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I have a problem understanding the following:

Let $A$ be an $m \times n$ matrix and let t $\in \mathbb{N}$. Prove that $\operatorname{rank}(A)\leq t$ if and only if there exists an $m \times t$ matrix $B$ and a $t \times n$ matrix $C$ so that $A = BC$.

I know what a rank is but I can't make a connection between the rank and the existence of two matrices such that $A = BC$.

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    Does the term "singular value decomposition" mean something to you? If the connection between an SVD and the rank is clear, the above statement is easy to conclude. – 2017-01-12

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Think of matrices as linear transformations: $A:F^n\to F^m$, $C:F^n\to F^t$, $B:F^t\to F^m$. If $\mathrm{rk}(A)\le t$, then $\mathrm{Im}(A)\le t$, so there exists a $t$-dimensional subspace $T$ of $F^m$, such that $\mathrm{Im}(A)\subset T$. Consequently, if we restrict $A$ to $A'=C:F^n\to T$, and let $B:T\to F^m$ be the canonical inclusion, we have $A=BC$. The converse is obvious: If $A=BC$, then \begin{equation} \mathrm{rk}(A)=\mathrm{rk}(BC)\le\mathrm{dim}(\mathrm{Im}(B))\le t, \end{equation} because the domain of $B$ is $t$-dimensional.

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There is a Theorem that states that Rank(XY)<=min(rankX, rankY) So Rank(BC)<=min(rankB, rankC) So if t is less than n and m. Then it could be the rank of either B or C Rank(BC)<=t. I hope this helps.