# $V=\left \{ C|\space C\in M_{q\times r}(R)\space such \space that\space ACB=O \right \}$, find the dimension of the vector space V

Let $$A$$ be a $$p\times q$$ matrix of rank $$\alpha$$ and $$B$$ a $$r\times s$$ matrix of rank $$\beta$$. Let $$V=\left \{ C|\space C\in M_{q\times r}(\mathbb{R}) \text{ and } ACB=0_{p\times s} \right \}$$, find the dimension of the vector space V.

Can I solve this by the following......

(1)the range space of B is the domain of C, where rank(B)=$$\beta$$

(2)the range space of C covers the whole null space of A, where nullity(A)=q-rank(A)=q-$$\alpha$$

Hence the dimension of C need not to be q$$\times r$$, it only takes (q-$$\alpha)\times \beta$$ in necessary.

• What you did doesn't work. For example if $\beta =0$ then $B=0$ and $\dim V =qr$. Jun 4 at 20:03
• I suggest starting by assuming $A, B = \Sigma_1, \Sigma_2$ which are Diagonal matrices (albeit not necessarily square). You can then recover the general case via SVD (or rank normal form). If you know about Kronecker Products, the fact that $\text{vec}\big(ACB\big) = \big( B^T \otimes A\big)\text{vec}\big(C\big)$ combined with rank-nullity gives an immediate answer. Jun 4 at 20:37
• @user8675309 sorry sir, I have known nothing about Kronecker Products yet. Is there any possible alternatives(perhaps more basically please?) to solve this problem? Jun 4 at 20:51
• Yes-- do my first suggestion of assuming $A=\Sigma_1$ and $B=\Sigma_2$.. After you solve that, you should be able to recover the general case with SVD. No Kronecker Products needed if you go this route. Jun 4 at 20:56
• @user8675309 Sir, may I ask that is the answer = $(q-\alpha) \times (r-\beta)$?? Jun 4 at 21:55

Let $$f : \mathbb{R}^q \rightarrow \mathbb{R}^p$$, and $$g : \mathbb{R}^s \rightarrow \mathbb{R}^r$$ be the linear transformations canonically associated to $$A$$ and $$B$$ respectively.

The set $$W=\lbrace h \in \mathcal{L}(\mathbb{R}^r,\mathbb{R}^q) \mid f \circ h \circ g= 0 \rbrace$$ is clearly isomorphic to $$V$$, so we shall find the dimension of $$W$$.

Let $$\mathcal{B}=(e_1, ..., e_r)$$ be a basis or $$\mathbb{R}^r$$ such that $$(e_1, ..., e_\beta)$$ is a basis of $$\mathrm{Im}(g)$$.

Then for $$h \in \mathcal{L}(\mathbb{R}^r,\mathbb{R}^q)$$, one has the following \begin{align*} h \in W & \Longleftrightarrow h(\mathrm{Im}(g)) \subset \mathrm{Ker}(f) \\ & \Longleftrightarrow h(e_i) \in \mathrm{Ker}(f) \text{ for every } i=1, ..., \beta \\ & \Longleftrightarrow (h(e_1), ..., h(e_n)) \in \mathrm{Ker}(f)^\beta \times (\mathbb{R}^q)^{r-\beta} \end{align*}

Since the map $$\varphi : \mathcal{L}(\mathbb{R}^r,\mathbb{R}^q) \rightarrow (\mathbb{R}^q)^r$$ defined by $$\varphi(h)=(h(e_1), ..., h(e_n))$$ is an isomorphism, one deduces that $$\dim(W)=\dim(\mathrm{Ker}(f)^\beta \times (\mathbb{R}^q)^{r-\beta})=\beta \times\dim(\mathrm{Ker}(f)) + (r-\beta)q$$

i.e., since $$\dim(\mathrm{Ker}(f))=q-\alpha$$, one has $$\dim(W)=\beta(q-\alpha) + (r-\beta)q$$, i.e.

$$\boxed{\dim(W)=rq-\alpha\beta}$$

• Sir, could you please explain more about your"\begin{align*} h \in W & \Longleftrightarrow h_{|\mathrm{Im}(g)} \subset \mathrm{Ker}(f) \\ & \Longleftrightarrow h(e_i) \in \mathrm{Ker}(f) \text{ for every } i=1, ..., \beta \\ & \Longleftrightarrow (h(e_1), ..., h(e_n)) \in \mathrm{Ker}(f)^\beta \times (\mathbb{R}^q)^{r-\beta} \end{align*}"? I dont get the point here. Jun 13 at 6:39
• @mlrofcloud Sorry, there was a typo, I meant $h(\mathrm{Im}(g)) \subset \mathrm{Ker}(f)$. The fact that $h$ belongs to $W$ means that it sends the image of $g$ inside the kernel of $f$ ; equivalently, it means that all the basis of the image of $g$ is sent, by $h$, in the kernel of $f$. So you get that the $\beta$ first vectors $h(e_1)$, ..., $h(e_\beta)$ must belong to $\mathrm{Ker}(f)$, whereas there are no restrictions on the other ones, which belong to $\mathbb{R}^q$. Jun 13 at 7:13
• Sir, i guess what you actually mean is that the isomorphism $\varphi : W \rightarrow \mathrm{Ker}(f)^\beta \times (\mathbb{R}^q)^{r-\beta}$, hence to see the dimension of $W$, we can equivalently see the dimension of $\mathrm{Ker}(f)^\beta \times (\mathbb{R}^q)^{r-\beta}$? Jun 13 at 8:08
• Yes, this is basically the argument I used : if you consider the isomorphism $\varphi : \mathcal{L}(\mathbb{R}^r,\mathbb{R}^q) \rightarrow (\mathbb{R}^q)^r$ that I defined in my answer, then what we proved is that it induces an isomorphism between $W$ and $\mathrm{Ker}(f)^\beta \times (\mathbb{R}^q)^{r-\beta}$ (i.e. the restriction of $\varphi$ to $W$ is an isomorphism between $W$ and $\mathrm{Ker}(f)^\beta \times (\mathbb{R}^q)^{r-\beta}$). So indeed, the dimension of $W$ is the same as the dimension of $\mathrm{Ker}(f)^\beta \times (\mathbb{R}^q)^{r-\beta}$. Jun 13 at 8:21

It is not hard to check that $$V$$ is a vector space.

Since $$A$$ is a $$p \times q$$ matrix of rank $$\alpha$$, an invertible $$p \times p$$ matrix $$F_1$$ and an invertible $$q \times q$$ matrix $$F_2$$ exist such that $$A = F_1 A_0 F_2,$$ in which $$A_0 = \begin{bmatrix} I_{\alpha} & 0_{\alpha \times (q - \alpha)} \\ 0_{(p - \alpha) \times \alpha} & 0_{(p - \alpha) \times (q - \alpha)} \\ \end{bmatrix}.$$

Since $$B$$ is an $$r \times s$$ matrix of rank $$\beta$$, an invertible $$r \times r$$ matrix $$G_1$$ and an invertible $$s \times s$$ matrix $$G_2$$ exist such that $$B = G_1 B_0 G_2,$$ in which $$B_0 = \begin{bmatrix} I_{\beta} & 0_{\beta \times (s - \beta)} \\ 0_{(r - \beta) \times \beta} & 0_{(r - \beta) \times (s - \beta)} \\ \end{bmatrix}.$$

Hence $$ACB = 0$$ becomes $$F_1 A_0 (F_2 C G_1) B_0 G_2 = 0_{p \times s},$$ which means $$A_0 (F_2 C G_1) B_0 = 0_{p \times s}.$$ Denote the $$q \times r$$ matrix $$F_2 C G_1$$ by $$X$$. It is not hard to find that $$A_0 X B_0 = 0$$ if and only if $$[X]_{i,j} =0 \qquad \text{for i \leq \alpha and j \leq \beta},$$ in which $$[X]_{i,j}$$ is the $$(i,j)$$-entry of $$X$$.

Let $$E_{u,v}$$ be the $$q \times r$$ matrix with the property that $$[E_{u,v}]_{i,j} = \begin{cases} 1, & \text{u = i and v = j}; \\ 0, & \text{else}. \end{cases}$$

Let $$J$$ be the set $$\{ F_2^{-1} E_{u,v} G_1^{-1} \mid \text{u > \alpha or v > \beta} \}.$$

(1) Every solution to $$ACB = 0$$ is some linear combination of the members of $$J$$. If $$A C_0 B = 0$$, then $$A_0 (F_2 C_0 G_1) B_0 = 0$$, which means $$[F_2 C_0 G_1]_{i,j} =0 \qquad \text{for i \leq \alpha and j \leq \beta}.$$ Hence \begin{aligned} F_2 C_0 G_1 = {} &\sum_{\substack{ 1 \leq i \leq q \\ 1 \leq j \leq r \\ }} {[F_2 C_0 G_1]_{i,j} E_{i,j}} \\ = {} & \sum_{\substack{ 1 \leq i \leq q \\ 1 \leq j \leq r \\ i \leq \alpha \,\text{and}\, j \leq \beta }} {[F_2 C_0 G_1]_{i,j} E_{i,j}} + \sum_{\substack{ 1 \leq i \leq q \\ 1 \leq j \leq r \\ i > \alpha \,\text{or}\, j > \beta }} {[F_2 C_0 G_1]_{i,j} E_{i,j}} \\ = {} & \sum_{\substack{ 1 \leq i \leq q \\ 1 \leq j \leq r \\ i > \alpha \,\text{or}\, j > \beta }} {[F_2 C_0 G_1]_{i,j} E_{i,j}}, \end{aligned} which means $$C_0 = F_2^{-1} (F_2 C_0 G_1) G_1^{-1} = \sum_{\substack{ 1 \leq i \leq q \\ 1 \leq j \leq r \\ i > \alpha \,\text{or}\, j > \beta }} {[F_2 C_0 G_1]_{i,j} (F_2^{-1} E_{i,j} G_1^{-1})}.$$

(2) The members of $$J$$ are linearly independent. Suppose that $$\sum_{\substack{ 1 \leq i \leq q \\ 1 \leq j \leq r \\ i > \alpha \,\text{or}\, j > \beta }} {k_{i,j} (F_2^{-1} E_{i,j} G_1^{-1})} = 0.$$ Hence $$F_2 \left( \sum_{\substack{ 1 \leq i \leq q \\ 1 \leq j \leq r \\ i > \alpha \,\text{or}\, j > \beta }} {k_{i,j} (F_2^{-1} E_{i,j} G_1^{-1})} \right) G_1 = 0.$$ Hence $$\sum_{\substack{ 1 \leq i \leq q \\ 1 \leq j \leq r \\ i > \alpha \,\text{or}\, j > \beta }} {k_{i,j} E_{i,j}} = 0.$$ Hence $$k_{i,j} = 0$$.

We have shown that the members of $$J$$ form a basis of $$V$$.

$$J$$ has $$rq - \alpha \beta$$ members, so the dimension of $$V$$ is the number of the members of $$J$$, $$rq - \alpha \beta$$.

I will describe my (almost) complete thought process that led me to a solution. $$\DeclareMathOperator{\im}{Im}\\\DeclareMathOperator{\ker}{Ker}$$ I find it helpful to "draw" the chain of operators involved in the condition imposed on your $$C$$'s like this:

$$\forall v\in\mathbb{R}^{s}: v\overset{B}{\to}Bv\overset{C}{\to}CBv\overset{A}{\to}ACBv=0$$

You're looking at such $$C$$'s that every $$v$$ going through the multiple stages ends up the zero vector. And this is the "going forward" point of view.

It is now natural (at least to me) to look at what happens if we go over this chain backwards:

1. Starting from the zero vector, it has been produced by $$A$$, hence $$CBv\in\ker A$$.
2. Going back, $$CBv$$ has been produced by $$C$$. But $$C$$ is arbitrary, in the sense that it is not "given" in the problem statement, so we can't say much at this point about this part of the chain.
3. We're at the last piece of the chain now, $$v\overset{B}{\to}Bv$$. Here, $$B$$ is fixed while $$v$$ is "floating" around all of $$\mathbb R^s$$. What we can say here is that (again, by definition) $$Bv \in \im B$$.

What I said in all these words so far is near trivial: the $$C$$'s in $$V$$ are obliged to send $$\im B$$ within $$\ker A$$: $$\im B\overset{C}{\to}\ker A.$$

Even more, this is a sufficient property. That is, $$\forall C: ACB=0\Leftrightarrow C(\im B)\subseteq\ker A.$$

In other words, we're looking at all the $$C$$'s whose restriction on $$\im B$$ is a linear map $$\im B\to\ker A$$: $$V=\left\{ C:C\vert_{\im B}:\hom(\im B, \ker A)\right\}$$ Now note that outside of $$\im B$$, we have no restrictions on $$C$$ - $$C$$ is "free" to send the "rest" of $$\mathbb{R}^{r}$$ (any vector space complement, say, $$\im B^\perp$$) in any way it wants to $$\mathbb R^q$$. Hence $$\dim V= \dim\hom\left(\im B^\perp,\mathbb{R}^{q}\right) + \dim\hom(\im B,\ker A).$$

• @downvoters I fixed the problem in my answer with the $\text{Im} B$ complement and the dimension multiplication at the end. Replacing all dim's in my final result with the givent constants I get exactly what TheSilverDoe got in his answer. Jun 12 at 21:56