I have a matrix $A = \begin{bmatrix} a & 0\\ 2(a-b) & b\end{bmatrix}$.

I found the eigenvalues to be $\lambda_1 = a$ and $\lambda_2 = b$.

For $\lambda_1$, I found an eigenvector $\begin{bmatrix}1\\ 2\end{bmatrix}$.
But when $\lambda_2 = b$, I don't know what the corresponding eigenvector is. I get equations $(a-b)x_1 + 0x_2 = 0$ and $2(a-b)x_1 + 0x_2 = 0$. I feel like the second eigenvector can be anything since $x_2$ is a free variable.

Can anybody explain? The whole purpose is to diagnolize $A$ such that $A = XDX^{-1}$ and $X$ is composed of the eigenvectors of $A$. I already found matrix $D$, it is $\begin{bmatrix} a & 0 \\ 0 & b\end{bmatrix}$.


If $a \neq b$, the problem is to find the solution for $x_1=0$. (that is the first coordinate of the eigenvector has to be $0$)

$x_2$ indeed can be anything non-zero (eigenvector cannot be the zero vector $0$). Hence, you can choose your eigenvector to be $\begin{bmatrix} 0 \\ 1\end{bmatrix}$

(remark: the same vector can be chosen for $a=b$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.