Image of the map that multiplies every matrix by a rank $r$ matrix

Question

Motivated by this question, this question.

Let $n\ge m, A \in \mathbb{R}^{m\times n}$ be fixed. Define $$T_A:\mathbb{R}^{m\times n}\to \mathbb{R}^{m\times m}:= T_A(B)=AB^T.$$

Assume $rank(A)=r.$ How do I show that the maximum rank of a matrix in the image $Im(T_A)$ of $T_A$ is also r? So I just need to get one $B$ so that $AB$ has rank $r.$

When $r:=m,T_A$ is surjective, so the claim is proved. To show that $T_A$ is surjective, one can work with the right inverse, say $C,$ of $A:$ so there is a $C \in \mathbb{R}^{m\times n}$ so that $AC^T=I_m.$ And then for any $D \in \mathbb{R}^{m\times m},$ we can take $B:=D^TC$ so that $AB^T=AC^TD=I_mD=D.$

But if $r<m,T_A$ won't be surjective, as all matrices will have rank $\le r <m.$ But how do we show that the maximal rank of $r$ is attained by $T_A, i.e. \exists B$ so that $rank(AB^T)=r?$

The way I'm trying this is:

We can pick the $r\times r $ minor/submatrix $E$ of $A,$ that's invertible, and the $r \times k$ block submatrix $F$ that 'contains' $E$ as a submatrix, so $F,$ as a $m\times r$ matrix, has full rank $r,$ and then we can consider $G,$ the $k\times r$ right inverse of $F$. Now $rank(G)=r$ as well and $FG=I_r.$ I tried to complete $G$ by adding some rows to it, but it didn't quite go anywhere...hints appreciated! It seems we may need to work with the above $E$ or $F,$ and reduce the problem to one where the restricted linear map has full rank, but am a bit stuck here.

I just looked at this answer, and it seems the answer should be $B:=A?!$ So if $rank(A)=r, rank(AA^T)=r$ as well? :)

Answer 1

You are asking tho prove that given a $nxm$ matrix $C$, then $AC$ has rank less or equal than $A$.

You have to think what are the columns (or equivalently the rows) of a matrix $AC$ in general, $C=(c_{ij})_{1\leq i \leq n, 1\leq j\leq m}$. Indeed, you can observe that if we denote by $D_i$ the $i-th$ column of a matrix $D$, then

$$ (AC)_i= c_{1i}A_1+\dots c_{ni}A_n$$

But then the subspace generated by the columns of the matrix $AC$ is contained in the space generated by the columns of $A$:

$$ \langle (AC)_1, \dots, (AC)_m\rangle \subseteq \langle A_1,\dots, A_n\rangle $$

By the definition of rank of a matrix, namely the dimension of the space generated by the columns of the matrix, this means

$$rk(AC)\leq rk(A).$$

You can also observe that the rank of $AC$ is always less or equal than the rank of $C$:

$$rk(AC)\leq \min(rk(A),rk(C)).$$

This means that in order to have $AC$ of rank $r$, then also the rank of $C$ has to be $r$.

Moreover, you can prove the following:

Let us consider the natural restriction map $C_{|\ker(AC)}\colon \ker(AC)\to \mathbb R^n$, $x\mapsto Cx$. Then

$$\ker(A)\cap Im(C)=Im_C(\ker(AC))$$

and

$$rk(AC)=rk(C)-\dim(\ker(A)\cap Im(C)).$$

In particular, if $\ker(A)\cap Im(C))=\{0\}$, then the rank of $AC$ is the rank of $C$.

$\textbf{Proof.}$ The first set equality is easy to prove.

Finally, from null rank theorem on the restriction map $C$ we have

$$ rk(AC)=m-null(AC)=m-(dim(\ker(A)\cap Im(C))+null(C))=(m-null(C))-dim(\ker(A)\cap Im(C))=rk(C)-dim(\ker(A)\cap Im(C)).$$

Thus, you have to choose $C$ such that $C$ has rank $r$ equal to the rank of $A$ and and such that $\ker(A)\cap Im(C))=\{0\}$.

As you already observed, you can choose $C=A^t$ since the rank of $A^t$ is still $r$ and the intersection is zero. Indeed, if $AA^tx=0$, then

$$0=\langle AA^tx,x\rangle=\langle A^tx,A^tx\rangle=\vert \vert A^tx\vert \vert^2 \implies A^tx=0.$$

Note that the latter holds only if we are using the field $\mathbb R$ but it fails on $\mathbb C$. For instance, if you take $A=\begin{pmatrix} 1 & i \end{pmatrix}$, then $AA^t=1-1=0$, that it has no rank $1$.

Answer 2

Let $A\in \mathrm{Mat}_{m,n}(\mathbb R)$ be fixed and let $T_A\colon \mathrm{Mat}_{m,n}(\mathbb R)\to \mathrm{Mat}_{m,m}(\mathbb R)$ be given by $T_A(B) = AB^{\intercal}$.

Claim 1: If $r(A)=\mathrm{rank}(A)=d$ then $r(T_A(B))\leq d$ for all $B \in \mathrm{Mat}_{m,n}(\mathbb R)$.

The rank of a matrix is the dimension of its row or column space, but here it will be easier to work with the column space of $T_A(B)$.

Let $A = (\mathbf a_1|\ldots|\mathbf a_n)$ and $B=(\mathbf b_1|\ldots|\mathbf b_n)$, so that $\mathbf a_i$ and $\mathbf b_i$ are the i-th columns of $A$ and $B$ respectively. Then if we write $(AB^{\intercal})_k$ for the $k$-th column of $T_A(B) = AB^{\intercal}$ ($1\leq k\leq m$), we have $$ (AB^{\intercal})_k = \sum_{j=1}^n b_{kj}\mathbf a_j $$

It follows that $\mathrm{Span}\{(AB^{\intercal})_k: 1\leq k\leq m\} \subseteq \mathrm{Span}\{\mathbf a_1,\ldots,\mathbf a_n\}$, and hence the inequality $$ \begin{split} \mathrm{rank}(AB^{\intercal}) &= \dim(\mathrm{Span}\{(AB^{\intercal})_k: 1\leq k\leq m\} )\\ &=\dim(\text{Span}\{\sum_{j=1}^n b_{kj}\mathbf a_j: 1\leq k\leq m\})\\ &\leq \dim(\mathrm{Span}\{\mathbf a_1,\ldots, \mathbf a_n\})\\ &= \mathrm{rank}(A) \end{split} $$ follows immediately.

Moreover equality holds, i.e. $r(B^{\intercal}A)=r(A)$, when the columns of $AB^{\intercal}$ have the same span as the columns of $A$. One way to ensure this is as follows: the columns $\{\mathbf a_1,\ldots,\mathbf a_n\}$ of $A$ contain a subset $\{\mathbf a_{i_1},\ldots,\mathbf a_{i_d}\}$ which is a basis of the column spaces of $A$, where as above $d= r(A)$. (You can find such a subset by row reducing $A^{\intercal}$ and taking the columns of $A$ corresponding to leading $1$s in the reduced matrix). Let $B = (b_{ij})$ where $b_{ij} = 0$ unless $i=j=i_k$ for some $k$, $1\leq k\leq d$. Then $AB^{\intercal}$ has exactly $d$ nonzero columns $\{\mathbf a_{i_1},\ldots,\mathbf a_{i_d}\}$ and these by construction span the column spaces of $A$ so that $r(AB^{\intercal})=d=r(A)$

Note that there are, in general, many matrices $B$ for which $r(AB^{\intercal}) = r(A)$: In the diagram below, $\ker(A)= \{v \in \mathbb R^n: Av=0\}$ is the kernel of $A$. $$\require{AMScd} \begin{CD} \mathbb R^m @>B^{\intercal}>> \mathbb R^n @>A>> \mathbb R^m \\ @. @AAA @.\\ @. \ker(A) @. \end{CD} $$

Claim 2: The composition $AB^{\intercal}$ has rank $d$ if and only if $\mathrm{im}(B^{\intercal})+\ker(A)=\mathbb R^m$.

Proof: Clearly, if $\mathrm{im}(B^{\intercal})+\ker(A)=\mathbb R^n$, then $A(\mathbb R^n) = A(\mathrm{im}(B^{\intercal})) = \mathrm{im}(AB^{\intercal})$ so that $r(AB^{\intercal})=d$. Conversely, if $r(AB^{\intercal})=r(A)$, then since $\mathrm{im}(AB^{\intercal}) \subseteq \mathrm{im}(A)$, they are equal once they have the same dimension. But then $A^{-1}(A(\mathrm{im}(B^{\intercal})) = \mathrm{im}(B^{\intercal})+\ker(A)$ must be equal to $\mathbb R^n$ as required.

Thus if we pick a basis of the kernel of $A$, say $\{w_1,\ldots,w_{n-r}\}$ and extend it to a basis $\{w_1,\ldots,w_{n-r}, w_{n-r+1},\ldots,w_n\}$ then if $C$ is the matrix of the linear map $v\mapsto B^{\intercal}v$ with respect to the standard basis of $\mathbb R^m$ and the basis $\{w_1,\ldots,w_n\}$ of $\mathbb R^n$ then $AB^{\intercal}$ has rank $d$ precisely when some $d\times d$ minor of $C$ obtained by taking the last $d$ rows of $C$ and any $d$ columns must be nonzero. Now if $P$ denotes the change of basis matrix from the standard basis of $\mathbb R^n$ to the basis $\{w_1,\ldots,w_n\}$ then $C = PB^{\intercal}$, and hence it follows that the matrices $B$ for which $AB^{\intercal}$ has rank $r$ are an open dense subset of $\mathrm{Mat}_{m,n}(\mathbb R)$ (given by the locus where at least one of the relevant minors of $PB^{\intercal}$ does not vanish, and those minors are polynomial functions in the matrix entries of $B$).

The OP notes that another possibility for a $B$ with $r(AB^{\intercal})=r(A)$ is $B=A$. You can check that this choice for $B$ works by using the criterion $\mathrm{im}(A^{\intercal})+\ker(A)=\mathbb R^n$. Since $r(A)=r(A^{\intercal})$, by rank-nullity it suffices to show that $\mathrm{im}(A^{\intercal})\cap \ker(A)=\{0\}$.

Suppose that $A^{\intercal}v \in \ker(A)$. Then $AA^{\intercal}v =0$, and hence $$ 0=v^\intercal AA^{\intercal}v=(A^{\intercal}v)^{\intercal}(A^{\intercal}v) = \|A^{\intercal}v\|^2, $$ so that $A^{\intercal}v$ has length zero, and hence $A^{\intercal}v =0$. Thus it follows that $\ker(A)\cap \mathrm{im}(A^{\intercal})= \{0\}$ as required.

Answer 3

If $A=0$ then there is nothing to prove. So let $\text{rank}(A) = r > 0$. By rank factorization (or by SVD) there is a $m \times r$ matrix $U$ and an $n \times r$ matrix $V$, both of rank $r$ such that $A=UV^T$. Note the $r \times r$ matrix $V^TV$ is invertible (because it is an $r \times r$ matrix of rank $r$). Choose $B^T = V(V^TV)^{-1}U^T$. Then, $AB^T =UU^T$. Since $\text{rank}(UU^T) = \text{rank}(U) = r$ we are done.

If you know about the Moore-Penrose inverse, you can choose $B^T$ to be the Moore-Penrose inverse of $A$