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I have a real square matrix $M$ that I'd like to express as $M=A+B$ such that $A^2=0$,$B^2=0$. $M$ has an additional property that $M^2$ is a scalar matrix : ($M^2=s^2I$); and it's dimension is a power of 2 : $dim(M)=2^n,n>0$; Any suggestions?

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4 Answers 4

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Regarding what such matrices $M$ can be like and in what dimensions they can exist, here is a relatively thorough (but still incomplete) answer:

  • If $s=0$, then such matrices $M$ exist in all dimensions. You can take $A=M$ and $B=0$.
  • If $s\neq0$ (this will be assumed for all remaining bullets), such a matrix $M$ is diagonalizable and the only possible eigenvalues are $\pm s$.
  • If the dimension $d$ is even, a matrix $M$ with $M^2=s^2I_d$ can be written as $A+B$ with $A^2=B^2=0$ if the eigenvalues $+s$ and $-s$ have the same multiplicity. In other words, up to change of basis, it is sufficient that $M$ is the diagonal matrix $\operatorname{diag}(s,\dots,s,-s,\dots,-s)$ with equal numbers of $s$ and $-s$.
  • A necessary condition is that both $A$ and $B$ have the same rank $d/2$, but it is unclear whether it implies that the eigenvalues $\pm s$ of $M$ have equal multiplicities.
  • If the dimension $d$ is odd, no such matrices $M$ exist (unless $s=0$).

These results follow from the following observations. $\DeclareMathOperator{\rank}{rank}$

Lemma: If a square matrix $M$ satisfies $M^2=s^2I$ for $s\neq0$, then $M$ is diagonalizable and the only possible eigenvalues are $\pm s$.

Proof: The properties you are after are independent of basis, so write $M$ in it's Jordan normal form. Then each upper triangular Jordan block $J$ of $M$ will have to satisfy $J^2=s^2I$. First, this implies that the diagonal value (eigenvalue) of the block is $\pm s$. Second, since there cannot be any non-zero off-diagonal entries in $J^2$ and $s\neq0$, the off-diagonals of $J$ must in fact vanish. This means that $M$ is in fact diagonalizable. $\square$

Remark: The lemma is false for $s=0$, which is why I excluded it.

Case where A and B exist: If the eigenvalues $+s$ and $-s$ have same multiplicity, then $s^{-1}M$ is a direct sum (block diagonal matrix) of copies of the matrix $$ N = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} = X+Y, $$ where the matrices $ X = \frac12 \begin{pmatrix} 1&-1\\ 1&-1 \end{pmatrix} $ and $ Y = \frac12 \begin{pmatrix} 1&1\\ -1&-1 \end{pmatrix} $ have zero square. This gives a way to construct matrices $A$ and $B$ (as $s$ times a direct sum of the matrices given above).

Details of construction of $A$ and $B$: Suppose $$ M = s \begin{pmatrix} N&\cdots&0\\ \vdots&\ddots&\vdots\\ 0&\cdots&N \end{pmatrix}. $$ Any $M$ expressible in the desired way is of this form after a change of basis. You can choose $A$ be the block diagonal matrix $$ s \begin{pmatrix} X&\cdots&0\\ \vdots&\ddots&\vdots\\ 0&\cdots&X \end{pmatrix} $$ and $B$ can be given the same way using $Y$ instead of $X$. Now it's evident that $A^2=B^2=0$ (since $X^2=Y^2=0$), $A+B=M$ (since $X+Y=N$), and $M^2=s^2I$ (since $N^2=s^2I$).

General case: Assume $s\neq0$. Due to the lemma, an $d\times d$ matrix $M$ with $M^2=s^2I_d$ can be written in the block form $$ M = s \begin{pmatrix} I_a&0\\ 0&-I_b \end{pmatrix} $$ with $a+b=d$. If $a=b$, we can construct the matrices $A$ and $B$ as above. Such $A$ and $B$ don't exist if $a\neq b$, as we will see next.

Observations with rank: It follows from Sylvester's rank inequality that if a $d\times d$ matrix $A$ has zero square, then $\rank(A)\leq d/2$. The inequality is strict if $d$ is odd. Also, $\rank(M)=\rank(A+B)\leq\rank(A)+\rank(B)$. Thus unless both $A$ and $B$ have rank $d/2$, $M$ can't have the full rank $d$. If $s\neq0$, the condition $M^2=s^2I_d$ requires that $M$ has full rank.

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  • $\begingroup$ nice answer...the eigenvalues of the matrices I'm interested in do in fact have the same multiplicity and are nonzero. In fact, I know their eigenvectors : define $D=diag(s,\cdots,s,-s,\cdots ,-s)$ and define $V=M+D$; then $MV=VD$ and $V$ is a complete set of eigenvectors. I don't have a proof of the completeness part but it should follow from $M(M\pm sI)=\pm s(M\pm sI)$...($V^2$ is also scalar); so according to what you have $A$,$B$ should always exist..I would really like to see an explicit construction if possible $\endgroup$
    – unknown
    Commented Jun 28, 2017 at 6:49
  • $\begingroup$ @unknown Thanks! I added an explicit formula for $A$ and $B$, assuming that $M$ is expressed in a particular form (which is always possible after change of basis if $M$ is of the desired type). $\endgroup$ Commented Jun 28, 2017 at 13:26
  • $\begingroup$ @unknown If you do know that $M$ is similar to $(sI)\oplus(-sI)$, you should state that in the question. $\endgroup$
    – user1551
    Commented Jun 28, 2017 at 15:58
  • $\begingroup$ @JoonasIlmavirta In your third bullet point, you claim that what the OP asks for is possible if and only if $s$ and $-s$ have the same multiplicity, but I only see the "if" part in your answer. Did I miss something? $\endgroup$
    – user1551
    Commented Jun 28, 2017 at 16:03
  • $\begingroup$ @user1551 I didn't want the problem the be too specific but you can edit the question if you think that will improve it...also the $dim(M)=2^n$ can be relaxed to $dim(M)=2n$ since it doesn't affect the answers $\endgroup$
    – unknown
    Commented Jun 28, 2017 at 16:57
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Take $$ A=\begin{pmatrix} 1 & -r^{-1}\cr r & -1 \end{pmatrix},\quad B=\begin{pmatrix} 1 & -s^{-1}\cr s & -1 \end{pmatrix} $$ for non-zero $r,s$, and $M=A+B$. Then $A^2=B^2=0$ and $$ M^2=(A+B)^2=\frac{-r^2+2rs-s^2}{rs}I $$

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  • $\begingroup$ A more general solution can be obtained by multiplying $A$ by $a$ & $B$ by $b$. But your answer is good. $\endgroup$ Commented Jun 27, 2017 at 19:13
  • $\begingroup$ nice suggestion...this moves the problem to that of finding $r$ and $s$ that satisfy certain equations (here $r$ and $s$ are square invertible matrices of half the dimension of $M$; there's also a set of equation $M=A+B$ which might not have solutions if for example the diagonal of $M$ is zero. $\endgroup$
    – unknown
    Commented Jun 27, 2017 at 19:50
  • $\begingroup$ Sorry, I don't follow. You seem to have demonstrated the possibility of writing $M=A+B$ in some cases, but isn't the $M$ in the OP pre-specified? $\endgroup$
    – user1551
    Commented Jun 28, 2017 at 16:06
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This can always be done for $M$ of even size $2n$, i.e. $M$ is a $2n \times 2n$ matrix.

Set

$P = \begin{bmatrix} 0 & I_n \\ 0 & 0 \end{bmatrix}, \tag{1}$

where $I_n$ is the $n \times n$ identity matrix. Then

$P^2 = \begin{bmatrix} 0 & I_n \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & I_n \\ 0 & 0 \end{bmatrix} = 0; \tag{2}$

likewise take

$Q = P^T = \begin{bmatrix} 0 & 0 \\ I_n & 0 \end{bmatrix}; \tag{3}$

then

$Q^2 = (P^T)^2 = P^TP^T = (PP)^T = (P^2)^T = 0 \tag{4}$

as well. Then

$P + Q = \begin{bmatrix} 0 & I_n \\ I_n & 0 \end{bmatrix}, \tag{5}$

and

$(P + Q)^2 = \begin{bmatrix} 0 & I_n \\ I_n & 0 \end{bmatrix} \begin{bmatrix} 0 & I_n \\ I_n & 0 \end{bmatrix} = \begin{bmatrix} I_n & 0\\ 0 & I_n \end{bmatrix} = I_{2n}, \tag{7}$

so setting

$A = sP, \tag{8}$

$B = sQ, \tag{9}$

we find

$(A + B)^2 = s^2(P + Q)^2 = s^2 I_{2n} = M. \tag{10}$

Also worth noting:

$(P + Q)^2 = P^2 + PQ + QP + Q^2 = PQ + QP; \tag{11}$

$PQ = \begin{bmatrix} 0 & I_n \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ I_n & 0 \end{bmatrix} = \begin{bmatrix} I_n & 0 \\ 0 & 0 \end{bmatrix}, \tag{12}$

and likewise

$QP = \begin{bmatrix} 0 & 0 \\ 0 & I_n \end{bmatrix}. \tag{12}$

Remark: Not sure if there is a solution for $M$ of odd size.

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    $\begingroup$ I don't think it works for odd, but this is more intuition rather than proof. In order to get $A^2=0$ half or more the eigvals should be 0 (and same for $B^2=0$. At the same time, rank$(A+B) \leq$ rank($A$) $+$ rank($B$). So the only way that works is if $A$ and $B$ have half 0 eigenvalues, which only works if $m$ is even. $\endgroup$
    – Y. S.
    Commented Jun 27, 2017 at 23:19
  • $\begingroup$ @whyyes: good points. I think your ideas might be turned into a proof with a little more work/investigation. $\endgroup$ Commented Jun 27, 2017 at 23:30
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    $\begingroup$ @whyyes It is indeed true that odd dimensions are impossible. Also, it turns out that up to similarity, the matrix $M$ has to be of the form given in this answer. I put the details in a separate answer. $\endgroup$ Commented Jun 28, 2017 at 5:59
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    $\begingroup$ Sorry, I don't follow. You seem to have demonstrated the possibility of writing $M=A+B$ in some cases, but isn't the $M$ in the OP pre-specified? $\endgroup$
    – user1551
    Commented Jun 28, 2017 at 16:06
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Another class of solutions: $A = \left[\begin{matrix}0 &sI \\ 0& 0\end{matrix}\right]$, $B = \left[\begin{matrix}0&0\\sI &0\end{matrix}\right]$.

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  • $\begingroup$ $(A + B)^2 = sI$ $\endgroup$ Commented Jun 27, 2017 at 19:16
  • $\begingroup$ this doesn't solve $M=A+B$...for example if the diagonal of $M$ is nonzero. $\endgroup$
    – unknown
    Commented Jun 27, 2017 at 19:56
  • $\begingroup$ A+B = \sqrt{s} [0 I; I,0]. (A+B)^2 = sI. $\endgroup$
    – Y. S.
    Commented Jun 27, 2017 at 20:32
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    $\begingroup$ This does not answer the OP's question. The OP wants $M=A+B$, but you only make $M^2=(A+B)^2$. $\endgroup$
    – user1551
    Commented Jun 28, 2017 at 15:25
  • $\begingroup$ Wait I'm really confused. If $M = A+B$ then $M^2 = (A+B)^2$? The OP didn't specify any properties for $M$, only $M^2$, which is satisfied. $\endgroup$
    – Y. S.
    Commented Jun 29, 2017 at 16:42

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