Ytukyg

Let $A$ be a fixed $ntimes n$ matrix over a field $F$. We can look at the subspace
$$W={Xin M_{n,n}(F); AX=XA=0}$$
of the matrices which fulfill both $AX=0$ and $XA=0$.

Looking a these equations we get that all columns of $X$ have to fulfill the equation $Avec c=vec 0$. (Let us say we're working with column vectors.) Similarly we get for the rows $vec r^T A=vec 0^T$. This tells us that if we are looking at the possible choices for columns/rows of the matrix $X$, they have to be in a subspace of dimension $n-operatorname{rank}A$ (in the right/left null space of $A$).

At least in some cases it is almost immediately possible to find $W$ or at least $dim W$.

• Obviously, if $A$ is invertible, then $W={0}$ and $dim W=0$.


• Another trivial case is when $A=0$, which gives us $W=M_{n,n}$ and $dim W=n^2$.


• Slightly less trivial but still simple case is when $operatorname{rank} A=n-1$. In this case the condition on rows/columns give us one-dimensional spaces, so there are non-zero vectors $vec r$, $vec c$ such that each row has to be multiple of $vec r^T$ and each column has to be a multiple of $vec c$. Up to a scalar multiple, there is only one way how to get such a matrix and we get that $W$ is generated by the matrix $vec cvec r^T$ and $dim W=1$.

The general case seems to be a bit more complicated. If we denote $k=n-operatorname{rank}A$, we can use the same argument to see that there are $k$ linearly independent vectors $vec c_1,dots,vec c_k$ such that the columns have to be linear combinations of these vectors. Similarly, row can be chosen only from the span of the linearly independent vectors $vec r_1,dots,vec r_k$. (This is again just a direct consequence of $Avec c=vec 0$ and $vec r^TA=vec 0^T$.)

Using these vectors we can get $k^2$ matrices $$A_{ij}=vec c_i vec r_j^T$$
for $i,jin{1,2,dots,k}$. Unless I missed something, it seems that showing that these matrices are linearly independent is not too difficult. So we should get that $$dim W ge k^2 = (n-operatorname{rank}A)^2.$$
It is not obvious to me whether these vectors actually generate $W$. (And perhaps something can be said about the dimension of $W$ without exhibiting a basis.)

You may notice that in the three trivial examples above (with $k=0,1,n$) we got the equality $dim W=(n-operatorname{rank}A)^2$.

Another possible way to look at this problem could be to use the linear function
$$fcolon Xto(AX,XA)$$
$fcolon M_{n,n} to M_{n,n}oplus M_{n,n}$, then we have $W=operatorname{Ker} f$, so we are basically asking for the dimension of the kernel of this map.
So to find $dim W$ it would be sufficient to find $dimoperatorname{Im} f$. However, this does not seem to be easier than the original formulation of the problem.

It is also possible to see this as a system of $n^2$ linear equations with $n^2$ unknowns $x_{11}, x_{12}, dots, x_{nn}$. If we try to use this line of thinking, the difficult part seems to be determining how many of those equations are linearly dependent.

Question: What can be said about the dimension of the subspace $W$? Is it equal to $(n-operatorname{rank}A)^2$? Is it determined just by the rank of $A$? If not, what are best possible bounds we can get, if we know only the rank of $A$ and have no further information about $A$?

Motivation for this question was working on an exercise which asked for calculating dimensions of spaces $W_1$, $W_2$, $W_1cap W_2$ and $W_1+W_2$, where the spaces $W_1$ and $W_2$ were determined by the conditions $AX=0$ and $XA=0$, respectively. Since the matrix $A$ was given, in this exercise it was possible to find a basis of $W_1cap W_2$ explicitly. (And the exercise was probably intended just to make the students accustomed to some basic computations such as finding basis, using Grassmann's formula, etc.) Still, I was wondering how much we can say just from knowing the rank of $A$, without going through all the computations.

There are invertible matrices $P$ and $Q$ such that $A=PJQ$ where
$J=pmatrix{I_r&0\0&0}$ with $I_r$ an identity matrix of size $r=text{rank}(A)$.
Then $AX=0$ iff $PJQX=0$ iff $J(QXP)=0$. Likewise $XA=0$ iff $(QXP)J=0$.
Let $Y=QXP$. Then $YJ=JY=0$ iff $Y=pmatrix{0&0\0&*}$. So the dimension
of admissible $Y$ (and so of admissible $X$) is $(n-r)^2$.

Yes, the dimension is always $(n - operatorname{rank}(A))^2$. Here's one justification.

For the convenience of eigenvalue stuff, I assume that $F$ is algebraically closed, or at least that we can appeal to the existence of its algebraic closure.

Let $V$ denote the subspace $V_0 = {X: AX = XA}$. That is, $V$ is the solution space to the Sylvester equation $AX - XA = 0$. By using some vectorization tricks, we can see that $V_0$ is spanned by the matrices of the form $xy^T$ such that $Ax = lambda x$ $A^Ty = lambda y$ for some $lambda in bar F$. We can see that $dim(V_0) = sum d_k^2$ where $d_k$ is the geometric multiplicity of the $k$th eigenvalue.

Some care is required in showing that this basis spans $V_0$ for a non-diagonalizable $A$. One way to show that this happens is to compute the kernel of $I otimes A - A^T otimes I$, taking $A$ to be in Jordan canonical form.

The space $W$ that you're looking for is the intersection $V_0$ with the kernel of $X mapsto AX$. This is spanned by the vectors $xy^T$ such that $x in ker(A)$ and $y in ker(A^T)$. Your conclusion follows.

Here is a generalized version where you may be dealing with infinite dimensional vector spaces. For a given linear map $T:Vto V$ on a vector space $V$, I have a description of all linear maps $S:Vto V$ such that $ST=TS=0$.

Let $V$ be a vector space over a field $F$ and let $T:Vto V$ be a linear transformation. Define $L_T:operatorname{End}_F(V)to operatorname{End}_F(V)oplus operatorname{End}_F(V)$ via
$$L_T(S)=(ST,TS).$$
We claim that there exists an isomorphism $varphi: ker L_Tto operatorname{Hom}_F(operatorname{coim} T,ker T)$ of vector spaces, where $operatorname{coim} T$ is the coimage of $T$: $$operatorname{coim} T=V/operatorname{im}T.$$

Observe that $operatorname{im}Ssubseteq ker T$ and $operatorname{im}Tsubseteq ker S$ for all $Sinker L_T$. Let $pi:Vto operatorname{coim}T$ be the canonical projection $vmapsto v+operatorname{im}T$. For $Sin ker L_T$, we see that $S:Vtoker T$ factors through $pi$, i.e., $S=tilde{S}circ pi$ for a unique linear map $tilde{S}:operatorname{coim}Ttoker T$.
We define $varphi:ker L_Tto operatorname{Hom}_F(operatorname{coim} T,ker T)$ in the obvious manner: $Smapsto tilde{S}$. This map is clearly an isomorphism with the inverse map $$varphi^{-1}(X)=Xcircpi$$ for all $Rin operatorname{Hom}_F(operatorname{coim} T,ker T)$. The claim is now justified.

The nullity $operatorname{null} T$ of $T$ is the dimension of the kernel of $T$. The corank $operatorname{cork}T$ of $T$ is the dimension of $operatorname{coim} T$. In the case $operatorname{null}T<infty$ or $operatorname{cork}T<infty$,
$$operatorname{Hom}_F(operatorname{coim} T,ker T)cong (ker T)otimes_F (operatorname{coim}T)^*,$$
where the isomorphism is natural, so
$$operatorname{null}L_T=dim_F ker L_T=(operatorname{null}T)big(dim_F(operatorname{coim}T)^*big)$$
in this case. In particular, if $operatorname{cork}T<infty$, we have $(operatorname{coim}T)^*cong operatorname{coim}T$, so that
$$operatorname{null}L_T=(operatorname{null}T)big(dim_F(operatorname{coim}T)^*big)=(operatorname{null}T)(dim_Foperatorname{coim}T)=(operatorname{null}T)(operatorname{cork}T).$$
Particularly, when $V$ is finite dimensional, we have $operatorname{cork}T<infty$, and by the rank-nullity theorem, we get $operatorname{cork}T=operatorname{null}T=dim_F V-operatorname{rank}T$, and so
$$operatorname{null}L_T=dim_F ker L_T=(dim_F V-operatorname{rank}T)^2$$
as the OP conjectures. (But if $V$ is infinite dimensional, for any pair $(m,k)$ of non-negative integers, there exists $Tinoperatorname{End}_F(V)$ with nullity $m$ and corank $k$.)

Here is example of $T:Vto V$ with nullity $m$ and corank $k$ when $V$ is infinite dimensional. Pick a basis $B$ of $V$. Since $B$ is infinite, it has a countable subset ${b_1,b_2,b_3,ldots}$. Let $Y$ be the span of ${b_1,b_2,b_3,ldots}$ and $Z$ the span of $Bsetminus{b_1,b_2,b_3,ldots}$. Then, $V=Yoplus Z$. Define $T:Vto V$ as follows: $$Tleft(sum_{i=1}^infty s_i b_i+zright)=sum_{i=1}^infty s_{m+i} b_{k+i}+z$$ for all $s_1,s_2,s_3,ldotsin F$ with only finitely many non-zero terms and for all $zin Z$. We have $ker T=operatorname{span}{b_1,b_2,ldots,b_m}$ and $V=(operatorname{im} T)oplus operatorname{span}{b_1,b_2,ldots,b_k}$, so $T$ has nullity $m$ and corank $k$.

The situation is not so straightforward when $T$ has infinite corank. If $operatorname{null}T<infty$, then we already know that
$$operatorname{null}L_T= (operatorname{null}T)big(dim_F(operatorname{coim}T)^*big),.$$
From this mathoverflow thread, $dim_F(operatorname{coim}T)^*=|F|^{operatorname{cork}T}$. So, we have two cases when $operatorname{null}T$ is finite but $operatorname{cork}T$ is infinite:
$$operatorname{null}L_T= begin{cases}0&text{if} operatorname{null}T=0,\
|F|^{operatorname{cork}T}&text{if} 0<operatorname{null}T<infty.end{cases}$$
If both $operatorname{null}T$ and $operatorname{cork}T$ are infinite, we can use the result from the same mathoverflow thread to prove that
$$operatorname{null}L_T=operatorname{Hom}_F(operatorname{coim} T,ker T)=maxleft{|F|^{operatorname{cork}T},(operatorname{null}T)^{operatorname{cork}T}right}.$$

Even more generally, let $U$ and $V$ be vector spaces over $F$. For $Rinoperatorname{End}_F(U)$ and $Tinoperatorname{End}_F(V)$, define $L_{R}^T:operatorname{Hom}_F(U,V)tooperatorname{Hom}_F(U,V)oplus operatorname{Hom}_F(U,V)$ by $$L_R^T(S)=(SR,TS).$$ (That is, when $U=V$, we have $L_T=L_T^T$.) Then, there exists an isomorphism of vector spaces
$$varphi:ker L_R^Tto operatorname{Hom}_F(operatorname{coim}R,ker T).$$
In particular, if $U$ and $V$ are both finite dimensional, then
$$operatorname{null} L_R^T=dim_Fker L_R^T=(operatorname{cork}R)(operatorname{null} T)=(dim_FU-operatorname{rank}R)(dim_FV-operatorname{rank}T).$$
In general,
$$operatorname{null}L_R^T=begin{cases}(operatorname{cork} R)(operatorname{null}T)&text{if} operatorname{cork}R<infty,\
0&text{if} operatorname{null} T=0,\
|F|^{operatorname{cork}R}&text{if} 0<operatorname{null} T<infty wedge operatorname{cork}R=infty,\
maxleft{|F|^{operatorname{cork}R},(operatorname{null} T)^{operatorname{cork}R}right}&text{if} operatorname{null}T=infty wedge operatorname{cork}R=infty.
end{cases}$$

This is my old proof that $operatorname{null}L_T=(operatorname{null}T)(operatorname{cork}T)$ when $T$ has finite nullity and finite corank.
Suppose that $T$ has finite nullity $m$ and finite corank $k$, I claim that $L_T$ also has finite nullity $mk$.

For $Sinker L_T$, we see that $operatorname{im} Ssubseteq ker T$ and $operatorname{im} Tsubseteq ker S$. Because $T$ has finite nullity $m$, it follows that $S$ has finite rank $rleq m$. Therefore,
$$S=v_1otimes phi_1+v_2otimes phi_2+ldots+v_rotimes phi_r$$
for some linearly independent $v_1,v_2,ldots,v_rin ker T$ and for some linearly independent $phi_1,phi_2,ldots,phi_rin V^*=operatorname{Hom}_F(V,F)$. Since $v_1,v_2,ldots,v_r$ are linearly independent, $$ker S=bigcap_{i=1}^rker phi_i.$$
Therefore, $operatorname{im} T$ must be contained in $ker phi_i$ for all $i=1,2,ldots,r$.

Since $T$ has finite corank $k$, $W=V/operatorname{im} T$ is a finite dimensional vector space of dimension $k$. Note that each $phi_i$ factors through $operatorname{im} T$. That is, $phi_i=psi_icirc pi$, where $pi:Vto V/operatorname{im} T=W$ is the canonical projection and $psi_iin W^*=operatorname{Hom}_F(W,F)$. We can now conclude that each $Sin ker L_T$ is of the form
$$sum_{i=1}^r v_iotimes (psi_icirc pi),$$
where $v_1,v_2,ldots,v_rin ker T$ are linearly independent and $psi_1,psi_2,ldots,psi_rin W^*=left(V/operatorname{im} Tright)^*$ are linearly independent.

Define the linear map $f:(ker T)otimes_F W^*toker L_T$ in the obvious manner:
$$votimes psimapsto votimes (psicircpi).$$
By the observation in the previous paragraph, $f$ is surjective. By choosing a basis of $ker T$, say ${x_1,x_2,ldots,x_m}$, we see that an element in $ker f$ must take the form
$$sum_{i=1}^m x_iotimes alpha_i$$
for some $alpha_iin W^*$. Since $x_1,ldots,x_m$ are linearly independent, we must have that $alpha_icirc pi=0$ for all $i$. But this means $alpha_i=0$ as $pi$ is surjective. Thus, $ker f={0}$, and so $f$ is injective. Hence,
$$ker L_Tcong (ker T)otimes_F W^*=(ker T)otimes_F (V/operatorname{im} T)^*.$$
This establishes the assertion that $L_T$ has nullity $mk$.

One can consider $U={(A,B)in M_ntimes M_n;AB=BA=0},V={(A,B)in M_ntimes M_n;AB=0}$.

$U,V$ are closed algebraic sets stratified by $rank(A)$.

Let $W_r$ be the algebraic set of matrices of rank $r$; from $dim(W_r)=r(2n-r)$, we deduce that the dimension of a stratum is $(n-r)^2+r(2n-r)=n^2$. In particular, the strata have same dimension and $dim(U)=n^2$.

You'd think $V$ has about the same dimension as $U$, for example, $dim(V)=dim(U)+O(n)$. This is not the case; recall that, when $AB=0$, we may have $rank(BA)=n/2$.

Using the Lord Shark the Unknown's post, we obtain that the dimension of a stratum is $d_r=[r(n-r)+(n-r)^2]+r(2n-r)=n^2+nr-r^2$ and depends on $r$.

Since $max(d_r)$ is obtained with $r=n/2$, we deduce that $dim(V)=floor(5n^2/4)$.

Now we can seek the singular locus of $U$ or $V$.