# Find the Eigenvalues and Eigenvectors of 4x4 Matrix.

## Homework Statement

X= 1st row: (0, 1, 0, 0), 2nd row: (1, 0, 0, 0), 3rd row: (0, 0, 0, 1-i), 4th row: (0, 0, 1+i, 0)
Find the eigenvalues and eigenvectors of the matrix X.

## Homework Equations

|X-λI|=0 (characteristic equation)
(λ is the eigenvalues, I is the identity matrix)
(X-λI)V=0 (V is the corresponding eigenvector V= (V1, V2, V3, V4))

## The Attempt at a Solution

Apologies firstly for my poor attempt at a matrix...

I have found the eigenvalues by solving the characteristic equation to be
λ1= +1,
λ2=-1
λ3= +√2
λ4=-√2
and then using the second equation above to find the eigenvectors of X.

The problem I'm having is that when trying to find the eigenvectors, I end up with 4 equations, 2 of which have components V1, V2 only, and the other 2 equations have components V3 and V4 only so there's no way to eliminate them and find the normalised eigenvectors. The subsequent parts of the question ask to find the exponential of X and so without the eigenvectors I cannot achieve this. I've only ever been taught how to do this for a 2x2 which is relatively simple but that procedure doesn't seem to be working for me here.

Mark44
Mentor

## Homework Statement

X= 1st row: (0, 1, 0, 0), 2nd row: (1, 0, 0, 0), 3rd row: (0, 0, 0, 1-i), 4th row: (0, 0, 1+i, 0)
Find the eigenvalues and eigenvectors of the matrix X.

## Homework Equations

|X-λI|=0 (characteristic equation)
(λ is the eigenvalues, I is the identity matrix)
(X-λI)V=0 (V is the corresponding eigenvector V= (V1, V2, V3, V4))

## The Attempt at a Solution

Apologies firstly for my poor attempt at a matrix...

I have found the eigenvalues by solving the characteristic equation to be
λ1= +1,
λ2=-1
λ3= +√2
λ4=-√2
and then using the second equation above to find the eigenvectors of X.
Your eigenvalues agree with the ones I got, so I think they're OK.
Freya said:
The problem I'm having is that when trying to find the eigenvectors, I end up with 4 equations, 2 of which have components V1, V2 only, and the other 2 equations have components V3 and V4 only so there's no way to eliminate them and find the normalised eigenvectors. The subsequent parts of the question ask to find the exponential of X and so without the eigenvectors I cannot achieve this. I've only ever been taught how to do this for a 2x2 which is relatively simple but that procedure doesn't seem to be working for me here.
For the eigenvalue ##\lambda = 1## I get this matrix:
##\begin{bmatrix} -1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1- i \\ 0 & 0 & 1 + i & -1\end{bmatrix}##
The matrix above is used to solve the matrix equation Ax = 0.
When I row-reduce this matrix I end up with an eigenvector of ##\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}##. It's always a good idea to verify that your eigenvalues and eigenvectors are correct.

Your eigenvalues agree with the ones I got, so I think they're OK.

For the eigenvalue ##\lambda = 1## I get this matrix:
##\begin{bmatrix} -1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1- i \\ 0 & 0 & 1 + i & -1\end{bmatrix}##
The matrix above is used to solve the matrix equation Ax = 0.
When I row-reduce this matrix I end up with an eigenvector of ##\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}##. It's always a good idea to verify that your eigenvalues and eigenvectors are correct.
Hi thank you for your response.

I saw before you had put 2 eigenvectors for lambda=1 so I was firstly wondering what that meant? I thought there is only one eigenvector to one eigenvalue? Also, is there a way of doing this without using row reduction? I haven't used it before to find eigenvectors, and I haven't come across them on the module I'm studying.
I can see where you got the (1,1,0,0) eigenvector from plugging in lambda and the first 2 equations this yields but I don't understand how you can lose the information from the other 2 rows and how that works?
Thanks

• RJLiberator
Mark44
Mentor
Hi thank you for your response.

I saw before you had put 2 eigenvectors for lambda=1 so I was firstly wondering what that meant?
It meant that I made an error, now corrected. That's why I said it was important to always check your work, by verifying that any eigenvector you find actually works. By that, I mean checking that ##A\vec{x} = \lambda \vec{x}##
Freya said:
I thought there is only one eigenvector to one eigenvalue? Also, is there a way of doing this without using row reduction? I haven't used it before to find eigenvectors, and I haven't come across them on the module I'm studying.
You could do it without using matrix row-reduction, by solving the system of equations. Using matrices in this context is really just a shortcut.
Freya said:
I can see where you got the (1,1,0,0) eigenvector from plugging in lambda and the first 2 equations this yields but I don't understand how you can lose the information from the other 2 rows and how that works?
Thanks
After row reduction, my final (and corrected) matrix was:
##\begin{bmatrix} 1 & -1 & 0 & 0 \\0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}##
This is really just shorthand for this system of equations:
##x_1 - x_2 = 0##
##x_3 = 0##
##x_4 = 0##

This can be rewritten as
##x_1 = x_2##
##x_2 = x_2##
##x_3 = 0##
##x_4 = 0##
Since ##x_2## is arbitrary, the entire set of eigenvectors can be written as ##r \begin{bmatrix} 1 \\ 1 \\ 0 \\0 \end{bmatrix}##, where r is any scalar.

Mark44
Mentor
I thought there is only one eigenvector to one eigenvalue?
Not necessarily. For some matrices one eigenvalue can be associated with multiple eigenvectors.

• RJLiberator
Not necessarily. For some matrices one eigenvalue can be associated with multiple eigenvectors.
Thanks for all your help, I managed to get all 4 eigenvectors using your help, I think I'm just going to try and remember to check them when it's something slightly tricky like this.