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

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 http://math.stackexchange.com/questions/264969/proof-that-eigenvalues-are-the-diagonal-entries-of-the-upper-triangular-matrix-i — dato datuashvili, Feb 16 '14 at 14:22

score 5 · Accepted Answer · answered Feb 16 '14 at 13:57

5

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

answered Feb 16 '14 at 13:57

Mark Bennet

100,194

I found it on somewhere that for matrix $A^n$, the Eigen Value is $\lambda^n$ – An SO User 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] ?? – An SO User Feb 16 '14 at 14:13
@LittleChild I get $[0, 1, 4]$ - $(\lambda-1)^2$ – Mark Bennet Feb 16 '14 at 14:16

score 2 · Answer 2 · answered Feb 16 '14 at 14:06

2

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

answered Feb 16 '14 at 14:06

Marlon Abeykoon

369

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

dato datuashvili · Answer 3 · 2014-02-16T14:24:20.023

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

Stuck in finding Eigen values

3 Answers3