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Assume A is a square $n \times n$ matrix and all its eigenvalues are either $1$ or $0$.

And I supposed to examine whether the following statement is true:

the number of linearly independent eigenvectors of $A^n$ is equal or less than the number of linearly independent eigenvectors of $A$

What I've done: suppose $x_1, \dots, x_s$ eigenvectors of $A$, and assume $x_1, \dots, x_k$ relate to the eigenvalue $1$, and $x_{k+1}, \dots, x_s$ relate to the eigenvalue $0$. So, $\forall \ i \in [1, k] \ \to Ax_i = x_i$ hence $A^nx_i = x_i$ and all $x_1, \dots, x_k$ are eigenvectors of $A^n$. Following the same way, we can conclude that $x_{k+1}, \dots, x_s$ are also eigenvectors of $A^n$.

But $\mathbf{that \ statement \ is \ said \ to \ be \ false}$ so, new eigenvectors can appear for $A^n$, but I was unable to think up such matrix.

Could you please help me? Thank you a lot for any help!

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    $\begingroup$ Try $\tiny{\begin{bmatrix}0&1&0\\0&0&0\\0&0&1\end{bmatrix}}$. $\endgroup$ – amd Apr 7 '18 at 8:50
  • $\begingroup$ @amd wow! Thank you a lot! $\endgroup$ – D F Apr 7 '18 at 8:53

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