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Let $A$ be an $n\times n$ matrix.

If $A$ is invertible then one of it's eigenvalues is 0.

If $A$ is diagonalizable then it has $n$ linearly independent eigenvectors.

  • Are these two statements true?
  • Is there any connection between them? May I infer one from the other?
  • 3
    if $A$ is invertible then none of its eienvalues is $0$. – 2017-01-21
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    it is possible to state invertibility in terms of eigenvalues, namely that $0$ is not an eigenvalue. – 2017-01-21

1 Answers 1

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One does not imply the other.

This matrix is invertible and not diagonalizable:

$\begin{pmatrix} 1 & 1 \\ 0 & 1\\ \end{pmatrix}$

This matrix is diagonalizable (in fact it is already a diagonal matrix) but not invertible:

$\begin{pmatrix} 0 & 0 \\ 0 & 0 \\ \end{pmatrix}$

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    Not only that: statement (1) is completely false, of course, whereas (2) is true... – 2017-01-21