# Solving for eigenvectors of an eigenvalues

$A=\begin{bmatrix} a & 0\\ 0& b \end{bmatrix}$

The eigenvalues are $\lambda_{1}=a,\lambda_{2}=b$

To solve for the corresponding eigenvectors for eigenvalue $\lambda_{1}=a$:

We solve $A\vec{v}_{1}=\lambda_{1}\vec{v}_{1}$

$\begin{bmatrix} 0 &0 \\ 0& d-a \end{bmatrix} \begin{bmatrix} v_{1,1}\\ v_{1,2} \end{bmatrix}$=$\begin{bmatrix} 0\\ 0 \end{bmatrix}$

I understand this is a simple question and I have no reasons to doubt the theory on hand but for some reason I am not yielding the correct eigenvectors.

Any help is appreciated.

• Even without going through the long process, what can you say happens to $[1,0]^T$ and $[0,1]^T$ when multiplied by $A$? Note: for the longer traditional way there are two different cases to consider, when $a=b$ and when $a\neq b$. In the first case $A-aI$ is the zero matrix and so rowreduces to the zero matrix implying both entries are free variables for solution vector. In the second case $A-aI$ row reduces to $\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$ implying the second entry is zero and implies nothing about the first entry meaning it is free to be any value. – JMoravitz Mar 19 '17 at 7:07
• $\left[\begin{smallmatrix}a&0\\0&b\end{smallmatrix}\right]\left[\begin{smallmatrix}1\\0\end{smallmatrix}\right] = \left[\begin{smallmatrix}a\\0\end{smallmatrix}\right]=a\left[\begin{smallmatrix}1\\0\end{smallmatrix}\right]$, and what is the definition of an eigenvector? You should be able to show rather straightforwardly that the eigenvectors of any diagonal matrix are the standard basis vectors. – JMoravitz Mar 19 '17 at 7:26
• The system you arrive at reduces to $(b-a)v_{1,2}=0$, that is, if $b\ne a$, $v_{1,2}=0$, that is, $\vec v_1$ colinear to $(1,0)^T$. Where is the mystery here? – Did Mar 19 '17 at 10:58