I am trying to find out if for any symmetric (Not necessarily self-adjoint), invertible matrix $A$ over $\mathbb{C}$, there is a square root of the matrix that is also symmetric. I was able to figure out that for invertible matrices there always exists a square root since I can explicitly do it for a general Jordan block and go from there but I was hoping that it would necessarily be symmetric under this construction (which seems either not obvious or not true). Any thoughts?
[Math] Symmetric Square Root of Symmetric Invertible Matrix
linear algebramatricesoperator-theory
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What about the matrix $(-1){}{}{}{}{}{}{}{}{}{}{}$?
I believe $I+N$ must indeed be "positive definite".
Apply induction on the dimension. We are given a nilpotent matrix $N'$ which by Schur decomposition we can assume is strictly upper triangular. As a block matrix, $N'=\begin{pmatrix}N&x\\0&0\end{pmatrix}$ with $N$ a square strictly upper triangular matrix and $x$ a column vector. By assumption $(I+N')^2+(I+N')^{2t}$ is SPD (SPD=symmetric positive definite; superscript $2t$ means square and transpose). We wish to show that $(I+N')+(I+N')^t$ is SPD.
Let $U=I+N$ and note $$\begin{pmatrix}U&x\\0&1\end{pmatrix}^2+\begin{pmatrix}U&x\\0&1\end{pmatrix}^{2t}=\begin{pmatrix}U^2+U^{2t}&(I+U)x\\x^t(I+U)^t&2\end{pmatrix}.$$
By the Schur decomposition characterization of SPD we have by assumption:
- $U^2+U^{2t}$ is SPD
- $2-x^t(I+U)^t(U^2+U^{2t})^{-1}(I+U)x>0$
And we need to show:
- $U+U^t$ is SPD
- $2-x^t(U+U^t)^{-1}x>0.$
$U+U^t$ is SPD by induction. The latter inequality would follow from $$x^t(I+U)^t(U^2+U^{2t})^{-1}(I+U)x\geq x^t(U+U^t)^{-1}x$$ Letting $y=(I+U)x,$ this is $$y^t(U^2+U^{2t})^{-1}y\geq y^t(I+U^t)^{-1}(U+U^t)^{-1}(I+U)^{-1}y$$
This would follow from this result and symmetric positive definiteness of $$(I+U)(U+U^t)(I+U^t)-(U^2+U^{2t})=U+U^t+2UU^t+U(U+U^t)U^t$$ Since $U+U^t$ is SPD, so is $U(U+U^t)U^t,$ and $UU^t$ is certainly SPD. So we are done.
See loup blanc's answer for a generalization of the above argument from unipotent matrices to all matrices with positive spectrum.
Here is a reference with a nice proof: Uniqueness of matrix square roots and an application by Charles R. Johnson, Kazuyoshi Okubo, Robert Reams, Theorem 7. It uses the following theorem:
Theorem (Lyapunov). Let $A\in\mathbb C^{n\times n}$ (not necessarily Hermitian) and let $X\in\mathbb C^{n\times n}$ be Hermitian. If the eigenvalues of $A$ all have positive real part and $AX+XA^*$ is positive definite, then $X$ is positive definite.
In particular if $A$ has eigenvalues with positive real part and $A(A+A^*)+(A+A^*)A^*=A^2+(A^*)^2+2AA^*$ is positive definite, then $A+A^*$ is positive definite.
Proof 1 (sketch) following Horn and Johnson's Topics in Matrix Analysis:
Suppose not. We can take a kind of Jordan normal form, but making any above-diagonal $1$'s arbitrarily small, so $S^{-1}AS+(S^{-1}AS)^*$ is positive definite for some non-singular $S.$ Setting $G=SS^*$ we find that $AG+GA^*$ is positive definite. For $0\leq \theta\leq 1$ define $X_{\theta}=\theta G+(1-\theta)X.$ Note:
- $X_0=X$ is not positive definite
- $X_1=G$ is positive definite
- $AX_\theta+X_\theta A^*$ is a convex combination of positive definite matrices so must be positive definite.
All the matrices $X_\theta$ have real eigenvalues, and by continuity some $X_\theta$ must have $0$ as an eigenvalue: $X_\theta v=0$ for some non-zero $v.$ This implies $v^*(AX_\theta+X_\theta A^*)v=0,$ contradicting positive definiteness of $AX_\theta+X_\theta A^*.$
Proof 2.
Consider the function defined by $f(W)=\int_0^\infty e^{-tA}We^{-tA^*}dt.$ We can compute
\begin{align*} W&=-\frac d{d\tau}\Bigr|_{\tau=0}\int_{\tau}^\infty e^{-tA}We^{-tA^*}dt\\ &=-\frac{d}{d\tau}\Bigr|_{\tau=0}\int_0^\infty e^{-\tau A}e^{-tA}We^{-tA^*}e^{-\tau A^*}dt\\ &=Af(W)+f(W)A^*. \end{align*}
This means $f$ is a left inverse of the map $X\mapsto AX+XA^*.$ Since $f$ is an $\mathbb R$-linear map from the space of Hermitian matrices to itself, and has a left inverse, it must be an isomorphism. So the solution $X$ of $AX+XA^*=W$ is unique. And if $W$ is positive definite, then so is $e^{-tA}W^{1/2}W^{1/2}e^{-tA^*},$ so $f(W)$ is positive definite by construction. This proves that if $AX+XA^*$ is positive definite then so is $X.$
Best Answer
If $\|A-I\|<1$ you can always define a square root with the Taylor series of $\sqrt{1+u}$ at $0$: $$ \sqrt{A}=\sqrt{I+(A-I))}=\sum_{n\geq 0}\binom{1/2}{n}(A-I)^n. $$ If $A$ is moreover symmetric, this yields a symmetric square root.
More generally, if $A$ is invertible, $0$ is not in the spectrum of $A$, so there is a $\log$ on the spectrum. Since the latter is finite, this is obviously continuous. So the continuous functional calculus allows us to define $$ \sqrt{A}:=e^\frac{\log A}{2}. $$ By property of the continuous functional calculus, this is a square root of $A$.
Now note that $\log$ coincides with a polynomial $p$ on the spectrum (by Lagrange interpolation, for instance). Note also that $A^t$ and $A$ have the same spectrum. Therefore $$ \log(A^t)=p(A^t)=p(A)^t=(\log A)^t. $$ Taking the Taylor series of $\exp$, it is immediate to see that $\exp(B^t)=\exp(B)^t$. It follows that if $A$ is symmetric, then our $\sqrt{A}$ is symmetric.
Now if $A$ is not invertible, certainly there is no log of $A$ for otherwise $$ A=e^B\quad\quad\Rightarrow \quad 0=\mbox{det}A=e^{\mbox{Tr}B}>0. $$ I am still pondering the case of the square root.