A is a nxn matrix with zero as an eigenvalue. Show that $A$ is not invertible.

Question

Suppose that $A$ is an $n\times n$ matrix with zero as an eigenvalue. Show that $A$ is not invertible.

Hint: Assume that $A$ is invertible and compute $A^-1*Av$ where v is an eigenvector of A corresponding to the zero eigenvalue.

Can you also explain what zero as an eigenvalue is telling me/ adding to te quesiton.

Please help!

Av=0 since the matrix has zero as an eigenvalue, right? – Elena Feb 11 '18 at 22:18 — Elena, Feb 11 '18 at 22:18

score 1 · Accepted Answer · answered Feb 11 '18 at 22:14

1

Suppose $A(u)=0, u\neq 0$ and $A$ invertible, $u=A^{-1}A(u)=A^{-1}(0)=0$ contradiction.

answered Feb 11 '18 at 22:14

Tsemo Aristide

score -1 · Answer 2 · answered Feb 12 '18 at 01:12

-1

Since $0$ is an eigenvalue of $A$ and $\det A$ is the product of eigenvalues of $A$, so $\det A$ must be zero. Thus $A$ is not invertible.

answered Feb 12 '18 at 01:12

SAHEB PAL

2 Answers2