For a matrix Q prove the eigenvectors of kI-Q are equal to that of Q
2026-03-29 13:40:29.1774791629
Eigenvectors of Matrix A
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I imagine that the quantity "$4jk/dz$" is just a strange way to write a number, say $q$. So you are asking if $A$ and $q I-A$ have the same eigenvectors.
Let $v$ be an eigenvector of $A$, namely $A v= a v$. But $q I v = q v$, so you can conclude that
$$(qI-A)v = qIv-Av=qv-av = (q-a)v$$
As you can see, $v$ is also an eigenvector of $qI-A$ but the relative eigenvalue changes from $a$ to $q-a$.