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Suppose we have a $3 \times 3$ real invertible matrix $A$. If $A$ has two distinct eigenvalues, so one of them has multiplicity $2$. Is it possible to have only two linearly independent eigenvectors? i.e., is it possible that the eigenvectors corresponding to the eigenvalue with multiplicity $2$ are linearly dependent?

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What about $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ ?

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  • $\begingroup$ Could you explain what the idea behind that is? How do we construct this kind of matrix? $\endgroup$ – user398843 May 19 at 22:04
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    $\begingroup$ I took a 2x2 matrix with only one eigenvalue that is not the identity, so it has only one eigenspace of dimension 1. I completed in an orthogonal basis so that it is an invertible matrix $\endgroup$ – DIdier_ May 19 at 22:06
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$$A=\begin{bmatrix}2&0&0\\0&1&1\\0&0&1\end{bmatrix}$$ For the eigenvalue 1,which has algebraic multiplicity 2, the dimension of the eigenspace is 1. Any eigenvector for eigenvalue 1 is a multiple of $$\begin{bmatrix}0\\1\\0\end{bmatrix}.$$

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  • $\begingroup$ Could you explain what the idea behind that is? How do we construct this kind of matrix? $\endgroup$ – user398843 May 19 at 22:06

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