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All the matrices below are square, complex matrices.

1) Is it true that, for every Hermitian matrix $H$, there exists $A$, that $A^*A=H$?

2) For any $A$, does $A^*A$ always have a square root? If it's not, is there any simple presumption of $A$ that makes $A^*A$ always have a square root?

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up vote 7 down vote accepted
  1. It is false, because $A^{\ast}A$ has non-negative eigenvalues, but $H$ only needs to have real eigenvalues. However, if we instead require that $H$ be positive semi-definite (i.e. non-negative eigenvalues), then the result is true - this is the Cholesky decomposition. However, such a decomposition is not unique.

  2. This is true. $A^{\ast} A$ is a self-adjoint, positive semidefinite matrix, so it can be diagonalized as $A^{\ast}A = P^{-1}\Lambda P$ with $\Lambda$ a diagonal matrix with non-negative entries. Then we can simply square-root the diagonals of $\Lambda$, which I will denote by $\sqrt{\Lambda}$, and then the matrix $S = P^{-1} \sqrt{\Lambda}P$ serves as a square root of $A^{\ast}A$.

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So actually any diagonalizable matrix has a square root? Because we are talking about complex matrices. – Voldemort Oct 2 '12 at 20:41
When you have a Hermitian (i.e. self-adjoint) matrix, then all the eigenvalues are real and positive. Since the "square-rooting" only occurs on the diagonal matrix, then it works. We can easily verify that $A^{\ast} A$ is self-adjoint, since $(A^{\ast}A)^{\ast} = A^{\ast}(A^{\ast})^{\ast} = A^{\ast}A$. Furthermore, we can verify it is positive semidefinite, since $\langle A^{\ast}Ax, x \rangle = \langle Ax, Ax \rangle = \|Ax\|^2 \geq 0$. – Christopher A. Wong Oct 2 '12 at 21:02
Now that I think about it, I suppose one can also take "square roots" of diagonalizable matrices where the eigenvalues are complex. The only issue there is that when you take a square root of a complex number, you need to choose a branch of $z^{1/2}$ to work with. – Christopher A. Wong Oct 2 '12 at 21:05
  1. This property is only true for matrices $H$ such that $x^*Hx\geq 0$ for all $x$ (non-negative definite). To see that, note that if it's the case $x^*A^*Ax=\lVert Ax\rVert^2\geq 0$. Conversely, if $x^*Hx\geq 0$ for all $x$, then we can diagonalize $H$, and each eigenvalue is non-negative (hence admits a square root).

  2. Yes, as $A^*A$ is non-negative definite and symmetric, we can find $P$ unitary and $D$ diagonal such that $A^*A=P^*DP$. As each diagonal entry of $D$ is non negative, take $D'$ diagonal such that $D'^2=D$. Then take $R:=P^*D'P$.

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The spectral theorem implies that any Hermitian matrix $H$ is diagonizable using an unitary base change $U$, so $U^* H U = D$, where $D$ is diagonal with real entries. For $D$ proposition 2) is true, since $\mathbb{C}$ is closed under squareroots, so $D = R^2$ for some diagonal $R$. Therefore, $H = U R^2 U^* = (U R U^*)^2$.

For any matrix $A$, the matrix $AA^*$ is Hermitian.

However, if you go to dimension $1$, $H = -1$ is Hermitian, but for any $A \in \mathbb{C}$ you have $AA^* = \lvert A \rvert ^2 > 0$, so 1) is false.

As pointed out by others, it is true for positive-semi-definite matrices $H$, let me add a construction: You can run through the argument above, noting $R^* = R$, and take $A$ to be $UR$. So $AA^* = UR (UR)^* = UR^2 U^* = H$.

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