If the eigenbasis of $A$ is independent, $A$ must have an inverse?

Question

"If the eigenbasis of $A$ is independent, $A$ must have an inverse."

This statement is false, but I don't understand why.

If the eigenbasis of $A$ is independent, then we have: $A = X \Lambda X^{-1}$.

Simple application of the inverse gives us: $A^{-1} = X \Lambda^{-1} X^{-1}$

I guess the reason this logic doesn't work is: I can't just apply the inverse operator on the left hand side and expect it to give me $A^{-1}$. $A^{-1}$ must exist before I can do that, and $A^{-1}$ existing is contingent on all non-zero eigenvalues...

Answer 1

You have to ensure that $\Lambda^{-1}$ exists. This is the case iff $A$ has not eigenvalue $0$.

Now take any diagonalizable $A$ with eigenvalue $0$ and you have a counterexample.

Hint: Consider $A = 0$.

Answer 2

The matrix $$ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$ has the eigenvalues $1$, with eigenvector $\left[\begin{smallmatrix}1\\0\end{smallmatrix}\right]$, and the eigenvalue $0$, with eigenvector $\left[\begin{smallmatrix}0\\1\end{smallmatrix}\right]$.

The listed eigenvectors form a basis, but the matrix is not invertible.

A necessary and sufficient condition for a matrix to be invertible is that zero is not an eigenvalue.