Show that if $\lambda$ is an eigenvalue of $A$, then 1+$\lambda$ is an eigenvalue of $I+A$.
What is the corresponding eigenvector?
What I have done so far (if it is correct at all…):
$(I+A)x=Ix+Ax=1x+Ax=1x+{\lambda}x=(1+\lambda)x$, thus, $1+{\lambda}$ is an eigenvector of $I+A$
But how do I find the corresponding eigenvector just from the information above?
Best Answer
If $x$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$, then your computation of
$$(A + I)x = (\lambda + 1) x$$
is exactly the definition that states $x$ is an eigenvector of $A + I$ corresponding to eigenvalue $\lambda + 1$.