characterize all matrices $X$ such that $BA = X$ whenever $AB = X$

Question

It is clear that if $A$ and $B$ are $n\times n$ matrices (over a field) with $AB = I$ then $BA = I$. I like to characterize all matrices $X$ such that $BA = X$ whenever $AB = X$.

score 3 · Answer 1 · answered Sep 07 '13 at 09:41

3

If $AB=X$ then we have $BAB=BX$. So $BA=X$ can only follow if $BX=XB$ for all $B$. Therefore $X$ has to be in the center of the general linear group which is all multiples of the identity matrix $I$.

answered Sep 07 '13 at 09:41

Stefan

418
3
12

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Why "for all $B$"? – njguliyev Sep 07 '13 at 09:48
Because I assumed the op is talking about arbitrary matrices $A$ and $B$. But rethinking this you might be right. $X$ should be in the centralizer of $A$ and $B$ then, not in the center. – Stefan Sep 07 '13 at 10:02

score 3 · Answer 2 · answered Sep 07 '13 at 12:45

3

Hint: Let $a\ne 0,a\ne 1, \ 1\le k\le n$, $A=(a_{ij})$ where $$a_{ij}=0 \ {\rm for} \ i\ne j,$$ $$a_{kk}=a, $$ $$a_{ii}=1 \ {\rm for} \ i\ne k.$$

We can find $B$ such that $X=AB$. Then from $X=BA$ we get that $X$ is diagonal.

answered Sep 07 '13 at 12:45

Boris Novikov

17,470

score 1 · Answer 3 · answered Sep 07 '13 at 14:27

Hint. Suppose $n>1$ (the case $n=1$ is trivial and the answer is a bit different from that for $n>1$). The given condition implies that $AX=XA$ whenever $AB=X$. Now consider $A=I+E_{ij}$, where $i\ne j$ and $E_{ij}$ is the matrix whose $(i,j)$-th entry is $1$ and all other entries are zero. From the above discussion, argue that $E_{ij}X=XE_{ij}$. Entrywise, this means $x_{ii}=x_{jj}$ and all off-diagonal entries on row $j$ or column $i$ of $X$ are zero. Hence conclude that $X$ is a scalar multiple of the identity matrix. Finally, identify which scalar multiples of $I_n$ satisfy the given condition (not all of them do). This should be easy.

score -1 · Answer 4 · answered Sep 07 '13 at 09:21

-1

From mathworld.wolfram "matrix multiplication is not, in general, commutative (although it is commutative if A and B are diagonal and of the same dimension)."

So if for example $$A=\left(\begin{array}{ccc} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{array}\right)$$ and $$B=\left(\begin{array}{ccc} 4 & 0 & 0\\ 0 & 5 & 0\\ 0 & 0 & 6 \end{array}\right)$$ then $$AB=BA=\left(\begin{array}{ccc} 4 & 0 & 0\\ 0 & 10 & 0\\ 0 & 0 & 18 \end{array}\right)$$

answered Sep 07 '13 at 09:21

Ömer

2,038

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I am afraid you did not understand my question. – kian Sep 07 '13 at 09:30
I think the OP is trying to find a different way of writting the set $\left{X\in \mathcal M(\Bbb F)\colon \left(\exists A,B\in \mathcal M(\Bbb F)\right)(AB=X\implies BA=X)\right}$. – Git Gud Sep 07 '13 at 09:48
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-1 This doesn't answer the question at all. – rurouniwallace Sep 07 '13 at 12:50

characterize all matrices $X$ such that $BA = X$ whenever $AB = X$

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