How to show the following matrix is positive definite? How to show the following matrix is positive definite.
\begin{equation}
 \sum_{i=1}^n \Big[(d_i^Tp)^2\left\{\left(
 \begin{array}{c}
   d_i\\
   A_ip
 \end{array}
\right)
\left(
 \begin{array}{c}
   d_i\\
   A_ip
 \end{array}
\right)^T
- (d_i^Tp)
\left(
 \begin{array}{cc}
   0 & A_i^T\\
    A_i&0
 \end{array}
\right)
\right\}\Big],  
\end{equation}
where $A_i$ is a $k\times m$ matrix which has at least non-zero element in one row or column and at most one non-zero element in other rows, $p$ and $d_i$ are $m\times 1$ vectors with $p_j \in (0,1)$ for all $j=1,\ldots,m$, $p^T{\bf 1}=1$ and only one non-zero element in $d_i$.

You cannot, because it is not always positive definite. Counterexample: with $n=1$ (i.e. there is only one summand), $d_1=\pmatrix{1\\ 1},\,p=\frac12d_1$ and $A_1=\pmatrix{2&1\\ 1&1}$, the resulting matrix in the pair of the curly braces is
$$
B_1=\pmatrix{1&1&-\frac12&0\\ 1&1&\frac12&0\\ -\frac12&\frac12&\frac94&\frac32\\ 0&0&\frac32&1}.
$$
$B_1$ is real symmetric, but when $x=(1,-1,1,-1)^T$, we have $x^TB_1x=-\frac74<0$. Hence $B_1$ is not even positive semidefinite.
By perturbing $A_1$ by a little, we can also get an infinity family of distinct matrices $B_1,\,B_2,\,\ldots$ such that for the aforementioned vector $x$, each $x^TB_ix$ is negative. Hence we can obtain a counterexample for an arbitrary $n$.
