A problem related to rank and nullity of matrices. Let $A$ be a 4$\times$ 7 real matrix and B be a 7$\times$4 real matrix such that $AB=I_4$, where $I_4$ is the $4 \times 4$ identity matrix.Which of the followings are true? $\DeclareMathOperator{\rank}{rank}$


*

*$\rank(A)=4$

*$\rank(B)=7$

*nullity$(B)=0$

*$BA=I_7$


I know $B$ is a 7$\times$4 matrix ,then its row rank and column rank are same.  rank(B) $\leq$ 4. So (2) is false, but I have no idea about others.please help. Thanks.
 A: Hint: $\DeclareMathOperator{\rank}{rank}$


*

*Note that $\rank(I_4) = 4$ and $\rank(I_7) = 7$

*In general, $\rank(AB) \leq \min\{\rank(A),\rank(B)\}$

*Consider the rank-nullity theorem (AKA the dimension theorem)

A: $\DeclareMathOperator{\rank}{rank}$here is a way to see that $\rank(A) = \rank(B) = 4.$
first we establish that $\ker(B) = \{0\}.$  suppose $x \in \ker(B).$  then $Bx = 0$ pre multiplying by $A,$  gives $x = ABx = A0 = 0.$ using the nullity theorem implies $\rank(B) = 4.$
next, for any $b \in R^4$ we have  $ABb = b.$ therefore the column space of $A$ contains $R^4.$ which implies that the $\rank(A) = 4.$
A: Hint:
Let $f\colon \mathbf R^7\to \mathbf R^4$ and $g\colon  \mathbf R^4\to \mathbf R^7$ the linear maps corresponding to $A$ and $B$. The hypothesis means $f\circ g=\operatorname{id_{ \mathbf R^4}}$, and it implies $g$ is injective and $f$ surjective. Hence 2 is false, 3 and 1 are true.
Furthermore, 4 is false, for it would imply, together with the hypothesis, that $f$ and $g$ are isomorphisms.
A: Since $\operatorname{rank}(A)=\min\{m,n\}$, hence $\operatorname{rank}(A)=4$. $1$ is correct.
Also $\operatorname{rank}(B)=\min\{7,4\}=4.$ $2$ is false.
By rank-nullity theorem, $\operatorname{nullity}(B)=0$
Also $BA\neq I_7$, can you see why?
