Consider matrix $A = \begin{bmatrix} 1&1&-1&1\\2&1&-2&1\\-1&1&1&1\\2&-2&0&0 \end{bmatrix}$
Without computing the characteristic polynomial of $A$, find all $a \in \mathbb{R}$ for which $(1, 1, 1, a)$ is an eigenvector of $A$.
How shall I proceed here? Thank you for any hint / help.
Best Answer
Let $v=(1,1,1,a)^{T}$. Then
$Av = (a+1,a+1,a+1,0)^{T}$
If $\lambda$ is the eigen value corresponding to eigen vector $v$.
$\lambda v= (\lambda,\lambda,\lambda,a\lambda)^{T}$.
Then you have $\lambda = a+1$ and $a\lambda =0$.
So if $a=0$ then $\lambda=1$ is an eigen value for which $(1,1,1,a)^{T}=(1,1,1,0)^T$ is an eigen vector.
If $a\neq 0$ . Then $\lambda =0$ and $a=-1$.
So for eigen value $1$ , $(1,1,1,a)^{T}$ is an eigen vector for $a=0$.
And for eigen value $0$ , the vector $(1,1,1,a)^{T}$ is an eigen vector for $a=-1$.