Finite matrix trace-norm ideal inequalities

Diagonalize $X^{\ast }X$ by [F1], with nonnegative eigenvalues $s_j^2$ and orthonormal eigenvectors $v_j$. For each $s_j>0$ put $u_j=X{v}_j\mathrm{/}{s}_j$; then $⟨u_i,u_j⟩={\delta }_{ij}$ because $⟨X{v}_i,X{v}_j⟩=⟨v_i,X^{\ast }X{v}_j⟩$. Extending both finite orthonormal families to bases gives the singular-value expansion $X={\sum }_{s_j>0}{s}_j{R}_{u_j,v_j}$, where $R_{u,v}x=⟨x,v⟩u$. The only nonzero singular value of $R_{u,v}$ is $\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$, so $\mathrm{\parallel }{R}_{u,v}{\mathrm{\parallel }}_1=\mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }$ and $\mathrm{\parallel }X{\mathrm{\parallel }}_1={\sum }_{s_j>0}{s}_j$.

The rank-one trace formula is $\mathrm{Tr}(R_{u,v})=⟨u,v⟩$, so the expansion in step 1.1 gives $\mathrm{\parallel }X{\mathrm{\parallel }}_1={\sup }_{U\text{ unitary}}\mathrm{\mid }\mathrm{Tr}(UX)\mathrm{\mid }$. For the upper bound, if $u_j,v_j$ are unit vectors, then $\mathrm{\parallel }U{u}_j-⟨U{u}_j,v_j⟩v_j{\mathrm{\parallel }}^2=1-\mathrm{\mid }⟨U{u}_j,v_j⟩{\mathrm{\mid }}^2\ge 0$, so $\mathrm{\mid }⟨U{u}_j,v_j⟩\mathrm{\mid }\le 1$. Equality is attained by a unitary mapping each $u_j$ to $v_j$, extended from these finite orthonormal families to bases. The supremum formula implies the triangle inequality $\mathrm{\parallel }X+Y{\mathrm{\parallel }}_1\le \mathrm{\parallel }X{\mathrm{\parallel }}_1+\mathrm{\parallel }Y{\mathrm{\parallel }}_1$ and homogeneity; positivity and definiteness follow from the singular values.

Applying step 1.1 to $YX={\sum }_j{s}_j{R}_{Y{u}_j,v_j}$ and using the triangle inequality of step 2.1 gives $\mathrm{\parallel }YX{\mathrm{\parallel }}_1\le {\sum }_j{s}_j\mathrm{\parallel }Y{u}_j\mathrm{\parallel }\mathrm{\parallel }{v}_j\mathrm{\parallel }\le \mathrm{\parallel }Y\mathrm{\parallel }{\sum }_j{s}_j=\mathrm{\parallel }Y\mathrm{\parallel }\mathrm{\parallel }X{\mathrm{\parallel }}_1$. Similarly $XY={\sum }_j{s}_j{R}_{u_j,Y^{\ast }{v}_j}$ gives $\mathrm{\parallel }XY{\mathrm{\parallel }}_1\le \mathrm{\parallel }Y\mathrm{\parallel }\mathrm{\parallel }X{\mathrm{\parallel }}_1$. The rank-one trace formula and step 1.1 also give $\mathrm{\mid }\mathrm{Tr}X\mathrm{\mid }\le {\sum }_j{s}_j\mathrm{\mid }⟨u_j,v_j⟩\mathrm{\mid }\le \mathrm{\parallel }X{\mathrm{\parallel }}_1$.

Define the Frobenius norm by $\mathrm{\parallel }X{\mathrm{\parallel }}_F=({\sum }_{i,j}\mathrm{\mid }{X}_{ij}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$ in an orthonormal basis. The finite singular-value expansion gives $\mathrm{\parallel }X{\mathrm{\parallel }}_F=({\sum }_j{s}_j^2{)}^{1\mathrm{/}2}$. Since $s_j\ge 0$, $({\sum }_j{s}_j^2{)}^{1\mathrm{/}2}\le {\sum }_j{s}_j$; also $d{\sum }_j{s}_j^2-({\sum }_j{s}_j{)}^2={\sum }_{i<j}(s_i-s_j{)}^2\ge 0$. Thus $\mathrm{\parallel }X{\mathrm{\parallel }}_F\le \mathrm{\parallel }X{\mathrm{\parallel }}_1\le \sqrt{d}\mathrm{\parallel }X{\mathrm{\parallel }}_F$. A trace-norm Cauchy sequence is therefore Cauchy in each matrix coordinate and converges to a matrix; the second inequality gives convergence in trace norm. This proves completeness. For unitary $U$, step 3.1 gives $\mathrm{\parallel }UX{\mathrm{\parallel }}_1\le \mathrm{\parallel }X{\mathrm{\parallel }}_1$; applying it to $U^{\ast }$ and $UX$ gives the reverse inequality. The same argument on the right proves $\mathrm{\parallel }UX{U}^{\ast }{\mathrm{\parallel }}_1=\mathrm{\parallel }X{\mathrm{\parallel }}_1$.