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If rank (B) = n, R(B) = $R^n$ then N(A) $\cap$ R(B) = N(A).

I am not sure why N(A) $\cap$ R(B) = N(A).

I am trying to prove if rank (B) = n, then rank (AB) = rank (A) and R(AB) = R(A), where A and B are matrices. I saw the proof in the book is N(A) ∩ R(B) = N(A) and I don't know why it is true.

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  • $\begingroup$ Yes. R(B) is range space of B and N(A) is null space of A $\endgroup$ – Nabil Dec 2 '17 at 2:37
  • $\begingroup$ Do you at least accept the equation $n = rank(B) + dim(N(B))$? If you do, then my answer below will make sense.... if you do not know this equation, you should find it in your text book.... or is this essentially your question? $\endgroup$ – Clclstdnt Dec 2 '17 at 2:45
  • $\begingroup$ It seems that $B$ is a linear transformation from some linear space $V$ to $\mathbb{R}^n$. But what is $A$? $\endgroup$ – Catalin Zara Dec 2 '17 at 2:56
  • $\begingroup$ A and B is a matrix. I try to prove if rank (B) = n, then rank (AB) = rank (A) and R(AB) = R(A). I saw in the proof is N(A) ∩ R(B) = N(A) and I don't know why it is true. $\endgroup$ – Nabil Dec 2 '17 at 3:00
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If the range is the full space then in particular that space contains the zero vector. On the other hand, $A0=0$ for every matrix $A$ (where $0$ is the zero vector). But if this is the case, then it will always be that $N(A) \cap R(B) \supset \{0\}$. The point here is that there is nothing else in the space since the range is full it follows the null space is minimized.

Suppose that there were something else in the null space, $v$; then it follows that $\lambda v$ is also in the null space for all $\lambda \in \mathbb{R}$. But if that were the case then the dimension of the null space is at least $1$ and since $n = rank(B) + dim(N(B))$ this would make $rank(B)< n$ contradicting our assumption that $rank(B) = dim(\mathbb{R}^n)$

Let me know if you need more details.

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  • $\begingroup$ Does N(A)∩R(B)⊃{0} mean N(A)∩R(B) = N(A)? $\endgroup$ – Nabil Dec 2 '17 at 3:01

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