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Admittedly, made the title a little funny, but this is a valid question.

I have come across the following equation $$ I x^2=AA $$ where $I$ is a unit matrix, $A$ is a square matrix of the same dimension and x is just a number. Can I conclude that $I x=A$?

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  • $\begingroup$ How is square root defined in your case? $\endgroup$
    – Dmitry
    Commented Jun 18 at 1:52
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    $\begingroup$ $A$ is certainly one valid root under any reasonable interpretation. $\endgroup$
    – Randall
    Commented Jun 18 at 1:53
  • $\begingroup$ we diagonalize the matrix, then take the square root of the each element. I know how to do it by brute force, but I was wondering if there are any identities. $\endgroup$
    – Saeed
    Commented Jun 18 at 1:54
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    $\begingroup$ If $B$ is positive-semidefinite, $\sqrt B$ is the unique positive semi-definite matrix $C$ such that $C^2=B$ (see Square root of a matrix on Wikipedia.) In your case, you need $A$ to be self-adjoint to apply this definition. $\endgroup$
    – Just a user
    Commented Jun 18 at 1:56
  • $\begingroup$ $AA$ is not always diagonalizable, e.g. for $A = \begin{pmatrix}1 & 1 \\ 0 & 0 \end{pmatrix}$. $\endgroup$
    – Dmitry
    Commented Jun 18 at 1:58

1 Answer 1

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You already can't conclude this for ordinary real or complex numbers; the best you could possibly ask for is that $A = \pm x$ (by which I mean $A = \pm x$ times the identity matrix).

But actually you can't even expect this. $A$ can be any diagonalizable matrix whose eigenvalues are $\pm \sqrt{x}$; even if we restrict only to diagonal matrices there are $2^n$ square roots (if $x \neq 0$), and without that restriction there are uncountably many square roots.

Worse, if $x = 0$ then $A$ need not be diagonalizable at all.

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