The spectral-norm Böttcher--Wenzel conjecture for rectangular matrices

Source and prior results. Motoyuki Nobori, A generalization of the Böttcher--Wenzel inequality for three rectangular matrices, Linear Algebra and its Applications 725 (2025), 135--144, Conjecture 3.1; freely available as arXiv:2506.17365v2. The square-matrix estimates used below are established prior results, stated explicitly as equations (2) and (3) in that paper; see also K. M. R. Audenaert, Linear Algebra and its Applications 432 (2010), 1126--1143, and A. Böttcher and D. Wenzel, Linear Algebra and its Applications 429 (2008), 1864--1885. The new step is to deduce the full rectangular three-matrix conjecture from those square-matrix theorems by an explicit convex decomposition into isometries or coisometries.

Let

$$A,C\in M_{m,n}(\mathbb C), \qquad B\in M_{n,m}(\mathbb C),$$

and write

$$\|A\|_{(2),2}^2 = \sigma_1(A)^2+\sigma_2(A)^2,$$

with a missing second singular value interpreted as zero. We prove, for every $m,n\ge1$,

$$\boxed{ \|ABC-CBA\|_F^2 \le 2\|B\|_2^2\|A\|_{(2),2}^2\|C\|_F^2. } \tag{1}$$

In particular, this proves Conjecture 3.1 for all complex rectangular matrices in its stated range $m,n\ge2$.

1. Every rectangular contraction is an explicit average of extremal isometries

The case $B=0$ is immediate. Otherwise, by homogeneity, replace $B$ by

$$D=\frac{B}{\|B\|_2}.$$

Put $p=\min\{m,n\}$ and choose a singular-value decomposition

$$D=U\Sigma V^*, \qquad U\in U(n), \qquad V\in U(m), \qquad \Sigma_{ii}=s_i\in[0,1] \quad(1\le i\le p), \tag{2}$$

where the remaining entries of the rectangular $n\times m$ matrix $\Sigma$ are zero.

For each sign vector

$$\varepsilon=(\varepsilon_1,\ldots,\varepsilon_p) \in\{-1,1\}^p,$$

let $J_{\varepsilon}$ be the rectangular diagonal matrix with diagonal entries $\varepsilon_i$, and set

$$\omega_{\varepsilon} = \prod_{i=1}^p\frac{1+\varepsilon_i s_i}{2}, \qquad Q_{\varepsilon}=UJ_{\varepsilon}V^*. \tag{3}$$

The weights are nonnegative and satisfy

$$\sum_{\varepsilon}\omega_{\varepsilon}=1, \qquad \sum_{\varepsilon}\omega_{\varepsilon}\varepsilon_i=s_i.$$

Consequently

$$D = \sum_{\varepsilon} \omega_{\varepsilon}Q_{\varepsilon}. \tag{4}$$

If $n\ge m$, every summand is an isometry:

$$Q_{\varepsilon}^*Q_{\varepsilon}=I_m. \tag{5}$$

If $n\le m$, every summand is a coisometry:

$$Q_{\varepsilon}Q_{\varepsilon}^*=I_n. \tag{6}$$

The square case satisfies both identities. Thus no approximation, closure argument, or existence theorem for extreme points is needed.

2. The isometric case

Suppose $n\ge m$, and fix one summand $Q=Q_{\varepsilon}$ from (4). Form the square $n\times n$ matrices

$$X=QA, \qquad Y=QC.$$

Their commutator is exactly

$$[X,Y] = QAQC-QCQA = Q(AQC-CQA). \tag{7}$$

Since $Q^*Q=I_m$, left multiplication by $Q$ preserves the Frobenius norm. It also preserves all singular values, because

$$(QA)^*(QA)=A^*A.$$

Hence

$$\|[X,Y]\|_F =\|AQC-CQA\|_F, \qquad \|X\|_{(2),2}=\|A\|_{(2),2}, \qquad \|Y\|_F=\|C\|_F. \tag{8}$$

Apply the established square-matrix inequality, equation (2) of the source,

$$\|[X,Y]\|_F^2 \le 2\|X\|_{(2),2}^2\|Y\|_F^2.$$

Using (8) gives

$$\|AQC-CQA\|_F \le \sqrt2\,\|A\|_{(2),2}\|C\|_F. \tag{9}$$

3. The coisometric case

Suppose $n\le m$ instead, so $QQ^*=I_n$. This time form the square $m\times m$ matrices

$$X=AQ, \qquad Y=CQ.$$

Then

$$[X,Y] = AQCQ-CQAQ = (AQC-CQA)Q. \tag{10}$$

Right multiplication by $Q$ preserves the Frobenius norm. All singular values are again preserved, now because

$$(AQ)(AQ)^*=AA^*.$$

Therefore

$$\|[X,Y]\|_F =\|AQC-CQA\|_F, \qquad \|X\|_{(2),2}=\|A\|_{(2),2}, \qquad \|Y\|_F=\|C\|_F.$$

Applying the same established square inequality proves (9) in the coisometric case as well.

4. Averaging proves the conjecture

The generalized commutator is linear in its middle factor. Therefore (4), the triangle inequality, and (9) give

$$\begin{aligned} \|ADC-CDA\|_F &= \left\| \sum_{\varepsilon} \omega_{\varepsilon} (AQ_{\varepsilon}C-CQ_{\varepsilon}A) \right\|_F \\ &\le \sum_{\varepsilon} \omega_{\varepsilon} \|AQ_{\varepsilon}C-CQ_{\varepsilon}A\|_F \\ &\le \sqrt2\,\|A\|_{(2),2}\|C\|_F. \end{aligned} \tag{11}$$

Multiplying by $\|B\|_2$ and squaring yields (1).

5. The stronger alternating-tensor bound also holds in every rank

The source also asks whether the stronger-looking consequence

$$\|ABC-CBA\|_F^2 \le \|B\|_2^2 \|A\otimes C-C\otimes A\|_F^2 \tag{12}$$

holds. It verifies this when $\operatorname{rank}B\le2$ but leaves the unrestricted case unresolved.

Use precisely the same square matrices $X,Y$ as above. Isometric or coisometric multiplication preserves not only singular values and Frobenius norms, but also the Frobenius inner product:

$$\langle X,Y\rangle_F = \langle A,C\rangle_F.$$

The established square commutator estimate, equation (3) of the source, therefore gives

$$\begin{aligned} \|AQC-CQA\|_F^2 &= \|[X,Y]\|_F^2 \\ &\le 2\left( \|X\|_F^2\|Y\|_F^2 -|\langle X,Y\rangle_F|^2 \right) \\ &= 2\left( \|A\|_F^2\|C\|_F^2 -|\langle A,C\rangle_F|^2 \right) \\ &= \|A\otimes C-C\otimes A\|_F^2. \end{aligned} \tag{13}$$

Averaging the unsquared inequality exactly as in (11), then restoring $\|B\|_2$, proves (12) for arbitrary complex rectangular matrices and every rank of $B$.

By additionally exchanging $A$ and $C$, one obtains the simultaneous refinement

$$\begin{aligned} \|ABC-CBA\|_F^2 \le 2\|B\|_2^2 \min\Bigl\{ &\|A\|_{(2),2}^2\|C\|_F^2, \\ &\|C\|_{(2),2}^2\|A\|_F^2, \\ &\|A\|_F^2\|C\|_F^2 -|\langle A,C\rangle_F|^2 \Bigr\}. \end{aligned} \tag{14}$$

The coefficient $2$ is sharp. In dimension $m=n=2$, take

$$A= \begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix}, \qquad B=I_2, \qquad C= \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}.$$

Then

$$\|ABC-CBA\|_F^2=4, \qquad \|B\|_2^2=1, \qquad \|A\|_{(2),2}^2=2, \qquad \|C\|_F^2=1,$$

so equality holds in both (1) and (12). Padding these matrices with zero rows and columns gives sharp examples in every rectangular dimension with $m,n\ge2$.