Showing there exists an A and a B such that M = AB

Question

If $M \in R^{n \times m}$ and r = rank(M). How can I show that there exists $A \in R^{n \times r} and B \in R^{r \times m}$ such that $M=AB$?

Answer 1

Since $rank(M)=r$ so its linear transformation image is a subspace with dimension $r$. So the onto transformation to it, that reduced by $M$, is $A \in R^{n \times r}$ or $A \in R^{r \times m}$, depends on you see $M$ or $M^t$ act from right or left. In each case the second one is the identity on its image.

Answer 2

It is always easier to just think of linear transformations in these types of questions: The rank of $M$ being $r$ means that the image $M(\mathbb{R}^m)$ is a space of dimension $r$, hence isomorphic to $\mathbb{R}^r$. Let $A\colon M(\mathbb{R}^m)\hookrightarrow \mathbb{R}^n$ be the inclusion, and $B\colon\mathbb{R}^m\to M(\mathbb{R}^m)$ be the correstriction of $M$. Then $M=A\circ B$. Take the canonical bases $\mathcal{B}_m$ and $\mathcal{B}_n$ or $\mathbb{R}^m$ and $\mathbb{R}^n$, and any basis $\mathcal{B}$ of $M(\mathbb{R}^m)$, and write down the matrices of $M=A\circ B$ in these bases: $[M]=[A][B]$.

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Here is how to do it in the "classical linear algebra" way: Take a basis $b_1,\ldots,b_r$ of $M(\mathbb{R}^k)$. Take elements $v_1,\ldots,v_r\in\mathbb{R}^m$ such that $Mv_i=b_i$. Take a basis $v_{r+1},\ldots,v_m$ of $\ker M$.

The $m\times m$ matrix $C=[v_1\cdots v_m ]$ is invertible, because its columns are LI. Consider its inverse $C^{-1}$. Now consider the $r\times m$ matrix $D$ which "erases the $m-r$ last rows: $$D=\begin{bmatrix}I_r&0_{r\times (m-r)}\end{bmatrix}$$ where $I_r$ is the $r\times r$ identity and $0_{r\times (m-r)}$ is the $r\times (m-r)$ zero matrix. Consider the $r\times m$ matrix $B=DC^{-1}$.

Consider the $n\times r$ matrix $A=[b_1\cdots b_r]$.

We claim that $M=AB$. Since $v_1,\ldots,v_m$ is a basis of $\mathbb{R}^m$, it suffices to prove that $Mv_i=ABv_i$.

• If $i\leq r$: Let $e^m_i$ be the $i$-th element of the canonical basis of $\mathbb{R}^m$: all entries of $e^m_i$ are $0$, except the $i$-th one which is $1$. Then $Ce^m_i=v_i$, which means that $e^m_i=C^{-1}v_i$. Applying $D$ (erasing the last rows) we obtain $$Bv_i=DC^{-1}v_i=De^m_i=e^r_i$$ Then we can compute $ABv_i=Ae^r_i=b_i=Mv_i$.

• If $i\geq r$, we proceed as above, but then notice that the non-zero entry of $e^m_i$ is after the row $r$, so $Bv_i=DC^{-1}v_i=De^m_i=0$, which implies $ABv_i=0$. On the other hand, $Mv_i=0$ because $v_{r+1},\ldots,v_m$ are in $\ker M$.