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)

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

1 Answer 1


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 $$ \tr(A_1+A_2+A_3+A_4)=\sum_{k=1}^{n/2}\frac4n=2. $$


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