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Let $A^*$ denote the complex conjugate transpose of the matrix $A$ and $\|\cdot\|=\|\cdot\|_2$ the norm induced by the Euclidean vector norm.

If $$ \|A^*A+AA^*\|-\|A^*A-AA^*\|=\|A^*A\| $$ what can be said about $A$? Note that $A$ does not have to be normal, in fact if $A$ is normal and satisfies the above condition, then $A=0$, as noted in the comments below.

My first hypothesis, after lots of numerical experiments was that $\det(A)=0$, since almost all random matrices with the above property fulfilled had zero determinant, but then I found $$ A=\begin{pmatrix}-2&2&0\\0&0&4\\-2&-2&0\end{pmatrix} $$ for which $\det(A)=-32$. This shattered my hypothesis, but the question remains. What can be said about such matrices?

EDIT: I'm putting a small bounty on this question. Let's define a class $\mathcal{A}$ of complex-valued matrices such that $$ \mathcal{A}=\{A\in\mathbb{C}^{n\times n}\;:\; \|A^*A+AA^*\|-\|A^*A-AA^*\|=\|A^*A\|\}\;. $$ Can you find any necessary or sufficient conditions for a matrix to be in this class? Special cases, $A\in\mathbb{R}^{n\times n}$ for a fixed $n$ are welcome.

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    $\begingroup$ If $A$ is normal (as indicated by the title), then the second term $\|A^*A-AA^*\|$ is zero. Then $\|A^*A+AA^*\|=\|A^*A\|$ is equivalent to $2\max_{1\leq i\leq n}|\lambda_i|^2=\max_{1\leq i\leq n}|\lambda_i|^2$ and hence all the eigenvalues of $A$ are zero. Consequently, $A=0$. $\endgroup$ – Algebraic Pavel Feb 3 '15 at 12:37
  • $\begingroup$ Yes, that is true. But what if $A$ is not normal? $\endgroup$ – PeterA Feb 3 '15 at 12:43
  • $\begingroup$ Perhaps I should edit the title, not to cause confusion, but I cannot come up with a suitable name. $\endgroup$ – PeterA Feb 3 '15 at 12:46
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    $\begingroup$ The matrix $A$ in your example is a sort of "special". It has both rows and columns mutually orthogonal (though not orthonormal), so both $A^*A$ and $AA^*$ are diagonal. Maybe it could somehow help. $\endgroup$ – Algebraic Pavel Feb 3 '15 at 12:59
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    $\begingroup$ Going to $4\times 4$ matrices you can find $$ A=\begin{pmatrix} 0 & -1 & 1 & 1\\ 1 & 0 & 1 & 0\\ -1 & -1 & 0 & 0\\ 0 & 1 & 1 & 0 \end{pmatrix} $$ with $A^*A$ and $AA^*$ non diagonal. $\endgroup$ – PeterA Feb 3 '15 at 13:16
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Note first that $A\in {\mathcal A}$ if and only if $\lambda A\in {\mathcal A}$ for any non-zero number $\lambda$. Hence there is no loss of generality if we are concerned with matrices $A$ which have norm $1$. Since $\mathbb{C}^{n\times n}$ is a $C^*$-algebra if it is endowed with the operator norm, one has $\| A^* A\|=\| A\|^2$. Hence we are looking for matrices $A\in \mathbb{C}^{n\times n}$ such that $$ \| A\|=1\quad \text{and}\quad \|A^*A+AA^*\|-\|A^*A-AA^*\|=1. $$

First we are concerned with matrices of rank $1$. We will use the notation from operator theory: $A=u\otimes v$, where $u, v \in {\mathbb C}^n$ are nonzero vectors. The action of $A$ on ${\mathbb C}^n$ is given by $$Ax=\langle x,v\rangle u\qquad (x\in {\mathbb C}^n), $$ where $\langle \cdot, \cdot \rangle$ is the inner product in ${\mathbb C}^n$ (linear in the first component and antilinear in the second). Since it is assumed that $\| A\|=1$ we can assume also that $\| u\|=\| v\|=1$. If $u$ and $v$ were linearly dependent, then $A$ would be normal and hence it would not be in ${\mathcal A}$. Hence we assume that $u$ and $v$ are linearly independent vectors of norm $1$. We denote $\omega=\langle u,v\rangle$. Observe that $|\omega|< 1$ (since $u$ and $v$ are linearly independent of norm $1$).

Note that $A^*=v\otimes u$, $AA^*=u\otimes u$ and $A^*A=v\otimes v$. Of course, $AA^*+A^*A=u\otimes u+v\otimes v$ is a positive semidefinite matrix of rank $2$. Hence its norm is equal to the largest eigenvalue. Since the range of $AA^*+A^*A$ is the linear span of vectors $u$ and $v$ the eigenvectors corresponding to non-zero eigenvalues of $AA^*+A^*A$ are of the form $x=\alpha u+\beta v$. Let $\lambda>0$ be an eigenvalue of $AA^*+A^*A$ and $x=\alpha u+\beta v$ be an eigenvector at this eigenvalue. Then $$[u\otimes u+v\otimes v](\alpha u+\beta v)=\lambda (\alpha u+\beta v)$$ gives $$ (\alpha +\beta \overline{\omega}-\lambda \alpha)u+ (\alpha \omega +\beta -\lambda \beta)v=0. $$ Since $u, v$ are linearly independent we have $$ \alpha +\beta \overline{\omega}-\lambda \alpha=0\quad \text{and}\quad \alpha \omega +\beta -\lambda \beta=0. $$ We have $\alpha \ne 0$ or $\beta \ne 0$ and therefore one can deduce that $\lambda=1\pm |\omega|$. It follows that $$ \|A^*A+AA^*\|=1+|\omega|. $$

Matrix $A^*A-AA^*=u\otimes u-v\otimes v$ is selfadjoint and of rank $2$. In a similar way as before we see that the nonzero eigenvalues of this matrix are $\pm \sqrt{1-|\omega|^2}$. Hence $\|A^*A-AA^*\|=\sqrt{1-|\omega|^2}$.

Finally, it follows from $$ 1=\|A^*A+AA^*\|-\|A^*A-AA^*\|=1+|\omega|-\sqrt{1-|\omega|^2}$$ that $$ |\omega|=\frac{\sqrt{2}}{2}. $$ It is not hard to check now that any $A=u\otimes v$ with $\| u\|=1=\| v\|$ and $|\langle u,v\rangle|=\frac{\sqrt{2}}{2}$ belongs to ${\mathcal A}$.

Let now $A\in \mathbb{C}^{n\times n}$ be a matrix of rank $1$ in ${\mathcal A}$, i.e., $A=u\otimes v$ with $\| u\|=1=\| v\|$ and $|\langle u,v\rangle|=\frac{\sqrt{2}}{2}$, and let $B\in \mathbb{C}^{m\times m}$ be any matrix such that $$\| B\| \leq 1,\quad \| BB^*+B^*B\| \leq 1+\frac{\sqrt{2}}{2}\quad \text{and}\quad \| BB^*-B^*B\| \leq \frac{\sqrt{2}}{2}. $$ Then, for $A\oplus B\in \mathbb{C}^{(n+m)\times (n+m)}$, one has $$\| A\oplus B\| =1,\quad \| (A\oplus B)(A\oplus B)^*+(A\oplus B)^*(A\oplus B)\| = 1+\frac{\sqrt{2}}{2}$$ and $$ \| (A\oplus B)(A\oplus B)^*-(A\oplus B)^*(A\oplus B)\| = \frac{\sqrt{2}}{2}, $$ which means that $A\oplus B$ belongs to ${\mathcal A}$ (in dimension $n+m$).

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  • $\begingroup$ Well, it's a start! Thank you for the contribution Janko. Since you are the only one with a complete answer you'll get the bounty. $\endgroup$ – PeterA Feb 11 '15 at 21:31

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