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Let $A$ be a square real (or complex) matrix. Then $A$ is invertible if and only if zero is not an eigenvalue. Proof or counterexample.
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The statement is true.

If $A$ is invertible, then its determinant is non-zero, and the characteristic polynomial of $A$ is non-zero. This implies that zero is not an eigenvalue of $A$.

Conversely, if zero is not an eigenvalue of $A$, then the dimension of the eigenspace associated with zero is zero. This means that the columns of $A$ are linearly independent, so the matrix $A$ has a non-zero determinant, and is thus invertible.
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