=[lambda_i]-[lambda_1]

=[lambda_i-lambda_1]_(1<=i<=n) {which is diagonal matrix)

thus eigen values are

lambad_1-lambda_1,lambda_2-lambda_1, ..............,lambda_n-lambda_1 .

0 ,lambda_2-lambda_1, ..............,lambda_n-lambda_1 .

proved

`P^(-1)AP=[lambda_i]_(1<=i<=n)` ,is a diagonal matrix ,diagonal elements are eigen values of A.

Now consider

`P^(-1)(A-lambda_1 I)P=P^(-1)AP-lambda_1(P^(-1)IP)`

`lambda_1` is scalar quantity ,matrix multiplication is distributive.

`=[lambda_i]_(1<=i<=n]- lambda_1[1]_(1<= i<=n)`

I is an identity matrix.

`=[lambda_i-lambda_1]_(1<=i<=n)` ,which is diagonal matrix and diagonal elements are eigen values of matrix `A-lambda_1I` .Thus eigen values are

`lambda_1-lambda_1,lambda_2-lambda_1,.........,lambda_n-lambda_1.` ``

`0,lambda_2-lambda_1,.........,lambda_n-lambda_1 .`

hence proved.