Matrix coefficient

Question

How can I solve the following problem:

$$P=X+Y\quad\quad Q=X\cdot Y,$$ where $P$, $Q$, $X$ and $Y$ are matrices. What is the condition that $P$ and $Q$ exist? What is the relation between $P$, $Q$, $X$ and $Y$?

I have been confused by this problem for weeks. I can't find a way to solve this problem because there is no commutative law.

Actually, are you looking for $X$ and $Y$ instead of $P$ and $Q$? By clearly specifying what look for you avoid the trivial answers given below. — Algebraic Pavel, Oct 13 '13 at 19:36
If you are solving for $X$ and $Y$, see Is there a unique solution for this quadratic matrix equation? — user1551, Oct 13 '13 at 20:51

score 1 · Answer 1 · answered Oct 13 '13 at 17:40

The question was what condition is required for $P$ and $Q$ to exist given $X$ and $Y$.

Since $P$ and $Q$ are defined by operations between $X$ and $Y$ we have to be sure those operations are well defined.

$$P=X+Y$$

$P$ is defined as the sum of $X$ and $Y$. These are matrices. The only way the can add is if they have the same number of rows and columns. This isn't particularly deep its just the rule/definition for adding matrices requires this. So if $X$ is an $m \times n$ matrix then so is $Y$.

$$ Q=X*Y$$

$Q$ is defined by the multiplication of $X$ and $Y$. What has to be true for this operation to be well defined? Look at the rule for multiplying matrices.

For the last part "what is the relations [between these matrices]", I feel that can mean a lot of things. The smart aleck answer would be to that the relationship is the equations. Nothing you can say about these will contain content not contained in the equations. This may require you to ask the writer of the question for clarification as to what they are looking for.

while I was doing this problem, because the matrices don't have the commutative law, so now I stopped here: X=QY^(-1),P=QY^(-1)+Y, so Y^2-P*Y+Q=0 and it is the problem that I couldn't solve: how to find the solution of Y? — Evan, Oct 13 '13 at 17:48
There probably isn't a way to get $Y$ in terms of $P$ and $Q$. You aren't even gauranteed that $X$ and $Y$ have inverses. Based on the reading of the question I don't think that is being asked for. — Spencer, Oct 13 '13 at 17:53
If you assumed they were commutative and invertable then it can be solved the same way you would solve the equation with real variables. That may be what your instructor is looking for if they asked you to solve for $Y$. — Spencer, Oct 13 '13 at 17:55

score 0 · Answer 2 · answered Oct 13 '13 at 17:45

For both to be well defined, $X$ and $Y$ need to be square matrices of the same size - let $X\in\mathbb R^{n\times m}$ and $Y\in\mathbb R^{k\times d}$. Since $P=X+Y$ must be well defined, $$n=k, m=d$$ Since $Q=X\cdot Y$ must be well defined, $$k=m$$ And thus $$n=k=m=d$$ so $$X, Y \in\mathbb R^{n\times n}$$

Matrix coefficient

2 Answers2