# Understanding the absolute value of a matrix.

I believe that the absolute value of a matrix is defined as $$|A|=\sqrt{A^{\dagger}A} \ .$$ But the square root of a matrix is not unique wikipedia gives a list of examples to illustrate this.

To understand this, how does one work out the absolute value of: $$A=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$$ Clearly $A^{\dagger}=A$ so $|A|=\sqrt{A^2}$, but this is not necessarily $A$. I want to pick the identity in this case, since then the eigenvalues of $|A|$ are both 1 (and they were $\pm1$ for $A$). But mathematics is not about what I want. So what is $|A|$? Is it well-defined? And how do I do this operation in general, since my application for this is of course far more complex.

• Does this from the wikipedia page you link to help? "However, a positive-semidefinite matrix has precisely one positive-semidefinite square root, which can be called its principal square root." Jun 23 '18 at 13:28
• yes if $|A|$ is "the positive definite square root $\sqrt{A^{\dagger}A}$", but I have not seen that stated in the papers I read. Perhaps it is obvious in the mathematical community?
– maor
Jun 23 '18 at 13:34
• I think it's safe to assume that's what's meant. You can always state clearly in your work that that is how you assume others are using it, and wait for an objection that will never come. Jun 23 '18 at 13:36
• For a positive diagonal matrix, the square root is simply the positive diagonal matrix of square roots of diagonal terms. Jun 23 '18 at 13:47
• Thanks you guys, I take it that $|A|$ is the positive square root (makes sense) and that then it is well defined for positive semidefinite matrices. Otherwise possibly not well defined. But that is alright since I was dealing with positive semidefinite matrices in my case.
– maor
Jun 25 '18 at 14:22

If $D$ is a diagonal matrix with positive terms, then the positive square root, $\sqrt D$ is uniquely determined by the diagonal matrix of positive square roots of diagonal terms.
If a matrix $A$ is diagonalizable with positive eigenvalues then $A= P^{-1}DP$ and we can define its positive square root as $\sqrt A= P^{-1} \sqrt D P$
Thus there is no confusion in finding the absolute value if $A$ if we consider only positive square roots.