Let $A$ be a square real matrix. We know that if $A$ is orthogonal matrix then $A^TA=I$. Consequently also in this case $(A^TA)^n=I$.

I would like to know whether it is possible to have expression of type $B=A^TA$ (I would call it transquare of $A$ - btw it is a little strange that such important expression, it seems, has no own name..) when $A$ is not orthogonal matrix, but for some natural $n$ equality $B^n=I$ is satisfied.

  • Is it possible?

Of course $B$ is full rank matrix, but on the other hand - in general - the equation $B^n=I$ can have even infinite number of solutions. Could one of them have decomposition $A^TA$ without $A$ being orthogonal?

  • Can the possible answer (probably negative) be also extended for the case of $ m \times n$ matrices ?
  • 1
    $\begingroup$ Thoughts: For sure, $\det(A)$ has to be equal to $(-1)^n$. $\endgroup$
    – Yanko
    Mar 12, 2018 at 10:59
  • $\begingroup$ @Yanko Yes, but it is still very broad condition.. $\endgroup$
    – Widawensen
    Mar 12, 2018 at 11:00
  • $\begingroup$ @Yes, in some case it can have, for example for $2 \times 2 $ matrices, other dimensions I have not analyzed.. $\endgroup$
    – Widawensen
    Mar 12, 2018 at 11:01
  • $\begingroup$ Concerning the name ..... I have found today that matrices of form $A^TA$ are called Gramian matrices en.wikipedia.org/wiki/Gramian_matrix#Examples $\endgroup$
    – Widawensen
    Mar 13, 2018 at 12:08
  • $\begingroup$ @Widawensen the result extends readily to non-square matrices using the SVD. $\endgroup$ Mar 13, 2018 at 12:18

1 Answer 1


It would a actually not be possible. If $B=A^TA$ with $A$ real (it doesn't even need to be square) and if $B^n=1$ for some $n$, then $B=I$.

Indeed, $B$ is symmetric and real, hence diagonalisable; moreover it is semidefinite positive, so all its eigenvalues, which I'll denote $\lambda_i$, are nonnegative. Now the eigenvalues of $B^n$ are exactly the $\lambda_i^n$; so $B^n=I$ implies that $\lambda_i^n=1$ for all $i$, and this also implies that $\lambda_i=1$ for all $i$, hence $B$ would actually be the identity.

  • $\begingroup$ Very clear argumentation. Thank you. $\endgroup$
    – Widawensen
    Mar 12, 2018 at 11:08

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