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Suppose I have a matrix: $$A \in \mathbb{R}^{n \times m}$$ and another one (same size): $$W \in \mathbb{R}^{n \times m}$$

  1. When is it possible to find a square matrix $L$ such that: $$L\cdot A=W$$ where the "$\cdot$" is the usual matrix multiplication? When have the matrix $L$ real values?
  2. The same as the previous point but with a right-matrix, that solves: $$A\cdot R=W$$
  3. Is it possible to find two square matrices $L'$ and $R'$ such that: $$L' \cdot A \cdot R' = W$$ and, if so, do you think that it will be easer, harder (or not easy to compare) than the decomposition proposed in the previous points?

Thank you very much.

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The answer to the question in the title is "no". Have you thought of examples where such $L$ and $R$ do not exist? –  Jonas Meyer Dec 30 '12 at 1:22
    
@JonasMeyer Do you mean when matrix $A$ is the zero-matrix, and $W$ isn't? –  Aslan986 Dec 30 '12 at 1:29
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@Aslan986 That's a good example. To think about what the sufficient / necessary conditions are, think about the range space and the null space of these matrices, and see what constraints are needed. –  Calvin Lin Dec 30 '12 at 1:32
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Well, obviously if $A$ is invertible then $L=W\cdot A^{-1}$ and $R=A^{-1}\cdot W$, but there are also solutions if $\operatorname{rank}(W)\leq\operatorname{rank}(A)<n$ as well (but a bit harder to describe). I'm not sure, but I don't think that generalization to two matrices $L',R'$ expands the set of solutions. I'd be interested to see a counterexample to my statement, though. –  Mario Carneiro Dec 30 '12 at 1:34
    
@Mario: Consider $A=\begin{bmatrix}1&0&0\\0&0&0\end{bmatrix}$ and $W=\begin{bmatrix}0&0&0\\0&0&1\end{bmatrix}$. –  Jonas Meyer Dec 30 '12 at 2:13
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1 Answer 1

up vote 3 down vote accepted

This is a complete description for the necessary and sufficient conditions for the matrices to exist. I'd leave 3 for you to do, since it's just combining 1 and 2.

  1. Think about the Kernel. If $v \in \operatorname{Null}(A)$, $Av = 0$, then $0=LA v= Wv $, so $v \in \operatorname{Null}(W)$. Hence, $\operatorname{Null}(A) \subseteq \operatorname{Null}(W)$.

    Conversely, if $\operatorname{Null}(A) \subseteq \operatorname{Null}(W)$, let $\{ v_1, \ldots, v_i, v_{i+1} , \ldots, v_j, v_j, \ldots v_m\}$ be a basis of $\mathbb{R}^m$ such that $\{v_k\}_{k=1}^i$ is a basis for $\operatorname{Null}(A)$, $\{v_k\}_{k=1}^j$ is a basis for $\operatorname{Null}(W)$. Notice that $\{ Av_k \}_{k=i+1}^m$ is a linearly independent set, and so is $\{ Wv_k\} _ {k = j+1}^m$. Define $L$ to be the linear transformation such that $L(Av_k) =0$ for $k = i+1 $ to $j$, $L(Av_k) = W_k $ for $k= j+1$ to $m$, and extend $L$ to $\mathbb{R}^n$ (it doesn't matter how you extend it). Then, $LA = W$ as linear transformation from $\mathbb{R} ^m \rightarrow \mathbb{R}^n$, by checking its action on the basis.

  2. Think about the Range Space. If $w \in \operatorname{Range}(W)$, then there exists $v \in \mathbb{R}^m$ such that $w = W v$, then we have $ A (R v) = w$, and so $w \in \operatorname{Range}(A)$. This shows that $\operatorname{Range}(W) \subseteq \operatorname{Range}(A)$.

    Conversely, if $\operatorname{Range}(W) \subseteq \operatorname{Range}(A)$, let $\{v_1, \ldots, v_i, v_{i+1}, \ldots, v_j, v_{j+1}, \ldots v_m\}$ be a basis of $\mathbb{R}^m$ such that $\{ W v_k\}_{k=1}^i$ is a basis for $\operatorname{Range}(W)$, $\{ A v_k\}_{k=1}^j$ is a basis for $\operatorname{Range}(A)$. Define $R$ to be the linear transformation such that $Rv_k$ is the vector which satisfies $A R v_k = W v_k$ for $k=1$ to $i$ (Why must this exist?), and $Rv_k = 0$ for $k=i+1$ to $m$. Then, $ARv_k = W v_k$ on the basis, hence $AR = W$.

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The markdown formatting on the site tries to "help" you create numbered lists. Putting something null after the period, like &nbsp; (non-breaking space) or $\,$ (small space in LaTeX) confuses it into not helping. –  Zev Chonoles Dec 30 '12 at 2:40
    
How do you get 3 by "just combining 1 and 2"? Let $A=\begin{pmatrix}1\\&0\end{pmatrix}$ and $W=I-A$. Then $Null(A)\not\subseteq Null(W)$ and $Range(W)\not\subseteq Range(A)$, but $LAR=W$ for some $L$ and $R$. –  user1551 Dec 30 '12 at 8:11
    
@Zev: The better solution is to indent the subsequent paragraphs that are supoosed to be part of the same bullet point, as Mario just did. –  Rahul Dec 30 '12 at 9:21
    
@user1551 That is because you didn't apply the problem at all. Treating $AR'=A'$, the existence of $L'$ implies that $Null (AR') \subset Null (W)$, and as such $\dim Ker(A) \leq \dim Ker (W)$. Similarly for $L'A = A'$, we have $\dim Range(A) \geq \dim Range (W)$. [This also follows from Rank-Nullity.] This is the necessary condition. Now show that it is sufficient by constructing $R'$ and $L'$, by defining their action on a suitable basis obtained from $W$ and $A$ respectively. –  Calvin Lin Dec 30 '12 at 18:14
    
@RahulNarain Can you explain how to indent the paragraphs? I tried to see your edits, but it doesn't show how to do it. –  Calvin Lin Dec 30 '12 at 18:15
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