A problem related to rank and nullity of matrices. [duplicate]

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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? $\DeclareMathOperator{\rank}{rank}$

1. $\rank(A)=4$
2. $\rank(B)=7$
3. nullity$(B)=0$
4. $BA=I_7$

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.

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4 Answers

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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)

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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.

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$\DeclareMathOperator{\rank}{rank}$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 $\rank(A) = 4.$

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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?

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