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By the fundamental theorem of invertible matrices, the null space of an invertible matrix has only the zero vector in it (it is trivial). Therefore there are no eigenvectors for the matrix. Is this correct? Are there any situations in which an invertible matrix has an eigenvalue?

Also, something I noticed is that even non-invertible matrices can not have any eigenvalues. So the invertibility of a matrix tells us it has no eigenvalues, but a lack of invertibility tells us nothing. Is this correct?

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    $\begingroup$ $A$ being invertible doesn't imply that $A-\lambda I$ is… $\endgroup$ – Bernard Feb 27 at 17:25
  • $\begingroup$ invertible <=> 0 not eigenvalue, not invertible <=> 0 eigenvalue $\endgroup$ – pyrogen Feb 27 at 17:26
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the null space of an invertible matrix has only the zero vector in it

Correct

Therefore there are no eigenvectors for the matrix. Is this correct?

No. That does not even make sense. Why do you conclude that?

From the fact $A {\bf x} = 0 \implies {\bf x}={\bf 0}$ you can only conclude that the equation $A {\bf x} = \lambda {\bf x} $ has no solution for non trivial (non null) ${\bf x}$... when $\lambda=0$. That is, that $\lambda=0$ cannot be eigenvalue of an invertible matrix.

even non-invertible matrices can not have any eigenvalues

Again, does not make sense. Every $n \times n$ matrix has $n$ (counting multiplicities) eigenvalues (in the complex domain).

The only relevant relationship is : a matrix is invertible if and only if it hasn't zero as eigenvalue. Put in other way: a matrix is singular if and only if $\lambda =0$ is eigenvalue.

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