Is it because invertibility of the matrix $A-\lambda I$ would imply that $A-\lambda I$ has a trivial null space, and thus there are no solutions to the equation $Ax-\lambda I = 0$ (other than the trivial solution $0$), and thus there are no eigenvalues? Does that line of thought make sense/are there mistakes with it?
2026-04-05 23:18:07.1775431087
Why does invertibility of the matrix $A-\lambda I$ imply that $\lambda$ is not an eigenvalue?
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