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Let $A$ and $B$ be two $3×3$ matrices with real entries such that $rank(A) = rank(B) =1$. Let $N(A)$ and $R(A)$ stand for the null space and range space of $A$. Define $N(B)$ and $R(B)$ similarly. Then which of the following is necessarily true ?

$(A) \dim(N(A) ∩ N(B)) ≥ 1$.

$(B) \dim(N(A) ∩ R(A)) ≥ 1.$

$(C) \dim(R(A) ∩ R(B)) ≥ 1.$

$(D) \dim(N(B) ∩ R(B)) ≥ 1.$

I am feeling that option A is true..Can anyone help me in this..

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From the dimension formula: $\dim N(A)+\operatorname{rank}(A)=3$

So applies to $B$: $\dim N(B)+\operatorname{rank}(B)=3$

Form that $\operatorname{rank}(A)=\operatorname{rank}(B)=1$ it follows: $\dim N(B)+\dim N(A)=4$

Hence they must have an intersection with a non zero dimension. and it follows that the correct answer is A.

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