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While reading a chapter on diagonalizable matrices, I found myself wondering:

Can a matrix $A \in \mathbb R^{n \times n}$ be invertible but not diagonalizable?

My quick Google search did not return a clear answer.

Garrett
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    Yes. A $2\times2$ shear matrix is the simplest example, e.g. the matrix whose first row is $(1,1)$ and whose second row is $(0,1)$. – symplectomorphic Mar 28 '17 at 15:57

2 Answers2

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After thinking about it some more, I realized that the answer is "Yes".

For example, consider the matrix

\begin{equation} A = \left [ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right ]. \end{equation}

It has two linearly independent columns, and is thus invertible.

At the same time, it has only one eigenvector:

\begin{equation} v = \left [ \begin{array}{c} 1 \\ 0 \end{array} \right ]. \end{equation}

Since it doesn't have two linearly independent eigenvectors, it is not diagonalizable.

Garrett
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    It is worth noting that there also exist diagonalizable matrices which aren't invertible, for example $\begin{bmatrix}1&0\\0&0\end{bmatrix}$, so we have invertible does not imply diagonalizable and we have diagonalizable does not imply invertible. There are matrices that are neither, that are both, or that are only one of either. – JMoravitz Mar 28 '17 at 15:35
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    @JMoravitz it seems to me that $\left[\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right]$ is the best example of a diagonalizable non-invertible matrix. Of course your example works too! – feralin Mar 28 '17 at 21:13
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    ...and to complete things: $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ is neither diagonalizable nor invertible. Garrett: you might want to look up Jordan blocks. – J. M. ain't a mathematician Mar 29 '17 at 01:47
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Geometric(-ish) answer. Take, in $\mathbb R^2$, a rotation of angle $0<\theta<2\pi$ with $\theta \neq \pi$. Then the associated matrix is invertible (the inverse being the rotation of $-\theta$) but is not diagonalisable, since no non-zero vector is mapped into a multiple of itself by a rotation of such angles.

Stefano
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