What is the dimension of the subspace $S$ of the vector space of all $8×10$ matrices?

Question

Let $A$ be a $8 \times 8$ matrix and the rank of $A$ is $5$.

Let $M_{m \times n}(\Bbb{R})$ denotes the the vector space of all $m \times n$ matrices with real entries.

$$S=\{B\in M_{8 \times 10}(\Bbb{R}) \mid AB=0\}$$

Question: What is the dimension of set $S$ ?

My try: As rank of $AB$ is $0$ and matrix $A$ only have to face the range of $B$ in product $AB$. So range space of $B$ must be contained in null space of $A$. So rank of $B \le 3$. I can't conclude forward from right here.

Answer 1

Hint: As you have correctly noted, $AB = 0$ if and only if the range of $B$ is a subset of the nullspace of $A$. Equivalently, $AB = 0$ if and only if each column of $B$ is an element of the nullspace of $A$. With that in mind, start with a basis $\{v_1,v_2,v_3\}$ of the nullspace of $A$ and use this to construct a basis of $S$ such that each element of the basis has exactly one non-zero column.

How many elements are in this basis?

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The basis I am hinting at is $\mathcal B = (M_1,M_2,M_3,\dots,M_k)$, where $$ M_1 = \pmatrix{v_1 & 0 & 0 & \cdots & 0},\\ M_2 = \pmatrix{v_2 & 0 & 0 & \cdots & 0},\\ M_3 = \pmatrix{v_3 & 0 & 0 & \cdots & 0},\\ M_4 = \pmatrix{0 & v_1 & 0 & \cdots & 0},\\ \vdots\\ M_k = \pmatrix{0 & 0 & 0 & \cdots & v_3}. $$

Answer 2

Remember the dimension theorem \begin{equation} \dim(V)= \dim \mbox{rank}\ (T) + \dim \ker (T). \end{equation} In this case $T$ is the transformation associated with matrix A and $\operatorname{rank}(A)=\operatorname{rank}(T)=5$.

Answer 3

$M_{m×n}(\mathbb{R}) $: Set of all $m×n$ matrices over $\Bbb{R}$.

$T:M_{8×8}(\Bbb{R})×M_{8×10}(\Bbb{R})\to M_{8×10}(\Bbb{R}) $ defined by $$T(A, B) =AB$$ is a Bilinear map.

For fixed $A\in M_{8×8}(\Bbb{R}) $ , $$T_A:M_{8×10}(\Bbb{R}) \to M_{8×10}(\Bbb{R}) $$ is a linear map.

Then $S=\ker(T_A) =\{B\in M_{8×10}(\Bbb{R}):T_A(B)=0\}$

Let $M=M(T_A) $ denote the matrix of $T_A$ with respect to standard basis $\mathcal{B}=\{E_{ij}:1\le i\le 8,1\le j\le 10\}$ where $E_{ij}$ has $1$ in $ij$-the place and zeros everywhere.

Then

$M=\begin{pmatrix}A& &&&\\&A&&&\\&&A&&\\&&&\ddots&&\\&&&&A\end{pmatrix}_{80×80}$

Matrix of $T_A$ is the block diagonal matrix with all diagonals $A$ and repeated $10$ times on the diagonal.

Hence $\operatorname{rank}(T_A) =10\times \operatorname{rank}(A) =10\times 5=50$

$\dim(\ker(T_A)) =80-50=30$

$$\text{or}$$

$\begin{align}\dim(\ker(T_A)) &=10\times \dim(\ker(A) \\&=10(8-\operatorname{rank}(A))\\&=10(8-5)\\&=30\end{align}$

Answer 4

We have that $AB=0$ if and only if $\operatorname{im}(B)\subset \ker(A)$. By the rank-nullity theorem, $\operatorname{rank}(A)+\operatorname{dim}\ker(A)=8$, so $\operatorname{dim}\ker(A)=3$.

We can decompose such a $B$ as the composition $$B:\mathbb R^{10}\stackrel{B'}{\longrightarrow} \ker(A)\stackrel{\iota}{\longrightarrow}\mathbb R^{8}$$

where $\iota$ is the inclusion of $\ker A$ into $\mathbb R^{8}$ Since the inclusion is injective, there is a unique $B'$ associated to each $B$ in our space. So the collection of $B$ such that $AB=0$ is in one to one correspondence with $\hom(\mathbb R^{10},\ker(A)).$

Given two vector spaces, the dimension of the space of linear maps between them $\hom(V,W)=\operatorname{dim}(V)\cdot \operatorname{dim}(W)$. So $\hom(\mathbb R^{10},\ker(A))=10\times 3=30.$