The Tang-Shu Zhang Schatten Norm Formula Fails Below the Hilbertian Exponent
Schatten norms, matrix absolute values, sharp constants, rank-one matrices, counterexamples, and a phase transition at \(p=2\)
1. Introduction
For \(A\in M_n(\mathbb C)\), write \[ |A|=(A^*A)^{1/2}, \] and let \(\|A\|_p=(\operatorname{Tr}|A|^p)^{1/p}\) be the Schatten \(p\)-norm for \(1\le p<\infty\); \(\|\cdot\|_\infty\) denotes the operator norm. For fixed \(m,n\ge1\), define \[ c^{\mathrm{abs}}_p(m,n) =\sup_{(A_1,\ldots,A_m)\ne0} \frac{\left\|\sum_{k=1}^{m}A_k\right\|_p} {\left\|\sum_{k=1}^{m}|A_k|\right\|_p}, \qquad A_k\in M_n(\mathbb C). \tag{1} \] We use the dimension-free notation \[ c_p(m)=\sup_{n\ge1}c^{\mathrm{abs}}_p(m,n). \tag{2} \] The all-zero family is excluded; the denominator otherwise cannot vanish.
Tang and Shu Zhang proved \[ c_1(m)=1,\qquad c_2(m)=\sqrt{\frac{1+\sqrt m}{2}},\qquad c_\infty(m)=\sqrt m, \] and proposed a formula for the remaining exponents. For finite \(p>1\), consider \[ g_{p,m}(x):=x^p-2x-(m-1)=0. \tag{3} \] This equation has exactly one root \(x_{p,m}>1\). Their conjectured value is \[ C^{\mathrm{TZ}}_{p,m} =\frac{\sqrt{x_{p,m}(x_{p,m}+m-1)}} {(x_{p,m}^{p}+m-1)^{1/p}}. \tag{4} \] The lower bound \(c_p(m)\ge C^{\mathrm{TZ}}_{p,m}\) is attained by a rank-one equiangular family.
The dimension parameter matters at fixed size. Writing \(d=\min\{m,n\}\), Teng Zhang subsequently obtained \[ c^{\mathrm{abs}}_1(m,n)=1,\qquad c^{\mathrm{abs}}_2(m,n)=\sqrt{\frac{1+\sqrt d}{2}},\qquad c^{\mathrm{abs}}_\infty(m,n)=\sqrt d, \] together with \(c^{\mathrm{abs}}_p(m,n)\le d^{1/2-1/(2p)}\) for \(1\le p\le\infty\). Bourin and Lee also highlighted the question for Schatten exponents other than two. Neither result asserts formula (4) for general \(p\).
2. Failure throughout \(1<p<2\)
Let \(U=(e,u)\) and \(V=(e,v)\). Their Gram matrices \[ L=U^TU=\begin{pmatrix}1&1-\delta\\1-\delta&1\end{pmatrix}, \qquad G=V^TV=\begin{pmatrix}1&s\\s&1\end{pmatrix} \] commute. Consequently, the squared singular values of \(A_1+A_2=UV^T\) are \[ (2-\delta)(1+s),\qquad \delta(1-s), \] whereas the eigenvalues of \(|A_1|+|A_2|=VV^T\) are \(1+s\) and \(1-s\). Hence, if \(R_p(\delta)\) denotes the attained ratio, then \[ R_p(\delta)^p =\frac{[(2-\delta)(1+s)]^q+[\delta(1-s)]^q} {(1+s)^p+(1-s)^p}. \tag{7} \]
At \(\delta=0\), this is the Tang-Shu Zhang value. Indeed, \[ 1+s=\frac{2x}{x+1},\qquad 1-s=\frac{2}{x+1},\qquad x^p+1=2(x+1), \] and direct substitution gives \[ R_p(0)^p=\frac12x^q(x+1)^{q-1} =(C^{\mathrm{TZ}}_{p,2})^p. \tag{8} \]
It remains to compare the numerators in (7). Their difference is \[ \Delta_p(\delta) =(1+s)^q\big((2-\delta)^q-2^q\big)+(1-s)^q\delta^q. \] For \(0<\delta\le1\), the mean value theorem and \(0<q<1\) give \[ 2^q-(2-\delta)^q\le q\delta\le\delta. \] Since \((1-s)/(1+s)=x^{-1}\), it follows that \[ \Delta_p(\delta) \ge(1+s)^q\delta\big(x^{-q}\delta^{q-1}-1\big). \tag{9} \] The choice \(\delta=\frac12x^{-q/(1-q)}\) satisfies \[ x^{-q}\delta^{q-1}=2^{1-q}>1. \] Thus \(\Delta_p(\delta)>0\), and (7)-(8) yield \(R_p(\delta)>C^{\mathrm{TZ}}_{p,2}\). □
3. An exact \(2\times2\) counterexample
Set
\[ e=\begin{pmatrix}1\\0\end{pmatrix},\qquad u=\begin{pmatrix}39/40\\\sqrt{79}/40\end{pmatrix},\qquad v=\begin{pmatrix}5/8\\\sqrt{39}/8\end{pmatrix}. \]These are real unit vectors. Define
\[ A_1=ee^T=\begin{pmatrix}1&0\\0&0\end{pmatrix}, \qquad A_2=uv^T= \begin{pmatrix} 39/64&39\sqrt{39}/320\\ \sqrt{79}/64&\sqrt{3081}/320 \end{pmatrix}. \tag{11} \]Both matrices have rank one and unique nonzero singular value equal to one. Consequently, \(|A_1|=ee^T\) and \(|A_2|=vv^T\).
We first prove \(R>207/200\) using rational arithmetic only. Positivity allows us to raise each proposed enclosure to the fourth or second power. The exact differences are \[ \left(\frac{1027}{320}\right)^3 -\left(\frac{11989}{5000}\right)^4 =\frac{29144230879351}{40000000000000000}>0, \tag{15} \] \[ \left(\frac{3}{320}\right)^3 -\left(\frac{301}{10000}\right)^4 =\frac{124819571}{40000000000000000}>0, \tag{16} \] \[ \left(\frac{4143}{2000}\right)^2 -\left(\frac{13}{8}\right)^3 =\frac{773}{8000000}>0, \tag{17} \] and \[ \left(\frac{2297}{10000}\right)^2 -\left(\frac{3}{8}\right)^3 =\frac{5543}{200000000}>0. \tag{18} \] Put \[ A_0=\frac{11989}{5000}+\frac{301}{10000}=\frac{24279}{10000}, \] \[ B_0=\frac{4143}{2000}+\frac{2297}{10000}=\frac{5753}{2500}, \qquad q=\frac{207}{200}. \] A final exact comparison gives \[ A_0^2-q^3B_0^2 =\frac{1172956601313}{50000000000000}>0. \tag{19} \] Equations (15)-(19) imply \(R>q\).
It remains to put the conjectured constant below the same rational separator. Let \(x>1\) solve \(x^{3/2}-2x-1=0\), and write \(x=t^2\). The unique positive root of \[ h(t)=t^3-2t^2-1=0 \tag{20} \] lies above \(4/3\), where \(h\) is strictly increasing. Moreover, \[ h\left(\frac{1103}{500}\right) =\frac{310727}{125000000}>0, \tag{21} \] so \(t<T:=1103/500\). Writing \(C=C^{\mathrm{TZ}}_{3/2,2}\) and using \(t^3=2t^2+1\), we obtain \[ C^6=\frac{t^6}{16(t^2+1)}=:H(t). \] The function \(H\) is strictly increasing for \(t>0\), since \[ H'(t)=\frac{t^5(2t^2+3)}{8(t^2+1)^2}>0. \] Finally, \[ q^6-H(T) =\frac{26731151399029597}{18772595200000000000}>0. \tag{22} \] It follows that \(C<q<R\). □
4. The sharp rank-one problem for \(p\ge2\)
Let \(K=L^{1/2}GL^{1/2}\). The nonzero eigenvalues of \(K\) are the squared singular values of \(UV^*\), while the nonzero eigenvalues of \(G\) are the eigenvalues of \(VV^*\). Therefore \[ \left\|\sum_kA_k\right\|_p^p=\operatorname{Tr}K^{p/2}, \qquad \left\|\sum_k|A_k|\right\|_p^p=\operatorname{Tr}G^p. \tag{25} \] Since \(p/2\ge1\), \[ \operatorname{Tr}K^{p/2}\le(\operatorname{Tr}K)^{p/2}. \tag{26} \] If \(\lambda=\lambda_{\max}(G)\), then \[ \operatorname{Tr}K=\operatorname{Tr}(LG) \le\lambda\operatorname{Tr}L=R\lambda. \tag{27} \]
List the \(m\) eigenvalues of \(G\), including zeros, as \(\lambda=\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_m\ge0\). Convexity gives \[ \operatorname{Tr}G^p \ge\lambda^p+\frac{(R-\lambda)^p}{(m-1)^{p-1}}. \tag{28} \] If \(\lambda=R\), the ratio is at most one. Otherwise put \[ y=\frac{(m-1)\lambda}{R-\lambda}. \] Since \(\lambda\ge R/m\), one has \(y\ge1\). Combining (25)-(28) yields \[ \frac{\left\|\sum_kA_k\right\|_p} {\left\|\sum_k|A_k|\right\|_p} \le F_{p,m}(y) :=\frac{\sqrt{y(y+m-1)}}{(y^p+m-1)^{1/p}}. \tag{29} \] Direct differentiation gives \[ \frac{d}{dy}\log F_{p,m}(y) =\frac{(m-1)(2y+m-1-y^p)} {2y(y+m-1)(y^p+m-1)}. \] For \(p\ge2\), the function \(y^p-2y-(m-1)\) has exactly one zero \(x_{p,m}>1\). Thus \(F_{p,m}\) has a unique maximum at \(x_{p,m}\), proving (5).
To track equality, put \(\beta=(R-\lambda)/(m-1)\). Equality in the scalar maximization and in (28) forces \[ \lambda=x_{p,m}\beta, \qquad \operatorname{spec}(G)=\{\lambda,\beta,\ldots,\beta\}. \] The top eigenspace is one-dimensional. Equality in (27) forces the range of \(L\) into this eigenspace. If \(w\) is its unit eigenvector, then \[ L=Rww^*, \qquad G=\beta I+(\lambda-\beta)ww^*. \tag{30} \] The common diagonal condition gives \(|w_k|^2=1/m\) and \(r_k=R/m\). After simultaneous phase changes, take \(w=m^{-1/2}(1,\ldots,1)^T\). Equation (30) then says that the \(u_k\) coincide and \[ \langle v_j,v_k\rangle =\frac{\lambda-\beta}{R} =\frac{x_{p,m}-1}{x_{p,m}+m-1} \qquad(j\ne k). \] Conversely, this family takes equality at every step. Its Gram matrix is positive definite, so equality requires \(n\ge m\). □
5. The full \(m=2,p=4\) problem
Introduce \[ e=\operatorname{Tr}H^3K, \qquad f=\operatorname{Tr}HK^3, \qquad g=\operatorname{Tr}HKHK. \] Cyclically collecting all words in the noncommutative expansion gives \[ \operatorname{Tr}(H+K)^4 =a+b+4(e+f)+4d+2g. \tag{32} \] Two lower bounds are needed. First, \[ e+f-2d =\operatorname{Tr}(H-K)H(H-K)K =\|H^{1/2}(H-K)K^{1/2}\|_2^2\ge0. \tag{33} \] Second, the three-factor Schatten Holder inequality gives \[ \sqrt d=\|HK\|_2 \le\|H^{1/2}\|_8\|H^{1/2}K^{1/2}\|_4\|K^{1/2}\|_8 =a^{1/8}g^{1/4}b^{1/8}. \] Thus, when \(ab>0\), \[ g\ge\frac{d^2}{\sqrt{ab}}. \tag{34} \] If \(ab=0\), one of \(H,K\) vanishes and the theorem is immediate. Combining (32)-(34), \[ \operatorname{Tr}(H+K)^4 \ge a+b+12d+\frac{2d^2}{\sqrt{ab}}. \tag{35} \]
Let \(A_0=\sqrt a\), \(B_0=\sqrt b\), and define \[ z=\frac{A_0+B_0}{\sqrt{A_0B_0}}\ge2, \qquad s=\sqrt{\frac{d}{A_0B_0}}\in[0,1]. \] The upper bound on \(s\) is the Hilbert-Schmidt Cauchy-Schwarz inequality \(d\le\sqrt{ab}\). From (31) and (35), \[ \frac{\|A+B\|_4^4}{\||A|+|B|\|_4^4} \le\Phi(z,s) :=\frac{(z+2s)^2}{z^2-2+12s^2+2s^4}. \tag{36} \]
If \(s\ge1/2\), then \[ 2(z^2-2+12s^2+2s^4)-(z+2s)^2 =(z-2s)^2+16s^2+4s^4-4\ge0, \] so \(\Phi(z,s)\le2\). Suppose \(0\le s\le1/2\). The sign of \(\partial\Phi/\partial z\) is the sign of \[ s^4+6s^2-zs-1. \] Since \(z\ge2\), \[ s^4+6s^2-zs-1 \le s^4+6s^2-2s-1\le2s-1\le0. \] Consequently, \[ \Phi(z,s)\le\Phi(2,s) =f(s):=\frac{2(1+s)^2}{1+6s^2+s^4}. \tag{37} \] The sign of \(f'(s)\) is the sign of \[ 1-6s-2s^3-s^4. \] Hence \(f\) has a unique maximizer \(s_0\in(0,1/2)\), characterized by \[ s_0^4+2s_0^3+6s_0-1=0. \tag{38} \] The first branch cannot dominate because \(f(1/6)>2\).
Set \(x=(1+s_0)/(1-s_0)\). Then \[ (x^4-2x-1)(1-s_0)^4 =2(s_0^4+2s_0^3+6s_0-1)=0, \] and \[ f(s_0) =\frac{x^2(x+1)^2}{x^4+1} =\frac{x^2(x+1)}{2}, \tag{39} \] where the last equality uses \(x^4+1=2(x+1)\). Equations (36)-(39) prove (6).
For sharpness, choose unit vectors \(r_1,r_2\) with \(\langle r_1,r_2\rangle=s_0\), choose a unit vector \(u\), and set \[ A=ur_1^*,\qquad B=ur_2^*. \] Then \[ \|A+B\|_4^4=4(1+s_0)^2 \] and \[ \||A|+|B|\|_4^4 =(1+s_0)^4+(1-s_0)^4 =2(1+6s_0^2+s_0^4). \] Their ratio is \(f(s_0)\), proving sharpness. □
Reproducibility and disclosure
The proof for the full interval \(1<p<2\) is analytic; a standard-library high-precision script checks the closed formulas at representative rational exponents. The fixed \(p=3/2\) certificate consists of the seven positive rational differences in (15)-(22) and uses exact arithmetic only. Both scripts accompany the manuscript. Low-dimensional numerical searches were used for exploration and adversarial testing only.
OpenAI Codex assisted with proof exploration, counterexample search, adversarial checking, and manuscript preparation. The submitting author is responsible for the correctness of every statement and for compliance with the target journal's authorship and disclosure policies.
References
- Jean-Christophe Bourin and Eun-Young Lee, Triangle inequalities for the operator symmetric modulus, Proceedings of the American Mathematical Society, Early View, July 2026.
- Quanyu Tang and Shu Zhang, Generalizing Lee's conjecture on the sum of absolute values of matrices, Linear Algebra and its Applications 731 (2026), 196-204.
- Teng Zhang, Operator symmetric moduli and sharp triangle inequalities, Journal of the London Mathematical Society 114(2) (2026), e70672.
Author information
Zijian Zeng
Institute of Computer Science and Digital Innovation, UCSI University,
Kuala Lumpur 56000, Malaysia
1002266693@ucsiuniversity.edu.my
Houde Liu
Tsinghua Shenzhen International Graduate School, Tsinghua University,
Shenzhen 518055, China
liu.hd@sz.tsinghua.edu.cn
Kuru Ratnavelu
Institute of Computer Science and Digital Innovation, UCSI University,
Kuala Lumpur 56000, Malaysia
Institute of Mathematical Sciences, University of Malaya,
Kuala Lumpur 50603, Malaysia
Kurunathan@ucsiuniversity.edu.my
