Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
@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
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
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 $AR = W$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.