Originally Posted by
Swlabr Do all non-bijective linear maps have an eigenvalue 0?
Hi Swlabr.
For $\displaystyle A=\begin{pmatrix}a&b\\c&d\end{pmatrix},$
$\displaystyle \det(A-\lambda I)\ =\ 0$
$\displaystyle \implies\ \left|\begin{array}{cc}a-\lambda&b\\c&d-\lambda\end{array}\right|\ =\ 0$
$\displaystyle \implies\ (a-\lambda)(c-\lambda)-bc\ =\ 0$
$\displaystyle \implies\ \lambda^2-(a+d)\lambda+ad-bc\ =\ 0$
$\displaystyle \implies\ \lambda(\lambda-(a+d))\ =\ 0$ if $\displaystyle ad-bc=0$
$\displaystyle \implies\ \lambda\ =\ 0\ \mbox{or}\ \lambda\ =\ a+d$
So the answer is no, they can have a nonzero eigenvalue if $\displaystyle a+d\ne0.$