# Trace of off-diagonal blocks of a positive semidefinite matrix

Consider the matrix

$$A=\begin{pmatrix} A_1 & A_2 \\ A_3 & A_4 \end{pmatrix}$$

Let's suppose that $$A$$ is a real $$n\times n$$ positive semidefinite and satisfies $$\|A\|\leq 1$$, i.e., the largest eigenvalue is $$1$$. Let $$B=A_1+A_2+A_3+A_4$$. Let's assume all these $$A_i$$ matrices are of the same dimension.

a) Then is it true that $$\mbox{Tr}(B)\leq n$$ (clearly $$\mbox{Tr}(A_1) + \mbox{Tr}(A_4) \leq n$$, but what happens when we add $$A_2, A_3$$?

b) In particular, i'd like to understand how does the trace of the off-diagonal elements of a PSD matrix behave?

c) Can they be arbitrarily large compared to the main-diagonal submatrices? Does this matrix $$A$$ necessarily need to be diagonally dominant (i don't think so...but I'm not sure)

• Are the $A_i$ square matrices? Mar 30 at 6:53
• What if $n=3$? You should say that $A$ is $2n \times 2n$, as $n$ isn't necessarily even Mar 30 at 10:54
• yup they are square (in my post, i meant same dimension to emphasize they are square) Mar 30 at 14:13

I will assume that you are considering symmetric as part of the definition of positive semi-definite; otherwise there are easy counterexamples like $$A=\begin{bmatrix}1&2\\0&1\end{bmatrix}.$$
When $$A$$ is positive semi-definite and symmetric, consider first the $$2\times2$$ case. We can write $$A=Q^TDQ$$, where $$Q$$ is orthogonal and $$D$$ is diagonal with diagonal given by the eigenvalues $$\lambda,\mu$$ of $$A$$. The positivity of $$A$$ together with $$\|A\|≤1$$ give us $$0≤\lambda,\mu≤1$$. Writing $$Q=\begin{bmatrix}\cos t&\sin t\\-\sin t&\cos t\end{bmatrix}$$ for an appropriate $$t$$ we get, after calculating the entries of $$Q^TDQ$$, $$a_{11}+a_{12}+a_{21}+a_{22}=\lambda+\mu+(\lambda-\mu)\cos t\sin t.$$ Assuming without loss of generality that $$\lambda≥\mu$$, the crude estimate $$-1≤\cos t\sin t≤1$$ gives us $$2\mu\leq a_{11}+a_{12}+a_{21}+a_{22}≤2\lambda.$$ Thus $$\tag1 a_{11}+a_{12}+a_{21}+a_{22}\leq2.$$
When $$A$$ is $$n\times n$$, we can write$$\def\tr{\operatorname{Tr}}$$ $$\tr(A_1+A_2+A_3+A_4)=\sum_{k=1}^{n/2}(A_1)_{kk}+(A_2)_{kk}+(A_3)_{kk}(A_4)_{kk}.$$ The $$2\times2$$ matrix $$\begin{bmatrix}(A_1)_{kk}&(A_2)_{kk}\\ (A_3)_{kk}& (A_4)_{kk}\end{bmatrix}$$ is a principal submatrix of $$A$$ and hence positive semi-definite (and it's norm is at most $$1$$, seen either by Cauchy Interlacing or by noticing that the submatrix is of the form $$PAP$$ with $$P$$ an orthogonal projection and hence $$\|PAP\|≤\|A\|$$). Then, using $$(1)$$, $$\tag2 \tr(A_1+A_2+A_3+A_4)\leq\sum_{k=1}^{n/2}2=n.$$
As for the diagonal dominance, it is not true in general. What can be said is the following: when $$A$$ is positive semi-definite and symmetric, there exists $$X$$, such $$A=X^TX$$. This gives us, using Cauchy-Schwarz, the inequality $$|a_{kj}|≤a_{kk}^{1/2}a_{jj}^{1/2},$$ which comes from $$|a_{kj}|=\Big|\sum_gx_{gk}x_{gj}\Big|≤\Big(\sum_gx_{gk}^2\Big)^{1/2} \Big(\sum_gx_{gj}^2\Big)^{1/2}= a_{kk}^{1/2}a_{jj}^{1/2}.$$ Since $$a_{kk}^{1/2}a_{jj}^{1/2}\leq\frac{a_{kk}+a_{jj}}2,$$ we get that $$\tr(A_2)\leq\frac{\tr(A_1+A_4)}2.$$ The inequality doesn't look to be sharp, though, as the maximum in $$(2)$$ seems to be achieved when $$A$$ is diagonal. For example, when $$a_{kj}=\frac1n$$, we have $$\tr(A_1+A_2+A_3+A_4)=\sum_{k=1}^{n/2}\frac4n=2.$$