I am not sure where to begin with a question like this:
Question: Let A be an n * n matrix. Prove that A is singular if and only if lambda = 0 is an eigenvalue of A.
Any guidance will be appreciated.
An alternative:
$\displaystyle A\;\textrm{singular}\Leftrightarrow\det A=0 \Leftrightarrow \det (A-0I)=0\Leftrightarrow 0\;\textrm{is\;eigenvalue\;of\;}A$
Fernando Revilla