Stuck in finding Eigen values

Question

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:
$$ A^2 -2A + I $$

Answer 1

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

Answer 2

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

Answer 3

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