Eigenvalues of $A^\dagger A$ With a $2\times 2$ matrix $A$, let $u$ be an eigenvector of $B=A^\dagger A$. My question is: when can the eigenvalue $\lambda = u^\dagger B u$ lie between $0$ and $1$, i.e., what are the conditions under which $\lambda \in [0,1]$? Here $\dagger$ denotes the Conjugate-Transpose. Also, $u$ is normalized i.e., $u^\dagger u = I$.
 A: Note that the eigenvalues of $A^\dagger A$ are necessarily real and non-negative, as
$$
\lambda=\lambda u^\dagger u= u^\dagger A^\dagger Au=(Au)^\dagger Au\geq0.
$$
The eigenvalues of $A$ can only provide necessary conditions for $\lambda\leq1$, but not sufficient. To see what can happen, consider
$$
A=\begin{bmatrix} 1&n\\0&1\end{bmatrix}. 
$$
Then the only eigenvalue of $A$ is $1$, but $A^\dagger A$ has eigenvalues
$$
\frac12\,\Big(2+n^2n\pm\sqrt{n^2+4}\Big)
$$
Some necessary conditions are that

*

*all eigenvalues of $A$ are in the unit disk;

*all entries of $A$ are in the unit disk.

A necessary and sufficient condition is that the operator norm of $A$ is at most 1. This means that
$$\tag1
\|Ax\|\leq\|x\|,\qquad\qquad x\in\mathbb C^2, 
$$
where $\|x\|=(|x_1|^2+|x_2|^2)^{1/2}$. The bad news is that, very often, the best method to check that $(1)$ holds is to show that the greatest eigenvalue of $A^\dagger A$ is at most $1$.
Among the countless examples of $A$ satisfying that $A^\dagger A$ has eigenvalues in $[0,1]$, we have the unitary matrices. Take any $u,v$ orthonormal and form $A=\begin{bmatrix} u&v\end{bmatrix}$. Then $A^\dagger A=I_2$. So for instance if
$$
A=\begin{bmatrix}\frac{1+i\sqrt3}3 & -\frac1{\sqrt3}\\ \frac1{\sqrt3}&\frac{1-i\sqrt3}3\end{bmatrix} 
$$
satisfies that $A^\dagger A$ has one $1$ as an eigenvalue.
A: Since we're working with 2x2 matrices, we can get the eigenvalues from the trace and determinant. Let $A$ be given by
$$
A=\begin{bmatrix} a&b\\c&d\end{bmatrix}. 
$$
$\mathrm{tr}(A^\dagger A) = |a|^2+|b|^2+|c|^2+|d|^2$ is just the matrix norm of $A$. Meanwhile, $\det(A^\dagger A) = |\det(A)|^2 = |ad - bc|^2$. For brevity, let $N$ be the norm of $A$ and $D$ be its determinant (the usual shorthands would result in $\left\|A\right\|^2$ and $\left|\left|A\right|\right|^2$, respectively, which would just lead to problems). Then we have
$$
\lambda = \frac{N\pm\sqrt{N^2-4|D|^2}}{2}.
$$
This is always positive, so we just need to check $\lambda < 1$, and in particular only need to check the $+$ sign. Doing a bit of algebra gives
$$
\frac{N+\sqrt{N^2-4|D|^2}}{2}<1\Longrightarrow N^2-4|D|^2<(2-N)^2\Longrightarrow N< 1 + |D|^2.
$$
So the matrix will have eigenvalues in the range $(0,1)$ if
$$
|a|^2+|b|^2+|c|^2+|d|^2 < 1 + |ad - bc|^2.
$$
