# 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., $$u^\dagger u = I$$.

• What is $u$ here? Commented Oct 12, 2022 at 18:31
• $u$ is defined as $B u = \lambda u$, i.e., an eigenvector of $B$ with eigenvalue $\lambda$. Commented Oct 12, 2022 at 18:33
• Notice that the eigenvalues of $A^\dagger A$ are always nonnegative, their square roots are usually called the singular values. Your question is not clear: you want a condition in terms of what? The coefficients of $A$? And there are two eigenvalues for $2\times 2$ matrices, are you looking for a condition such that all the eigenvalues are smaller than $1$, or at least one eigenvalue? Commented Oct 12, 2022 at 19:07
• I want some conditions on $A$ such that all the eigenvalues of $A^\dagger A$ lie in the interval [0,1]. Commented Oct 12, 2022 at 19:10
• What you want is equivalent to $\|A\| \leq 1$. For $2 \times 2$ matrices this can be done explicitly, but it is not particularly pretty. Commented Oct 12, 2022 at 19:56

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.

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 $$|a|^2+|b|^2+|c|^2+|d|^2 < 1 + |ad - bc|^2.$$