lajos/General trace inequalities

We fix a positive integer \(s\ge 2\) and apply the following notation: if \(A\) is an \(s\)-by-\(s\) matrix, then \(\tau:=\)tr\((A)\) and \(\tau_k:=\)tr\((A^k)\) for any integer \(k\ge 2\).

Proposition. Fix a positive integer \(n\ge 2\) and suppose that \(A\) is an \(s\)-by-\(s\) (componentwise) non-negative matrix. Then

\[ s^{n-1} \tau_n \ge \tau^{n}. \] Moreover, in the \(n=2\) case equality holds if and only if \(a_{k,k}=a_{1,1}\) for \(k=2,3,\ldots,s\) and \(a_{i,j}\cdot a_{j,i}=0\) for all \(i\neq j\), further, if \(n\ge 3\) and \(s^{n-1} \tau_n = \tau^{n}\), then \(a_{k,k}=a_{1,1}\) for \(k=2,3,\ldots,s\).

Proof. If \(A\) and \(B\) are non-negative \(s\)-by-\(s\) matrices, then the element of \(AB\) in the \(k^\mathrm{th}\) row and \(k^\mathrm{th}\) column is \(\ge a_{k,k} b_{k,k}\). (Mathwiki TeX rendering error: I could not enter the expression parenthesis AB parenthesis subscript something.) Applying this recursively, we get that tr(\(A^n)\ge \sum_{k=1}^s a_{k,k}^n\). Then the inequality between the \(n^\mathrm{th}\) power mean and the arithmetic mean shows that

\[ \sqrt[n]{\frac{\mathrm{tr}(A^n)}{s}}\ge \sqrt[n]{\frac{\sum_{k=1}^s a_{k,k}^n}{s}}\ge \frac{\sum_{k=1}^s a_{k,k}}{s}=\frac{\mathrm{tr}(A)}{s}, \] proving the trace inequalities.

Now if \(n\ge 3\) and \(s^{n-1} \tau_n = \tau^{n}\), then we have equality in the power mean inequality, which implies \(a_{k,k}=a_{1,1}\) for \(k=2,3,\ldots,s\).

Finally, suppose that \(n=2\). Then \(s \tau_2 = \tau^{2}\) holds if and only if we have equality in the power mean inequality and tr(\(A^2)= \sum_{k=1}^s a_{k,k}^2\). But the former holds if and only if \(a_{k,k}=a_{1,1}\) for \(k=2,3,\ldots,s\), while the latter holds if and only if \(a_{i,j}\cdot a_{j,i}=0\) for all \(i\neq j\).

Some remarks. (Constant \(n\ge 2\) is again a fixed positive integer.)

\(-\) If \(s=2\), \(n=3\), \(a_{1,1}=a_{2,2}=0\), \(a_{1,2}=a_{2,1}=1\), then \(s^{n-1} \tau_n = \tau^{n}\), but \(a_{1,2}\cdot a_{2,1}\ne 0\).

\(-\) The constant \(s^{n-1}\) is the best possible (as shown by \(A=I\)).

\(-\) We can not expect a “converse” trace inequality, since for the matrix \(A\) with 0’s in the diagonal and with all other entries 1, we have tr\(^2(A)=0\) and tr\((A^2)>0\).

\(-\) From the spectral mapping theorem we know that \[ \textrm{tr}(A^n)=\sum_{k}\lambda_k^n. \]

\(-\) If spr denotes the spectral radius and \(A\ge 0\), then

\[ 0\le \textrm{tr}(A^n)=\sum_{k}\lambda_k^n=|\sum_{k}\lambda_k^n|\le \sum_{k}|\lambda_k|^n \le s\cdot \textrm{spr}(A)^n. \]

The constant \(s\) is again the best possible.

\(-\) Notice that for the conjectured optimal \(A\) matrix, we have equality in all trace inequalities as well as in the spectral radius inequality above.