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.

  • 7
    $\begingroup$ 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. $\endgroup$
    – Philip M
    Apr 3 at 3:02
  • 2
    $\begingroup$ > A good way to avoid misunderstanding is to highlight your actual question in some way. $\endgroup$
    – Aig
    Apr 5 at 1:19

1 Answer 1


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.


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