# Solving Xa=b for an unknown matrix X

I'm interested in studying the solutions of $$Xa=b$$ for an unknown square matrix $$X$$, and given (known) column vectors $$a$$ and $$b$$ in $$\mathbb{R}^n$$.

For any numerical $$a, b$$, one can directly attempt solving the aforementioned system. But I'm interested in understanding the general setting to answer questions similar to the ones below.

(1) Under what conditions a solution $$X$$ exists?

(2) When does a symmetric solution exist? Under what conditions, no symmetric solution exists?

(3) When does a unique, invertible, symmetric solution exist?

The answer to (1) is easy: a solution exists whenever $$a\ne \vec{0}$$, or both $$a, b$$ are zero vectors.

$$Xa=b$$ is, of course, a system of $$n$$ linear equations in $$n^2$$ variables if there are no additional constraints on $$X$$, in which case the equations are "decoupled" because of having disjoint set of variables. But, for instance, the number of variables is cut down to $$n(n-1)/2$$ if we require X to be symmetric. So, if $$n(n-1)/2=n$$, that is, $$n=3$$ I expect (3) to be likely than when $$n>3$$.

What are some good ways to think about problems like this involving $$Xa=b$$? Any suggestions, or references are appreciated.

• To the user who voted to close: If you read the post more carefully, you notice that I've not asked multiple questions. The main question is at the bottom whereas (1)-(3) are there to clarify. Apr 3 at 3:02
• > A good way to avoid misunderstanding is to highlight your actual question in some way.
– Aig
Apr 5 at 1:19

If you are asking questions about real symmetric matrices and finding solutions to equations like this, a good start may be to note that all real symmetric matrices are diagonalisable. So under a suitable change in variables, you can reduce the case to one of a diagonal matrix. In terms of your "number of variables" view, you are going from $$n^2$$ to $$\frac{n(n-1)}{2}$$ to at most $$n$$ (the eigenvalues), which is a drastic decrease.