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Assume we have square matrix $A_{N \times N}$.

Can we say that the maximum number of independent eigen vectors, with nonzero eigenvalues, equals $\text{Rank}(A)$?

We can say that such number must be smaller equal than $\text{Rank}(A)$, because an eigenvector with nonzero eigenvalue means that it must belongs to $\text{Range}(A)$, so the independent set of such vectors must be a basis of some subspace of $\text{Range}(A)$, thus not exceeding $\text{dim} \{ \text{Range}(A) \} = \text{Rank} (A)$.

But my question is, does the equality holds?

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No. Let $A=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. $1$ is the only eigenvalue of $A$ and $ \dim ker(A-I)=1$. But $rank(A)=2$

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  • $\begingroup$ Oh thnaks, that solve the problem $\endgroup$ – Taylor Huang Feb 3 '17 at 12:49

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