Let $A$ be square or rectangular and $x\in \mathbb{R}$. Can you point me to equations/areas out there where $(AA^T)^x$ or $(A^TA)^x$ or their eigenvalues are used as applications? e.g. we find them in the polar decomposition when $x=\frac{1}{2}$.

I'm particular about cases when $x\ne 0, \frac{1}{2}, 1$. Thanks in advance.

$A$ may also be taken to be complex i.e. $(AA^*)^x$ or $(A^*A)^x$.

  • $\begingroup$ what does $A^x$ mean for real $x$? $\endgroup$ – Pink Panther Mar 17 at 14:14
  • $\begingroup$ Would you be willing to use a formula in terms of the eigenvalues of $A^TA$? There's also a formula using a power series. $\endgroup$ – Omnomnomnom Mar 17 at 14:36
  • $\begingroup$ @Omnomnomnom. Yes, I will consider their eigenvalues too. I have edited the post to include that $A$ can also be rectangular. $\endgroup$ – Kay Mar 17 at 15:47
  • $\begingroup$ @Omnomnomnom, please expantiate on the formula using a power series, and that which considers the eigenvalues. Thanks in advance. $\endgroup$ – Kay Mar 17 at 18:01

I'll stick to the real case, but the situation for complex matrices is similar.

$A^TA$ is a symmetric and positive semidefinite matrix. By the spectral theorem, there exists an orthogonal $U$ such that $U^T(A^TA)U$ is diagonal, that is $$ U^T(A^TA)U = D = \pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_n} $$ where the $\lambda_k$ are the eigenvalues of $A^TA$ (i.e. the singular values of $A$). Because $A^TA$ is positive semidefinite, the eigenvalues $\lambda_k$ are non-negative, that is $\lambda_k \geq 0$ for all $k$.

We may compute $$ (A^TA)^x = [U D U^T]^x = UD^x U^T = U \pmatrix{\lambda_1^x \\ & \ddots \\ && \lambda_n^x}U^T $$ Because $\lambda_k\geq 0$ for all $k$, $\lambda_k^x$ makes sense for all $x > 0$. If we also know that $A^TA$ is invertible (i.e. that $A$ has full column rank), then we have $\lambda_k > 0$ which means that $\lambda_k^x$ makes sense for all $x \in \Bbb R$.

  • $\begingroup$ thanks for the response! I actually meant 'applications' when I said areas/equations where those relations 'are found'. I have edited the post accordingly. I'm sorry about the ambiguity! $\endgroup$ – Kay Mar 17 at 19:21

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