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Let $A \in M_{2006}(\Bbb F)$ be a matrix, $U \subseteq \Bbb F^{2006}$ the row space of $A$ and $W \subseteq \Bbb F^{2006}$ the coloumn space of A. Suppose $U \oplus W = \Bbb F^{2006}$, and let $B \in M_{2006}(\Bbb F)$ be a matrix such that $A \cdot B= 0$.

Prove that $rank(B) \le 1003$.

I understand why it's true, and yet I can't find a way to write a formal proof. First I understood that $rank(A) = 1003$, but I'm kind of stuck from here, any help will be appreciated, thanks.

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    Your notation is pretty confusing. For me. $\;A\in M_n(F)\iff A\;$ is a **square** matrix of order $\;n\;$ , and thus both the row and columns spaces are contained in $\;F^n\;$ . – 2017-01-28
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    I fixed the question. does it make sense now? – 2017-01-28
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    Yes, now it does, yet the claim fails without further info. For example, what if $\;A=0\;$ ? Then we could have rank$\,B=2006\;$ , say with $\;B=I\;$ ... – 2017-01-28
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    In my answer below, I interpreted your condition as just $U + W = F^{2006}$. – 2017-01-28

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$\DeclareMathOperator{\rank}{rank}$Since $A B = 0$, the image of $B$ is contained in the kernel of $A$, so that by rank-nullity $$ \rank(B) \le 2006 - \rank(A) = 2006 - \dim(W), $$ or $$ \dim(W) \le 2006 - \rank(B). $$ But then, looking at transposed matrices, we have $B^{\top} A^{\top} = 0$, so that the image of $A^{\top}$ is contained in the kernel of $B^{\top}$, and thus $$ \dim(U) = \rank(A) \le 2006 - \rank(B). $$ The condition $U + W = F^{2006}$ tells you that $\dim(U) + \dim(W) \ge 2006$. You get $$ 2006 \le \dim(U) + \dim(W) \le 2 \cdot (2006 - \rank(B)), $$ and there you are.

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    In this answer, I interpreted your condition as just $U + W = F^{2006}$. – 2017-01-28
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By the row space of $A$ you probably mean the column space of $A^T$, otherwise it can be in the same space as the column space of $A$.

I'll denote by $C(A)$ the column space of $A$, so your setting can be stated as $C(A^T)\oplus C(A)=V$, where $V=F^{2n}$, being $A$ and $B$ matrices in $M_{2n}(F)$.

The rank of $A$ is the same as the rank of $A^T$, so both $C(A)$ and $C(A^T)$ have dimension $n$.

Since $AB=0$, you have $C(B)\subseteq N(A)$ (the null space of $A$). This means $$ \dim C(B)\le\dim N(A)=2n-n=n $$