# Thread: finding eigenvectors of a diagno matrix

1. ## finding eigenvectors of a diagno matrix

I have a three by three diagnol matrix
[1,2,-4;0,2,1;0,0,3]
$\lambda=$ 1,2,3
i found the eigenvectors for 2 and 3 but i am having finding the eigenvector for when lambda equals 1.
When I try to solve $(A-\lambda I)x=0$ I can only find the vector x=0 which it cannot be.

I realize that this would be a problem with any diagonal matrix and was wondering how to go about finding the eigenvector.

2. Originally Posted by mathleticman
I have a three by three diagnol matrix
[1,2,-4;0,2,1;0,0,3]
$\lambda=$ 1,2,3
i found the eigenvectors for 2 and 3 but i am having finding the eigenvector for when lambda equals 1.
When I try to solve $(A-\lambda I)x=0$ I can only find the vector x=0 which it cannot be.

I realize that this would be a problem with any diagonal matrix and was wondering how to go about finding the eigenvector.
$\displaystyle\begin{bmatrix}0&2&-4\\0&1&1\\0&0&2\end{bmatrix}\Rightarrow\text{rref} =\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$

$\displaystyle\left\{\begin{bmatrix}1\\0\\0\end{bma trix}\right\}$

3. That is NOT a "diagonal" matrix- it is an "upper triangular" matrix!

Personally, I have always prefered to find eigenvectors directly from the definition. An eigenvector of linear operator A, corresponding to eigenvalue $\lambda$ must satisfy $Av= \lambda v$

So any eigenvector of this matrix corresponding to eigenvalue 2 must satisfy
$\begin{bmatrix}1 & 2 & -4 \\ 0 & 2 & 1 \\ 0 & 0 & 3\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix} x + 2y- 4z \\ 2y+ z \\ 3z\end{bmatrix}= \begin{bmatrix}2x \\ 3y \\ 2z\end{bmatrix}$
which is the same as the three equations x+ 2y- 4z= 2x, 2y+ z= 2y, and 3z= 2z. (Since the matrix is upper triangular, those three equations are "partially uncoupled")- the last involves only z and it is easy to see that z= 0. then the second equation says 2y= 2y which is true for all y. The third is the same as x+ 2y- 4(0)= 2x or 2y= x. That is, such an eigenvector can be written <2y, y, 0>= y<2, 1, 0> so any multiple of <2, 1, 0> is an eigenvector corresponding to eigenvalue 2.