# Does a positive definite matrix have any power of real number?

For example, if $A$ is positive definite there exists a square root matrix $A^\frac{1}{2}$ for which $A^\frac{1}{2}A^\frac{1}{2}=A$. Proof. Let $A$ be a positive definite matrix with positive eigenvalues. Using the spectral decomposition, we have $$A=U^Tdiag(\lambda_1,…,\lambda_n)U=(U^Tdiag(\sqrt{\lambda_1},…,\sqrt{\lambda_n})U)(U^Tdiag(\sqrt{\lambda_1},…,\sqrt{\lambda_n})U) =A^\frac{1}{2}A^\frac{1}{2} (\lambda_i>0)$$ original text

My Question is: Since the eigenvalues of a positive definite matrix are all positive, does a positive definite matrix have any power of real number, $i.e.$ does $$A^x, x\in R$$exist? Such as $A^\frac{1}{3}, A^{-\frac{1}{2}} or A^{-1}$?

Please give me a proof, rather than just examples.

• Well, $A^{-1}$ is the inverse of $A$ and it exists when $\det\left(A\right)\neq 0$. Sep 10, 2015 at 7:20
• Yes, it is. Because $A$ is positive definite and $det(A)=\prod_{i=1}^n\lambda_i>0$ Sep 10, 2015 at 7:24
• The answer to your question is yes. Sep 10, 2015 at 7:30
• Also, I think we can generalize $a^z=e^{z \log\left( a \right)}$ for $a,z \in \mathbb C$ to something like $A^z = \exp \left( z \log \left( A \right) \right)$ with $\exp$ and $\log$ functions defined as power series of matrices, and choosing a branch of the complex logarithm. I think this would work for any invertible matrix. Sep 10, 2015 at 7:31
• go to canonical jordan form and there you may raise to any power then go back. Sep 10, 2015 at 7:37

If you proceed exacly in the same way as you found the square root replacing the $1/2$ by $s>0$, then you will obtain a matrix $U^T\operatorname{diag}(\dots,\lambda_i^s,\dots)U$ which deserves to be called the $s$th power of $A$.
Indeed, any sensible notion of $s$th power will be compatible with conjugation, so that $(C^{-1}AC)^s=C^{-1}A^sC$, and act in the obvious way on diagonal matrices.