# Stuck in finding Eigen values

The given matrix A is $$\left[\begin{matrix} 2 & 1 & -2 \\ 0 & 1 & 4 \\ 0 & 0 & 3 \\ \end{matrix}\right]$$
I know that the Eigen values are the diagonals (2, 1, 3) as it is an upper triangular matrix (wouldn't matter if it was a lower triangular matrix). However, what is the Eigen values of:
$$A^2 -2A + I$$

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It's eigenvalues not Eigen values. Contrary to somewhat popular belief, there's no mathematician named Eigen behind it. :) – Hagen von Eitzen Feb 16 '14 at 14:06
@Little Child in your case matrix is upper diagonal,so it's Eigenvalue are diagonal entries math.stackexchange.com/questions/264969/… – dato datuashvili Feb 16 '14 at 14:22

Note that if $v$ is a vector such that $Av=\lambda v$ then $$(A^2-2A+I)v=\lambda^2v-2\lambda v +v=(\lambda^2-2\lambda+1)v$$

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I found it on somewhere that for matrix $A^n$, the Eigen Value is $\lambda^n$ – Little Child Feb 16 '14 at 14:01
yes after diagonalization,eigenvalue of $A^n$ is exactly $\lambda^{n}$ – dato datuashvili Feb 16 '14 at 14:08
@LittleChild you can show generally that $p(A)v=p(\lambda)v$ for $v$ an eigenvector with eigenvalue $\lambda$ and $p$ a polynomial. – Mark Bennet Feb 16 '14 at 14:11
@MarkBennet The eigenvalues are [1,-2,4] ?? – Little Child Feb 16 '14 at 14:13
@LittleChild I get $[0, 1, 4]$ - $(\lambda-1)^2$ – Mark Bennet Feb 16 '14 at 14:16
A=[2 1 -2;0 1 4;0 0 3]

A =

2     1    -2
0     1     4
0     0     3

>> [V D]=eig(A)

V =

1.0000   -0.7071         0
0    0.7071    0.8944
0         0    0.4472

D =

2     0     0
0     1     0
0     0     3


$D$ matrix contains eigenvalues of $A$,related to your comment

B=A*A;
>> [V1 D1]=eig(B)

V1 =

1.0000   -0.7071         0
0    0.7071    0.8944
0         0    0.4472

D1 =

4     0     0
0     1     0
0     0     9


as you see eigenvalue of $A^2$ is simple $D^2$ and eigenvectors are not changed,but also please note that Lin your case matrix is upper diagonal,so it's Eigenvalue are diagonal entries

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Simply solve the quadratic equation $A^2−2A+I$ . Remember to take $I$ as one. This is just an easy way to remember.

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I'm not sure how this helps you to find the eigenvalues of $p(A)$ - if $A$ has eigenvalues which are roots of $p(x)$ then the corresponding eigenvectors are part of the null space of $p(A)$ – Mark Bennet Feb 16 '14 at 14:09