0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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: $\pm i$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.