A Counterexample to the Tang–Zhang Schatten Norm Conjecture and Sharp Positive Results

1. Introduction

For $A\in {\mathbb{M}}_n\left(\mathbb{C}\right)$, write

$$\left|A\right|={\left(A^{*}A\right)}^{1/2}$$

and let ${\parallel A\parallel }_p={(Tr|A|}^p{)}^{1/p}$ be the Schatten p-norm for $1\le p<\infty$; ${\parallel ·\parallel }_{\infty }$ denotes the operator norm. For fixed $m,n\ge 1$, first define

$$c_p^{\mathrm{abs}}(m,n)=\underset{(A_1,\dots ,A_m)\ne 0}{sup}{\displaystyle \frac{{∥{\sum }_{k=1}^m{A}_k∥}_p}{{∥{\sum }_{k=1}^m\left|A_k\right|∥}_p}},A_k\in {\mathbb{M}}_n\left(\mathbb{C}\right).$$

(1)

We use the dimension-free notation

$$c_p\left(m\right)=\underset{n\ge 1}{sup}{c}_p^{\mathrm{abs}}(m,n).$$

(2)

The all-zero family is excluded; the denominator otherwise cannot vanish.

Tang and Zhang [1] proved

$$c_1\left(m\right)=1,c_2\left(m\right)=\sqrt{{\displaystyle \frac{1+\sqrt{m}}{2}}},c_{\infty }\left(m\right)=\sqrt{m},$$

and proposed a formula for the remaining exponents. For finite $p>1$, let $x_{p,m}>1$ be the unique solution of

$$x^p-2x-(m-1)=0.$$

(3)

Their conjectured value is

$$C_{p,m}^{\mathrm{TZ}}={\displaystyle \frac{\sqrt{x_{p,m}(x_{p,m}+m-1)}}{{(x_{p,m}^p+m-1)}^{1/p}}}.$$

(4)

The lower bound $c_p\left(m\right)\ge C_{p,m}^{\mathrm{TZ}}$ is attained by a rank-one equiangular family.

The dimension parameter matters at fixed size. Writing $d=min\{m,n\}$, Zhang [2] subsequently obtained

$$c_1^{\mathrm{abs}}(m,n)=1,c_2^{\mathrm{abs}}(m,n)=\sqrt{{\displaystyle \frac{1+\sqrt{d}}{2}}},c_{\infty }^{\mathrm{abs}}(m,n)=\sqrt{d},$$

together with $c_p^{\mathrm{abs}}(m,n)\le d^{1/2-1/\left(2p\right)}$ for $1\le p\le \infty$. Bourin and Lee [3] also highlighted the question for Schatten exponents other than two. Neither result asserts the explicit formula (4) for general p.

Our first result shows that this lower bound is not the sharp constant in general.

Theorem 1 

(Exact counterexample). For $m=n=2$ and $p=3/2$, there are real rank-one matrices $A_1,A_2$ such that

$${\displaystyle \frac{\parallel A_1+A_2{\parallel }_{3/2}}{\parallel |A_1|+|A_2{|\parallel }_{3/2}}}>{\displaystyle \frac{207}{200}}>C_{3/2,2}^{\mathrm{TZ}}.$$

Consequently, the Tang–Zhang conjecture is false.

The failure occurs inside the rank-one class, but on the opposite side of the Hilbertian exponent from the natural positive result.

Theorem 2 

(Sharp rank-one bound). Let $m\ge 2$, $2\le p<\infty$, and let $A_1,\dots ,A_m\in {\mathbb{M}}_n\left(\mathbb{C}\right)$ have rank at most one and not all vanish. Then

$${∥{\displaystyle \sum _{k=1}^m}{A}_k∥}_p\le C_{p,m}^{\mathrm{TZ}}{∥{\displaystyle \sum _{k=1}^m}\left|A_k\right|∥}_p.$$

(5)

The constant is sharp in the dimension-free rank-one problem and is attained whenever $n\ge m$. Equality holds precisely as follows, up to common input and output unitaries, a common positive scale, and the harmless phase changes in rank-one factorizations:

$$A_k=ru{v}_k^{*},\langle v_j,v_k\rangle =\left\{\begin{array}{cc}1,\hfill & j=k,\hfill \\ s_{p,m},\hfill & j\ne k,\hfill \end{array}\right.$$

where $r>0$, u is a unit vector, and

$$s_{p,m}={\displaystyle \frac{x_{p,m}-1}{x_{p,m}+m-1}}.$$

In particular, equality requires $n\ge m$.

For $p=\infty$, the same rank-one argument gives the sharp constant $\sqrt{m}$, with the right vectors orthonormal. Our third result leaves the rank-one restriction entirely.

Theorem 3 

(The full case $m=2,p=4$). Let $A,B\in {\mathbb{M}}_n\left(\mathbb{C}\right)$, and let $x>1$ be the solution of

$$x^4-2x-1=0.$$

Then

$${\parallel A+B\parallel }_4^4\le {\displaystyle \frac{x^2(x+1)}{2}}\parallel \left|A\right|+{\left|B\right|\parallel }_4^4.$$

(6)

The constant is sharp in the dimension-free problem, is attained for every $n\ge 2$, and equals ${\left(C_{4,2}^{\mathrm{TZ}}\right)}^4$.

The paper is organized as follows. Section 2 gives the exact counterexample and its rational certificate. Section 3 proves the sharp rank-one theorem and its equality statement. Section 4 proves Theorem 3.

2. An Exact $2\times 2$ Counterexample

Set

$$e=\left(\begin{array}{c}1\\ 0\end{array}\right),u=\left(\begin{array}{c}39/40\\ \sqrt{79}/40\end{array}\right),v=\left(\begin{array}{c}5/8\\ \sqrt{39}/8\end{array}\right).$$

These are real unit vectors. Define

$$A_1=e{e}^T=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),A_2=u{v}^T=\left(\begin{array}{cc}39/64& 39\sqrt{39}/320\\ \sqrt{79}/64& \sqrt{3081}/320\end{array}\right).$$

(7)

Both matrices have rank one and unique nonzero singular value equal to one. Consequently,

$$|A_1|=e{e}^T,\left|A_2\right|=v{v}^T.$$

Proof 

(Proof of Theorem 1). Let $U=(e,u)$ and $V=(e,v)$. Then $A_1+A_2=U{V}^T$, and the two Gram matrices are

$$L=U^{T}U=\left(\begin{array}{cc}1& 39/40\\ 39/40& 1\end{array}\right),G=V^{T}V=\left(\begin{array}{cc}1& 5/8\\ 5/8& 1\end{array}\right).$$

The squared singular values of $U{V}^T$ are the eigenvalues of $LG$. The two Gram matrices are simultaneously diagonalized by ${(1,1)}^T$ and ${(1,-1)}^T$, so those squared singular values are

$$(1+39/40)(1+5/8)={\displaystyle \frac{1027}{320}},(1-39/40)(1-5/8)={\displaystyle \frac{3}{320}}.$$

(8)

On the other hand, the eigenvalues of $|A_1|+|A_2|=V{V}^T$ are

$${\displaystyle \frac{13}{8}},{\displaystyle \frac{3}{8}}.$$

(9)

Thus, if

$$R={\displaystyle \frac{\parallel A_1+A_2{\parallel }_{3/2}}{\parallel |A_1|+|A_2{|\parallel }_{3/2}}},$$

then

$$R^{3/2}={\displaystyle \frac{{(1027/320)}^{3/4}+{(3/320)}^{3/4}}{{(13/8)}^{3/2}+{(3/8)}^{3/2}}}.$$

(10)

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

$$\begin{array}{cc}\hfill {\left({\displaystyle \frac{1027}{320}}\right)}^3-{\left({\displaystyle \frac{11989}{5000}}\right)}^4& ={\displaystyle \frac{29144230879351}{40000000000000000}}>0,\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill {\left({\displaystyle \frac{3}{320}}\right)}^3-{\left({\displaystyle \frac{301}{10000}}\right)}^4& ={\displaystyle \frac{124819571}{40000000000000000}}>0,\hfill \end{array}$$

(12)

$$\begin{array}{cc}\hfill {\left({\displaystyle \frac{4143}{2000}}\right)}^2-{\left({\displaystyle \frac{13}{8}}\right)}^3& ={\displaystyle \frac{773}{8000000}}>0,\hfill \end{array}$$

(13)

$$\begin{array}{cc}\hfill {\left({\displaystyle \frac{2297}{10000}}\right)}^2-{\left({\displaystyle \frac{3}{8}}\right)}^3& ={\displaystyle \frac{5543}{200000000}}>0.\hfill \end{array}$$

(14)

Put

$$A_0={\displaystyle \frac{11989}{5000}}+{\displaystyle \frac{301}{10000}}={\displaystyle \frac{24279}{10000}},B_0={\displaystyle \frac{4143}{2000}}+{\displaystyle \frac{2297}{10000}}={\displaystyle \frac{5753}{2500}},$$

and $q=207/200$. A final exact comparison gives

$$A_0^2-q^3{B}_0^2={\displaystyle \frac{1172956601313}{50000000000000}}>0.$$

(15)

Equations (11)–(15) imply

$$R^{3/2}>{\displaystyle \frac{A_0}{B_0}}>q^{3/2},$$

and hence $R>q$.

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 t of

$$h\left(t\right)=t^3-2t^2-1=0$$

(16)

lies above $4/3$, where h is strictly increasing. Moreover,

$$h\left({\displaystyle \frac{1103}{500}}\right)={\displaystyle \frac{310727}{125000000}}>0,$$

(17)

so $t<T:=1103/500$.

Writing $C=C_{3/2,2}^{\mathrm{TZ}}$ and using $t^3=2t^2+1$, we obtain

$$C^6={\displaystyle \frac{t^6}{16(t^2+1)}}=:H\left(t\right).$$

The function H is strictly increasing for $t>0$, since

$$H^{\prime }\left(t\right)={\displaystyle \frac{t^5(2t^2+3)}{8{(t^2+1)}^2}}>0.$$

The remaining exact difference is

$$q^6-H\left(T\right)={\displaystyle \frac{26731151399029597}{18772595200000000000}}>0.$$

(18)

It follows that $C^6<H\left(T\right)<q^6$, hence $C<q<R$. □

Remark 1. 

Numerically,

$$R=1.0364136587048904\dots ,C_{3/2,2}^{\mathrm{TZ}}=1.0346539518514341\dots .$$

These decimals play no role in the proof.

3. The Sharp Rank-One Problem for $p\ge 2$

We now prove Theorem 2. The proof reduces the matrix problem to a single scalar variable.

Proof 

(Proof of Theorem 2). Write

$$A_k=r_k{u}_k{v}_k^{*},r_k\ge 0,$$

where $u_k,v_k$ are unit vectors whenever $r_k>0$. Form the column matrices

$$U=(\sqrt{r_1}{u}_1,\dots ,\sqrt{r_m}{u}_m),V=(\sqrt{r_1}{v}_1,\dots ,\sqrt{r_m}{v}_m)$$

and their Gram matrices

$$L=U^{*}U,G=V^{*}V.$$

Then

$$\sum _k{A}_k=U{V}^{*},\sum _k\left|A_k\right|=V{V}^{*},$$

(19)

and

$$diagL=diagG=(r_1,\dots ,r_m),TrL=TrG=:R>0.$$

(20)

Let

$$K=L^{1/2}G{L}^{1/2}.$$

The nonzero eigenvalues of K are the squared singular values of $U{V}^{*}$; the nonzero eigenvalues of G are the eigenvalues of $V{V}^{*}$. Therefore

$${∥\sum _k{A}_k∥}_p^p=Tr{K}^{p/2},{∥\sum _k\left|A_k\right|∥}_p^p=Tr{G}^p.$$

(21)

Since $p/2\ge 1$, the nonnegative eigenvalues of K give

$$Tr{K}^{p/2}\le {(TrK)}^{p/2}.$$

(22)

If $\lambda ={\lambda }_{max}\left(G\right)$, then

$$TrK=Tr\left(LG\right)\le \lambda TrL=R\lambda .$$

(23)

List the m eigenvalues of G, including zeros, as $\lambda ={\lambda }_1\ge {\lambda }_2\ge \dots \ge {\lambda }_m\ge 0$. Convexity gives

$$Tr{G}^p\ge {\lambda }^p+{\displaystyle \frac{{(R-\lambda )}^p}{{(m-1)}^{p-1}}}.$$

(24)

If $\lambda =R$, then (21)–(23) give a ratio at most one, which is strictly smaller than the desired sharp constant. Suppose henceforth that $\lambda <R$, and put

$$y={\displaystyle \frac{(m-1)\lambda }{R-\lambda }}.$$

Since $\lambda \ge R/m$, one has $y\ge 1$. Combining (21)–(24) yields

$${\displaystyle \frac{\parallel {\sum }_k{A}_k{\parallel }_p}{\parallel {\sum }_k|A_k{|\parallel }_p}}\le F_{p,m}\left(y\right):={\displaystyle \frac{\sqrt{y(y+m-1)}}{{(y^p+m-1)}^{1/p}}}.$$

(25)

Direct differentiation gives

$${\displaystyle \frac{d}{dy}}log{F}_{p,m}\left(y\right)={\displaystyle \frac{(m-1)(2y+m-1-y^p)}{2y(y+m-1)(y^p+m-1)}}.$$

For $p\ge 2$, the function $y^p-2y-(m-1)$ is strictly increasing on $[1,\infty )$, apart from an inessential zero derivative at the left endpoint when $p=2$. It has exactly one zero $x_{p,m}>1$. Thus $F_{p,m}$ has a unique maximum at $x_{p,m}$, and (5) follows from (4).

We next track equality. Put

$$\beta ={\displaystyle \frac{R-\lambda }{m-1}}.$$

Equality in the scalar maximization and in (24) forces

$$\lambda =x_{p,m}\beta ,spec\left(G\right)=\{\lambda ,\beta ,\dots ,\beta \}.$$

The top eigenspace is one-dimensional. Equality in (23) forces the range of L into this eigenspace. If w is its unit eigenvector, then

$$L=Rw{w}^{*},G=\beta I+(\lambda -\beta )w{w}^{*}.$$

(26)

The common diagonal condition (20) now gives

$$R|w_k{|}^2=\beta +(\lambda -\beta ){\left|w_k\right|}^2.$$

Because $R=\lambda +(m-1)\beta$, it follows that

$$|w_k{|}^2={\displaystyle \frac{1}{m}},r_k={\displaystyle \frac{R}{m}}(1\le k\le m).$$

After simultaneous phase changes in the factorizations of the $A_k$, we may take $w=m^{-1/2}{(1,\dots ,1)}^T$. Equation (26) then says that the $u_k$ coincide and

$$\langle v_j,v_k\rangle ={\displaystyle \frac{\lambda -\beta }{R}}={\displaystyle \frac{x_{p,m}-1}{x_{p,m}+m-1}}(j\ne k).$$

Conversely, this family takes equality at every step. The displayed Gram matrix is positive definite, so equality requires $n\ge m$. □

Corollary 1 

(Rank-one endpoint). If $A_1,\dots ,A_m$ have rank at most one, then

$${∥\sum _k{A}_k∥}_{\infty }\le \sqrt{m}{∥\sum _k\left|A_k\right|∥}_{\infty }.$$

The constant is sharp in the dimension-free sense. Equality is possible only when $n\ge m$, and then holds precisely for equal nonzero singular values, a common one-dimensional range, and pairwise orthogonal right vectors, modulo the same unitary and phase symmetries as above.

Proof. 

Use the notation in the preceding proof and set $\lambda ={\lambda }_{max}\left(G\right)$. Then

$$\parallel U{V}^{*}{\parallel }_{\infty }\le {\parallel U\parallel }_{\infty }{\parallel V^{*}\parallel }_{\infty }=\sqrt{{\parallel L\parallel }_{\infty }\lambda }\le \sqrt{R\lambda }.$$

Since $R=TrG\le m\lambda$, while $\parallel V{V}^{*}{\parallel }_{\infty }=\lambda$, the asserted inequality follows. Equality requires L to have rank one and $G=\lambda I_m$. The common diagonal condition then gives exactly the stated configuration. □

4. The Full $m=2,p=4$ Problem

Proof 

(Proof of Theorem 3). Put $H=\left|A\right|$ and $K=\left|B\right|$. Extend the partial isometries in the polar decompositions to unitaries, and write $A=UH,B=VK$. With $W=U^{*}V$, unitary invariance reduces the numerator to

$$S=H+WK.$$

Set

$$a=Tr{H}^4,b=Tr{K}^4,d=Tr{H}^2{K}^2.$$

The Hilbert–Schmidt triangle inequality applied to

$$S{S}^{*}=H^2+HK{W}^{*}+WKH+W{K}^2{W}^{*}$$

gives

$${\parallel S\parallel }_4^4\le {(\sqrt{a}+\sqrt{b}+2\sqrt{d})}^2.$$

(27)

Introduce

$$e=Tr{H}^{3}K,f=TrH{K}^3,g=TrHKHK.$$

Cyclically collecting all words in the noncommutative expansion gives

$$Tr{(H+K)}^4=a+b+4(e+f)+4d+2g.$$

(28)

Two lower bounds are needed. First,

$$e+f-2d=Tr\left((H-K)H(H-K)K\right)={\parallel H^{1/2}(H-K)K^{1/2}\parallel }_2^2\ge 0.$$

(29)

Second, the three-factor Schatten Hölder inequality gives

$$\begin{array}{cc}\hfill \sqrt{d}={\parallel HK\parallel }_2& =\parallel H^{1/2}\left(H^{1/2}{K}^{1/2}\right)K^{1/2}{\parallel }_2\hfill \\ \hfill & \le \parallel H^{1/2}{\parallel }_8\parallel H^{1/2}{K}^{1/2}{\parallel }_4{\parallel K^{1/2}\parallel }_8=a^{1/8}{g}^{1/4}{b}^{1/8}.\hfill \end{array}$$

Thus, when $ab>0$,

$$g\ge {\displaystyle \frac{d^2}{\sqrt{ab}}}.$$

(30)

If $ab=0$, one of $H,K$ vanishes and the theorem is immediate. Combining (28)–(30), we obtain

$$Tr{(H+K)}^4\ge a+b+12d+{\displaystyle \frac{2d^2}{\sqrt{ab}}}.$$

(31)

Let $A_0=\sqrt{a},B_0=\sqrt{b}$, and define

$$z={\displaystyle \frac{A_0+B_0}{\sqrt{A_0{B}_0}}}\ge 2,s=\sqrt{{\displaystyle \frac{d}{A_0{B}_0}}}\in [0,1].$$

The upper bound on s is the Hilbert–Schmidt Cauchy–Schwarz inequality $d\le \sqrt{ab}$. From (27) and (31),

$${\displaystyle \frac{{\parallel A+B\parallel }_4^4}{\parallel \left|A\right|+{\left|B\right|\parallel }_4^4}}\le \Phi (z,s):={\displaystyle \frac{{(z+2s)}^2}{z^2-2+12s^2+2s^4}}.$$

(32)

If $s\ge 1/2$, then

$$2(z^2-2+12s^2+2s^4)-{(z+2s)}^2={(z-2s)}^2+16s^2+4s^4-4\ge 0,$$

so $\Phi (z,s)\le 2$.

Suppose $0\le s\le 1/2$. The sign of $\partial \Phi /\partial z$ is the sign of

$$s^4+6s^2-zs-1.$$

Since $z\ge 2$,

$$s^4+6s^2-zs-1\le s^4+6s^2-2s-1\le 2s-1\le 0;$$

the middle inequality follows from $s^3+6s-4\le 1/8+3-4<0$. Consequently,

$$\Phi (z,s)\le \Phi (2,s)=f\left(s\right):={\displaystyle \frac{2{(1+s)}^2}{1+6s^2+s^4}}.$$

(33)

The sign of $f^{\prime }\left(s\right)$ is the sign of

$$1-6s-2s^3-s^4.$$

Hence f has a unique maximizer $s_0\in (0,1/2)$, characterized by

$$s_0^4+2s_0^3+6s_0-1=0.$$

(34)

The first branch cannot dominate, because $f(1/6)>2$.

Set $x=(1+s_0)/(1-s_0)$. A direct calculation gives

$$(x^4-2x-1){(1-s_0)}^4=2(s_0^4+2s_0^3+6s_0-1)=0,$$

and

$$f\left(s_0\right)={\displaystyle \frac{x^2{(x+1)}^2}{x^4+1}}={\displaystyle \frac{x^2(x+1)}{2}},$$

(35)

where the last equality uses $x^4+1=2(x+1)$. Equations (32)–(35) 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

$$A=u{r}_1^{*},B=u{r}_2^{*}.$$

Then

$${\parallel A+B\parallel }_4^4=4{(1+s_0)}^2$$

and

$$\parallel \left|A\right|+{\left|B\right|\parallel }_4^4={(1+s_0)}^4+{(1-s_0)}^4=2(1+6s_0^2+s_0^4).$$

Their ratio is $f\left(s_0\right)$, proving sharpness. □

Remark 2. 

Orthogonal direct sums of the two-dimensional extremal block give higher-dimensional equality examples. The proof above does not attempt a complete classification of all equality cases for Theorem 3.