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Let $A \in M_{m\times n}(\mathbb{R})$ and $B \in M_{n\times p}(\mathbb{R})$.

Prove that if $\operatorname{rank}(A)=n$ then $\operatorname{rank}(AB)=\operatorname{rank}(B)$.

I tried to start with definitions finding that $n \le m$, but didn't know what to do with $AB$.

Please help, thank you!

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  • $\begingroup$ how about seeing for square matrices first? $\endgroup$ – user87543 Feb 14 '14 at 13:31
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First observation is that $A^tA$ is non-singular. If $A^tAx=0$, for some $x=(x_1,\ldots,x_n)\in\mathbb R^n$, with $x\ne 0$, then $$ 0=\langle A^tAx,x\rangle=\langle Ax,Ax\rangle, $$ which implies that $Ax=0$. In $A=(a_1,\ldots,a_n)$, with $a_j$'s the columns of $A$, then $Ax=x_1a_1+\cdots+x_na_n=0$, means that the $n$ columns of $A$ are linearly dependent, and hence its rank is less than $n$.

Let $$b_1,\ldots,b_p,$$ be the columns of $B$, i.e., $B=(b_1,\ldots,b_p)$, and assume that rank$(B)=k$, and $b_{i_1},\ldots,b_{i_k}$ are linearly independent. We shall show that $Ab_{i_1},\ldots,Ab_{i_k}$ are also linearly independent. If not, then $$ c_1Ab_{i_1}+\cdots+c_kAb_{i_k}=0\quad\Longrightarrow\quad c_1A^tAb_{i_1}+\cdots+c_kA^tAb_{i_k} =0, $$ and thus $A^tA(c_1b_{i_1}+\cdots+c_kb_{i_k})=0$, and as $A^tA$ is non-singular, then $c_1b_{i_1}+\cdots+c_kb_{i_k}=0$, which is a contradiction. Thus $$ \mathrm{rank}\,(AB)\ge\mathrm{rank}\,(B). $$ The converse is as easy, since if rank$(AB)=k$, and $Ab_{i_1},\ldots,Ab_{i_k}$ are linearly independent, then $b_{i_1},\ldots,b_{i_k}$ are also linearly independent, for if they were not, $c_1b_{i_1}+\cdots+c_kb_{i_k}=0$, for some $c_j$'s not all zero. But then $$ 0+A(c_1b_{i_1}+\cdots+c_kb_{i_k})=c_1Ab_{i_1}+\cdots+c_kAb_{i_k}, $$ which is a contradiction.

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Hint: $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ is easy.

If $A$ is $m \times n$ and $\operatorname{rank}(A)=n$, then $A$ has a left inverse.

Call this inverse $C$ and use

$$\operatorname{rank}(CAB) \leq \operatorname{rank}(AB).$$

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We know that $L_A: \mathbb{R}^n \rightarrow \mathbb{R}^m$, $L_A x\mapsto Ax$ is an injection since $\text{rank} (A)=\text{rank} (L_A)= n$, (by dimension theorem $n-\text{rank}(L_A)= \dim(\ker(L_A))=0$). Let $L_B: \mathbb{R}^p \rightarrow \mathbb{R}^n$, $L_B x= Bx$. Clearly $L_B(\mathbb{R}^p)$ is subspace of $\mathbb{R}^n$.

Now since $L_A$ is 1-1 then clearly is injective when is restricted on the subspace $L_B(\mathbb{R}^p)$. Then $L_A\restriction_{L_B(\mathbb{R}^p)}: L_B(\mathbb{R}^p) \rightarrow L_A(L_B(\mathbb{R}^p))$ is a bijection. So $\dim(L_B(\mathbb{R}^p))=\dim(L_A(L_B(\mathbb{R}^p)))$.

[We can twist the argument by showing that $\dim(L_B(\mathbb{R}^p))<\dim(L_A(L_B(\mathbb{R}^p)))$ and $\dim(L_B(\mathbb{R}^p))>\dim(L_A(L_B(\mathbb{R}^p)))$ leads a contradiction. For the former contradicting the surjectivity and for the latter the injectivity ]

Since $\dim(L_A(L_B(\mathbb{R}^p)))= \dim(L_AL_B(\mathbb{R}^p))=\text{rank} (L_{A}L_{B})=\text{rank} (L_{AB})= \text{rank} (AB)$ and $\dim(L_B(\mathbb{R}^p))=\text{rank} (L_B)= \text{rank} (B)$. Hence $\text{rank} (B)=\text{rank} (AB)$ as desired.

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It is not hard to see that $\operatorname{rank}(B)\geq \operatorname{rank}(AB)$. Now, we can use Sylvester rank inequality $\operatorname{rank}(AB)+n\geq \operatorname{rank}(A)+\operatorname{rank}(B)$. One can see that $\operatorname{rank}(AB)+n\geq n+\operatorname{rank}(B)$. We can conclude that $\operatorname{rank}(AB)=\operatorname{rank}(B)$.

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    $\begingroup$ How do you show that $r(B) \ge r(AB)$? $\endgroup$ – Galc127 Feb 14 '14 at 13:55
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Remember, if:

(a) $rk(A + B) \leq rk(A) + rk(B)$ for any two $mxn$ matrices $A,B$;

(b) $rk(AB) \leq \min (rk(A),\ rk(B))$ for any $k\times l$ matrix $A$ and $l\times m$ matrix $B$;

(c) if an $n\times n$ matrix $M$ is positive definite, then $rk(M) = n$.

So that, for your question: Prove that if $rk(A)=n$ then $rk(AB)=rk(B)$

since $A \in Mat_{m\times n}(\mathbb{R})$ which is matrix $A$ is positive definite, and let for $n \le m$ the maximum number of linearly independent columns is $n$, hence $rk(A) = n$. similar for $rk(B)$, For $p \le n$ the maximum number of linearly independent columns is $p$, hence $rk(B) = p$.

Further since we have $A \in Mat_{m\times n}(\mathbb{R})$ and $B \in Mat_{n\times p}(\mathbb{R})$. we take $A,B$ such that $A $ is $mxn$ and $B$ is $nxp$ and $AB=I_{mp}$

So, $rk(AB)\le rk(A) \le n \lt p $ hence $rk(I_{mp})=mp$ and $rk(AB) \leq min (rk(A); rk(B))$

And as $rk(I_{mp})=mp$, then $rk(AB)\ne rk(I_{mp}) \Rightarrow AB \ne I_{mp}$, thus If $A$ is an $m × n$ as matrix of $rk(A) = n$ and , then $rk(B)$, For $p \le n$ the maximum number of linearly independent columns is $p$, and $rk(B) = p$, therefore $rk(AB)=rk(B)$.

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