A spectral-norm generalization of the Böttcher-Wenzel inequality for rectangular matrices

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Let Mm,n(C)M_{m,n}(\mathbb{C}) denote the space of complex m×nm\times n matrices. For a matrix, write ∥⋅∥F\|\cdot\|_F for the Frobenius norm, ∥⋅∥2\|\cdot\|_2 for the spectral norm, and ∥A∥(2),2\|A\|_{(2),2} for the norm appearing in the stated inequality. Let A,C∈Mm,n(C)A,C\in M_{m,n}(\mathbb{C}) and B∈Mn,m(C)B\in M_{n,m}(\mathbb{C}), where m,n≥2m,n\geq 2. Spectral-norm generalization.

∥ABC−CBA∥F2≤2∥B∥22∥A∥(2),22∥C∥F2.\|ABC-CBA\|^{2}_{F}\leq 2\|B\|^{2}_{2}\|A\|^{2}_{(2),2}\|C\|_{F}^{2}.

This conjecture would generalize the Böttcher-Wenzel-type inequalities considered in the paper to the case in which the norm of BB is the spectral norm. It is based on numerical experiments; its general validity remains open.

References

Primary source

Motoyuki Nobori, “A Generalization of the Böttcher-Wenzel inequality for three rectangular matrices”, arXiv:2506.17365 (2025).

Progress summary

Refreshed
Claimed solved

A reader-written complete-proof claim would settle the conjecture in all rectangular sizes, but it has not been independently checked.

Motoyuki Nobori posed this conjecture in 2025 for rectangular matrices with m,n≥2m,n\geq 2, based on numerical experiments. It asks whether the Böttcher--Wenzel-type bound remains valid when BB is measured by the spectral norm.

Known results

  • Nobori, 2025: the related bound with ∥C∥22∥B∥F2\|C\|_2^2\|B\|_F^2 is proved for all rectangular sizes.
  • Nobori, 2025: the target-form inequality is proved when m=1m=1 or n=1n=1.
  • Nobori, 2025: the related tensor bound is established when rank⁡B≤2\operatorname{rank}B\leq 2, but this does not settle the conjecture for m,n≥2m,n\geq 2.

Posted attempt

A reader-written argument claims a complete proof for every m,n≥1m,n\geq 1, by decomposing a normalized BB into isometries or coisometries and applying square-matrix inequalities. It also claims a stronger tensor bound, but neither claim has been independently verified.

Current status (as of August 2026): Nobori’s edge cases and related rank-restricted results are settled, while the full conjecture has only an unverified complete-proof claim and remains mathematically unconfirmed.

Sources

Solutions 1

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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∈Mm,n(C),B∈Mn,m(C),A,C\in M_{m,n}(\mathbb C), \qquad B\in M_{n,m}(\mathbb C),

and write

∥A∥(2),22=σ1(A)2+σ2(A)2,\|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≥1m,n\ge1,

∥ABC−CBA∥F2≤2∥B∥22∥A∥(2),22∥C∥F2.(1)\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≥2m,n\ge2.

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

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

D=B∥B∥2.D=\frac{B}{\|B\|_2}.

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

D=UΣV∗,U∈U(n),V∈U(m),Σii=si∈[0,1](1≤i≤p),(2)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×mn\times m matrix Σ\Sigma are zero.

For each sign vector

ε=(ε1,…,εp)∈{−1,1}p,\varepsilon=(\varepsilon_1,\ldots,\varepsilon_p) \in\{-1,1\}^p,

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

ωε=∏i=1p1+εisi2,Qε=UJεV∗.(3)\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

∑εωε=1,∑εωεεi=si.\sum_{\varepsilon}\omega_{\varepsilon}=1, \qquad \sum_{\varepsilon}\omega_{\varepsilon}\varepsilon_i=s_i.

Consequently

D=∑εωεQε.(4)D = \sum_{\varepsilon} \omega_{\varepsilon}Q_{\varepsilon}. \tag{4}

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

Qε∗Qε=Im.(5)Q_{\varepsilon}^*Q_{\varepsilon}=I_m. \tag{5}

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

QεQε∗=In.(6)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≥mn\ge m, and fix one summand Q=QεQ=Q_{\varepsilon} from (4). Form the square n×nn\times n matrices

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

Their commutator is exactly

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

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

(QA)∗(QA)=A∗A.(QA)^*(QA)=A^*A.

Hence

∥[X,Y]∥F=∥AQC−CQA∥F,∥X∥(2),2=∥A∥(2),2,∥Y∥F=∥C∥F.(8)\|[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]∥F2≤2∥X∥(2),22∥Y∥F2.\|[X,Y]\|_F^2 \le 2\|X\|_{(2),2}^2\|Y\|_F^2.

Using (8) gives

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

3. The coisometric case

Suppose n≤mn\le m instead, so QQ∗=InQQ^*=I_n. This time form the square m×mm\times m matrices

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

Then

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

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

(AQ)(AQ)∗=AA∗.(AQ)(AQ)^*=AA^*.

Therefore

∥[X,Y]∥F=∥AQC−CQA∥F,∥X∥(2),2=∥A∥(2),2,∥Y∥F=∥C∥F.\|[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

∥ADC−CDA∥F=∥∑εωε(AQεC−CQεA)∥F≤∑εωε∥AQεC−CQεA∥F≤2 ∥A∥(2),2∥C∥F.(11)\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\|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∥F2≤∥B∥22∥A⊗C−C⊗A∥F2(12)\|ABC-CBA\|_F^2 \le \|B\|_2^2 \|A\otimes C-C\otimes A\|_F^2 \tag{12}

holds. It verifies this when rank⁡B≤2\operatorname{rank}B\le2 but leaves the unrestricted case unresolved.

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

⟨X,Y⟩F=⟨A,C⟩F.\langle X,Y\rangle_F = \langle A,C\rangle_F.

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

∥AQC−CQA∥F2=∥[X,Y]∥F2≤2(∥X∥F2∥Y∥F2−∣⟨X,Y⟩F∣2)=2(∥A∥F2∥C∥F2−∣⟨A,C⟩F∣2)=∥A⊗C−C⊗A∥F2.(13)\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\|B\|_2, proves (12) for arbitrary complex rectangular matrices and every rank of BB.

By additionally exchanging AA and CC, one obtains the simultaneous refinement

∥ABC−CBA∥F2≤2∥B∥22min⁡{∥A∥(2),22∥C∥F2,∥C∥(2),22∥A∥F2,∥A∥F2∥C∥F2−∣⟨A,C⟩F∣2}.(14)\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 22 is sharp. In dimension m=n=2m=n=2, take

A=(100−1),B=I2,C=(0100).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∥F2=4,∥B∥22=1,∥A∥(2),22=2,∥C∥F2=1,\|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≥2m,n\ge2.