A real matrix with no eigenvalues is not invertible? True or False

Question

I know that determinant of a matrix is equal to its product of eigenvalues. If a matrix has no eigenvalues, does it mean the determinant is zero since there is nothing to multiply?

I am guessing no, but I want to be sure.

Answer 1

No. Take, for instance, $A=\left[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right]$. It has no (real) eigenvalues, but it is invertible; its inverse is $\left[\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right]$.

Of course, it has non-real eigenvalues: $\pm i$.

Answer 2

think about \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix} which haven't any eigenvalues$(\text{real})$ but have determinant $1$

Answer 3

A matrix with no real eigenvalues does not have the real eigenvalue $0$, and thus is always invertible.