# Conditions for Symmetric Square Root of a Real Matrix

What are necessary and sufficient conditions that a matrix in $\mathbb R^{n\times n}$ must satisfy in order for its square root to exist, be symmetric, and also belong to $\mathbb R^{n\times n}$? If the square root is restricted to belong to $\mathbb R^{n\times n}_+$ is it possible to guarantee uniqueness?

Sufficient conditions are simple to obtain, i.e. if the matrix is symmetric, invertible [edit: and has positive eigenvalues], but I have not been able to find necessary conditions.

• @kimchilover Thank you. I edited the condition for sufficiency to address your point. – mzp Dec 6 '17 at 14:38

If the square root $S$ of $A$ is symmetric then $A$ is symmetric as $A=S^2$ then $A^T=S^TS^T=S^2=A$.

Then if $A$ symmetric it is diagonalisable and it has a square root if and only if a diagonal matrix associated has one also and it does if and only if its coefficients (ie eigenvalues of $A$) are positive.

The if part is clear, not the only if (which doesn't take the diagonal matrix into account by the way). For this, suppose $A$ has a square root $S$ which is symmetric and let's consider $a$ and $s$ canonic linear applications associated (symmetric for the usual scalar product then). Then let's take $\lambda\in\text{Sp}(A)=\text{Sp}(a)$. The eigenspace $E_{\lambda}(a)$ is stable by $s$ because $a$ and $s$ commute (because $A$ and $S$ do). Then $s$ being symmetric (or self-adjoint) it has an eigenvector $x$ in that space with an eigenvalue $\alpha\in\mathbb{R}$ (its restriction is even diagonalisable with real eigenvalues), $s(x)=\alpha x$ then $a(x)=s^2(x)=\alpha^2 x$ but in the same time $a(x)=\lambda x$ on $E_{\lambda}(a)$ so, given $x\neq 0$, we have $\lambda=\alpha^2\ge 0$ ($\alpha\in\mathbb{R}$).

So a necessary and sufficient condition of existence is to be symmetric with spectrum in $\mathbb{R}_+$ this space is $S_n^+(\mathbb{R})$.

You have not uniqueness even on $\mathbb{R}_+^{n\times n}$, for example $I_2$ has these two square roots : $I_2$ and $\begin{pmatrix}0&1\\1&0 \end{pmatrix}$.

You have actually uniqueness of a square root in $S_n^+(\mathbb{R})$, it is a classical result on positive self-adjoint endomorphism that transpose itself on their matrices : any matrix of $S_n^+(\mathbb{R})$ (symmetric positive) has a unique square root in $S_n^+(\mathbb{R})$.

Do you need this proof ?

• Thank you. I edited the question to correct the sufficiency condition. I don't understand your final sentence, could you clarify? – mzp Dec 6 '17 at 14:46

On order for linear operator $A$ to have a (symmetric) square root, it has to be "positive", i.e. $\forall x \in V:(Ax,x)\geq0$. Just for the record $(Ax,x):=x^tAx$.

Proof: Take any $x\in V$. Let $B$ be a (symmetric) square root of $A$. $$0\leq||Bx||^2=(Bx,Bx)=(B^tBx,x)=(BBx,x)=(Ax,x).$$

A has to be symmetric as well: $$(Ax,y)=(BBx,y)=(x,BBy)=(x,Ay).$$

I am not very familiar about matrixes (frankly, all those indices only confuse me), but you can always transfer to them by choice of a basis.