let $A$ be an $n \times n$ matrix and $\lambda$ a particular eigenvalue of $A$ with $x$ an eigenvector corresponding to $\lambda$. prove that for any scalars $c$ and $k$, $c\lambda + k$ is an eigenvalue for $cA + kIn$ and $x$ is an eigenvector corresponding to $ c \lambda + k$.
I think I may have an idea of how to show this proof but I'm not sure how to express it. I kind of want to write $Ax = \lambda x$, multiply $(cA + kIn)$ by $x$ all the way through, then replace $A$ with $\lambda$, then factor $c \lambda + k$ out, since that follows a similar example my teacher did. but this seems to make not much of an explanation to me, it seems like all that does is show that matrices are commutative.
So is my method correct? if not, can someone explain in detail or show me the correct method? And if it is, can someone explain how this actually proves it?
Best Answer
$A\mathbf x = \lambda\mathbf x \implies cA\mathbf x =c\lambda \mathbf x$
Note that $kI_n \mathbf x = k\mathbf x$ so add this to the above equation to get:
$cA\mathbf x + kI_n \mathbf x = c\lambda \mathbf x + k\mathbf x \iff (cA + kI_n) \mathbf x = (c\lambda + k)\mathbf x$
Therefore, $c\lambda +k$ is an eigenvalue of $cA + kI_n$ corresponding to eigenvector $\mathbf x$. $\square$