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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 $\begin{bmatrix} a & 0 \\ 0 & b\end{bmatrix}$.

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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 $a=b$)

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