# How can we be sure that for every $A$, $A^\dagger A$ has a positive square root?

In the Polar Decomposition section in Nielsen and Chuang (page 78 in the 2002 edition), there is a claim that any matrix $$A$$ will have a decomposition $$UJ$$ where $$J$$ is positive and is equal to $$\sqrt{A^\dagger A}$$.

Firstly, how can we be sure that every matrix $$A^\dagger A$$ will have a square root, and secondly, that the square root will be a positive operator?

A matrix is positive if and only if it is Hermitian (and thus unitarily diagonalizable) and all its eigenvalues are positive (that they are real follows automatically from it being Hermitian). If this is not the way you define a positive semidefinite (PSD) operator, then you need to specify how you do so that we can prove the equivalence.

In other words, $$A$$ is positive, $$A\ge0$$, iff it can be written as $$A=\sum_k \lambda_k v_k v_k^*\equiv \sum_k \lambda_k \lvert v_k\rangle\!\langle v_k\rvert, \quad \lambda_k\ge0,$$ with $$\langle v_k\rvert v_j\rangle\equiv v_k^* v_j=\delta_{jk}$$. I used here both diadic notation and bra-ket notation just to point out that these are simply two equivalent ways to write the same thing.

# Why is $$A^\dagger A\ge0$$?

One way to show this is to start from the fact that unitarily diagonalizable is equivalent to normal. This means that a matrix $$B$$ can be written as $$B=UDU^\dagger$$ for some unitary $$U$$ and diagonal matrix $$D$$ if and only if $$BB^\dagger=B^\dagger B$$.

As you can readily verify, it is the case that $$(A^\dagger A)^\dagger (A^\dagger A)=(A^\dagger A)(A^\dagger A)^\dagger$$. It follows that we can write $$A^\dagger A=\sum_k s_k\lvert v_k\rangle\!\langle v_k\rvert$$ for some (generally complex as far as we know now) $$s_k$$, satisfying $$A^\dagger A\lvert v_k\rangle=s_k\lvert v_k\rangle$$.

That $$s_k\in\mathbb R$$, and more precisely $$s_k\ge0$$, now follows from $$A^\dagger A\lvert v_k\rangle=s_k\lvert v_k\rangle \Longrightarrow s_k=\langle v_k\rvert A^\dagger A\lvert v_k\rangle=\|Av_k\|^2\ge0.$$

# Why $$\sqrt{A^\dagger A}\ge0$$?

The square root of a PSD operator is (can be) defined through the square root of its eigenvalues. In other words, if $$A=\sum_k s_k\lvert v_k\rangle\!\langle v_k\rvert$$ with $$s_k\ge0$$, then we define $$\sqrt A=\sum_k \sqrt{s_k}\lvert v_k\rangle\!\langle v_k\rvert.$$ Clearly, $$s_k\ge0\Longrightarrow \sqrt{s_k}\ge0$$, and thus $$A\ge0\Longrightarrow \sqrt A\ge0$$.

Actually, more accurately, the square root of a matrix is not uniquely defined, and does not need to be positive. In part for the same reason the square root of a real or complex number is only defined up to a sign. In the case of matrices you have even more freedom whenever there's degenerate eigenvalues: if a matrix has degenerate eigenvalues, then you can choose arbitrary bases within each degenerate eigenspace, and then choose arbitrarily the signs to assign taking the square roots of the corresponding eigenvalues. As trivial examples, consider $$A^\dagger A=I$$, and observe that both $$(-I)^2=I$$ and $$X^2=I$$ , even though clearly $$-I$$ and $$X$$ are not PSD. Nonetheless, when saying that the square root of a PSD is PSD, you are choosing a positive square root, similarly to what you do when taking the algebraic square root of real numbers, hence the statement.

# Polar decomposition from SVD

This is not directly related to the question asked, but as we are talking about this in connection with the polar decomposition, let me show another way to get to the polar decomposition.

Start from the previously shown fact that $$A^\dagger A\ge0$$. This is equivalent to it being writable as $$A^\dagger A=\sum_k s_k\lvert v_k\rangle\!\langle v_k\rvert$$ for $$s_k\ge0$$. But then again, note that $$A^\dagger A v_k=s_k v_k \Longleftrightarrow \langle A v_k,Av_j\rangle=\delta_{jk}s_k \Longleftrightarrow Av_j=\sqrt{s_j}w_j$$ for some orthonormal set of vectors $$w_j$$. This is nothing but the singular value decomposition of $$A$$.

Now, to get to the polar decomposition, we just rewrite this as $$A=\sum_k \sqrt{s_k}\lvert w_k\rangle\!\langle v_k\rvert= \Bigg(\underbrace{\sum_k \lvert w_k\rangle\!\langle v_k\rvert}_{U}\Bigg)\Bigg(\underbrace{\sum_k \sqrt{s_k} \lvert v_k\rangle\!\langle v_k\rvert}_{\sqrt{A^\dagger A}}\Bigg),$$ which is nothing but the polar decomposition of $$A$$.