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Let $A$ be a square real or complex matrix. We’ll call $A$ special if among its rows (or among its columns) there are two identical ones, different from the zero vector, (Added:) and if it has no nilpotent Jordan block.

Obviously, if $A$ is special, then so are the powers $A^2,A^ 3,\dots$. I am wondering if the converse is true, more precisely:

Is there a $k\in\mathbb N$ and a singular matrix $A$ such that $A^k$ is special, but not $A$ itself ?

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  • $\begingroup$ Is there a special motivation for this question? (Not that I don't like it for its own sake :) $\endgroup$ Oct 24 '13 at 13:59
  • $\begingroup$ I changed "Zeilen/lines" to "rows", hope you don't mind. $\endgroup$
    – Suvrit
    Oct 24 '13 at 14:20
  • $\begingroup$ @Felix: I stumbled about that question when looking at the matrices in mathoverflow.net/questions/52370/…. Sure enough, singular matrices tend to seem much "less interesting" than invertible ones. $\endgroup$
    – Wolfgang
    Oct 24 '13 at 15:11
  • $\begingroup$ @suvrit: Thank you. Dieser Fehler passiert mir erstaunlich oft. :( $\endgroup$
    – Wolfgang
    Oct 24 '13 at 15:13
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    $\begingroup$ I'm confused by the added part of the question. Isn't a matrix without nilpotent Jordan blocks necessarily non-singular? $\endgroup$ Oct 24 '13 at 16:59
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Here is a non nilpotent example.

Take $A = \begin{pmatrix} 2 & 1 & 1 \\ 1 & 1 & -1 \\ 3 & 2 & 0 \end{pmatrix}$. Then $A$ is not special, but $A^2 = \begin{pmatrix} 8 & 5 & 1 \\ 0 & 0 & 0 \\ 8 & 5 & 1 \end{pmatrix}$ is. Moreover, $A^{r+2} = 3^r\cdot A^2$ for any $r\geq 1$, so $A$ is not nilpotent.


Note that for $2\times 2$ matrices, if a power of a matrix is special, then the matrix itself is. Indeed, assume that the two rows of $A^k$ are equal and non-zero. Since any special matrix is singular, then there is a scalar $\lambda$ such that the second row of $A$ is $\lambda$ times the first. This is still true for any power of $A$; thus the second row of $A^k$ is $\lambda$ times the first. Since they are equal and non-zero, $\lambda = 1$, so $A$ is special.


Added to fit the comments to the question:

If you assume that only one of the eigenvalues of the $n\times n$ matrix $A$ is $0$, then the answer becomes negative. Indeed, in that case, $A^r$ has rank $n-1$ for any $r\geq 1$. Without loss of generality, we can assume that the first $n-1$ rows of $A$, say $R_1, \ldots, R_{n-1}$, are linearly independent. The last row is then some linear combination $R_n = L(R_1, \ldots, R_{n-1})$ of the others.

But then, if $R_1^{(r)}, \ldots, R_n^{(r)}$ are the rows of $A^r$, we still have that $R_n^{(r)}=L(R_1^{(r)}, \ldots, R_{n-1}^{(r)})$, so the first $n-1$ rows of $A^r$ are still linearly independent (since $A^r$ has rank $n-1$). If $A^r$ is special, then $R^{(r)}_n$ has to be equal to some $R_i^{(r)}$; this means that the coefficients in $L$ are all zero, except the one in front of $R_i^{(r)}$, which is $1$. Thus $R_n = R_i$, and $A$ is special.

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  • $\begingroup$ Yes, it is not nilpotent and nice. But sure enough, its Jordan decomposition is $M=SJS^{-1}$ where $S=\pmatrix{-2&-5&1\\3&8&0\\1&0&1},J=\pmatrix{0&1&0\\0&0&0\\0&0&3}$. And $J$ contains a nilpotent 2x2 Jordan block... $\endgroup$
    – Wolfgang
    Oct 24 '13 at 17:24
  • $\begingroup$ I'm still confused by your added condition. Is it not true that a matrix without nilpotent Jordan blocks is non-singular? Thus it cannot be special, nor can any of its powers... $\endgroup$ Oct 24 '13 at 17:27
  • $\begingroup$ Thank you for the addition. I think you have settled all which can be said about this problem! $\endgroup$
    – Wolfgang
    Oct 24 '13 at 18:41
  • $\begingroup$ ......very nice! $\endgroup$
    – Suvrit
    Oct 24 '13 at 20:48
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Take $A= \begin{pmatrix} 0 & -1 & \,\,0 \\ 0 & -1 & -1 \\ 0 & \,\,\,1 & \,\,1 \end{pmatrix}$.

Then $A$ is not special, however $A^2=\begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},$ which is special since it has two identical non-zero columns.

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  • $\begingroup$ but this $A$ is nilpotent, and the OP asks for non nilpotent matrices...although by the orig. definition of special, it is indeed valid...so the question needs to be refined a bit. $\endgroup$
    – Suvrit
    Oct 24 '13 at 15:33
  • $\begingroup$ Well, he said that a priori nilpotent matrices don't yield counterxamples, "as the all-zero vector is not considered". Maybe I'm misreading the question, but it seems to me that this is a counterexample, since even if $A$ is nilpotent, $A^2$ has two nonzero identical columns, so it is special in the OP's sense. $\endgroup$ Oct 24 '13 at 15:38
  • $\begingroup$ anyway, I agree that maybe the question should be a bit refined. $\endgroup$ Oct 24 '13 at 15:40
  • $\begingroup$ Yes, I agree. I have refined the question to exclude matrices with nilpotent Jordan blocks. You have indeed provided a counterexample, and this suspected possibility was the very reason I had said "a priori"... This remark is moot now, so I have removed it, too. Obviously I would like to push it a bit further by excluding matrices that feature "too many zeros", so to speak. $\endgroup$
    – Wolfgang
    Oct 24 '13 at 16:49

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