Positive operators and square roots

We have $\hat M^\dagger=\hat A^\dagger(\hat A^\dagger)^\dagger=\hat A^\dagger\hat A=\hat M$. For every $\ket v$, $\bra v\hat M\ket v=\braket{\hat Av}{\hat Av}=\norm{\hat Av}^2\geq0$. If $\hat M\ket v=\mu\ket v$ with $\ket v$ normalised, then $\mu=\bra v\hat M\ket v\geq0$. Finally, if $\hat M\ket v=0$, then $\norm{\hat Av}^2=\bra v\hat M\ket v=0$, so $\hat A\ket v=0$; conversely, $\hat A\ket v=0$ implies $\hat M\ket v=0$. The two operators have the same kernel, and in finite dimensions one is invertible if and only if the other is.

Write the spectral decomposition $\hat M=\sum_k\mu_k\hat P_k$, with $\mu_k\geq0$ and mutually orthogonal projectors $\hat P_k$ whose sum is $\mathbf{1}$. Set $\hat R=\sum_k\sqrt{\mu_k}\,\hat P_k$, which is self-adjoint, with eigenvalues $\sqrt{\mu_k}\geq0$, and hence positive. Moreover, $\hat R^2=\sum_k\mu_k\hat P_k=\hat M$, since $\hat P_k\hat P_l=\delta_{kl}\hat P_k$. One can also show that no other positive operator has this property. The eigenvalues of $\sqrt{\hat A^\dagger\hat A}$ are called the singular values of $\hat A$.

We calculate

$$\begin{aligned} M=\begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix}=\begin{pmatrix}1&1\\1&2\end{pmatrix}. \end{aligned}$$

Its trace is $3$ and its determinant is $1$. The eigenvalues are the roots of $\mu^2-3\mu+1=0$, namely $\mu_\pm=\frac{3\pm\sqrt5}{2}$. Now $\phi^2=\phi+1=\frac{3+\sqrt5}{2}$, and $\phi^{-2}=\frac{2}{3+\sqrt5}=\frac{2(3-\sqrt5)}{9-5}=\frac{3-\sqrt5}{2}$. Thus $\mu_+=\phi^2$ and $\mu_-=\phi^{-2}$, both positive, as expected.

The Cayley—Hamilton theorem states that every $2\times2$ matrix satisfies its characteristic polynomial: $R^2-(\tr R)R+(\det R)\mathbf{1}=0$. With $R^2=M$ and $\det R=\sqrt{\mu_+\mu_-}=\sqrt{\det M}$, we obtain $(\tr R)\,R=M+\sqrt{\det M}\,\mathbf{1}$. It remains to calculate $\tr R=\sqrt{\mu_+}+\sqrt{\mu_-}$. Its square is $\mu_++\mu_-+2\sqrt{\mu_+\mu_-}=\tr M+2\sqrt{\det M}$, which is strictly positive if $M\neq0$. This gives the stated formula. For our matrix, $\det M=1$ and $\tr M=3$, hence

$$\begin{aligned} \sqrt M=\frac{1}{\sqrt5}\begin{pmatrix}2&1\\1&3\end{pmatrix}. \end{aligned}$$

Verification: $\frac15\begin{pmatrix}2&1\\1&3\end{pmatrix}^2=\frac15\begin{pmatrix}5&5\\5&10\end{pmatrix}=\begin{pmatrix}1&1\\1&2\end{pmatrix}=M$. The eigenvalues of $\sqrt M$ are $\phi$ and $1/\phi$: these are the singular values of $A$.

For normalised $\ket v$, $\norm{Av}^2=\bra vM\ket v$. Expanding $\ket v$ in an orthonormal eigenbasis of $M$, $\ket v=\sum_kc_k\ket{m_k}$ with $\sum_k|c_k|^2=1$, gives $\bra vM\ket v=\sum_k\mu_k|c_k|^2\leq\mu_{\max}$, with equality for an eigenvector associated with $\mu_{\max}$. Thus $\norm A^2=\mu_{\max}$, and $\norm A=\sqrt{\mu_{\max}}$ is the largest singular value. For our example, $\norm A=\phi\simeq1{,}618$. Yet the only eigenvalue of $A$ is $1$: for a nonnormal matrix, the norm may be strictly larger than the largest modulus of the eigenvalues. For a normal matrix, by contrast, the singular values are the moduli of the eigenvalues.