How to show this matrix is positive semidefinite? Let 
$$K=\begin{pmatrix}
K_{11} & K_{12}\\ 
K_{21} & K_{22}
\end{pmatrix}$$ 
be a symmetric positive semidefinite real matrix (PSD) with $K_{12}=K_{21}^T$. Then, for $|r| \le 1$, 
$$K^*=\begin{pmatrix}
K_{11} & rK_{12}\\ 
rK_{21} & K_{22}
\end{pmatrix}$$
is also a PSD matrix. Matrices $K$ and $K^*$ are $2 \times 2$ and $K_{21}^T$ denotes the transpose matrix. How do I prove this? 
 A: There is already a great answer by @whuber, so I will try to give an alternative, shorter proof, using a couple theorems.


*

*For any $A$ - PSD and any $Q$ we have $Q^TAQ$ - PSD

*For $A$ - PSD and $B$ - PSD also $A + B$ - PSD

*For $A$ - PSD and $q > 0$ also $qA$ - PSD


And now:
\begin{align*} 
K^* &= 
\begin{pmatrix}
  K_{1,1} & rK_{1,2} \\
  rK_{2,1} & K_{2,2} \\
 \end{pmatrix} \\
&= 
\begin{pmatrix}
  K_{1,1} & rK_{1,2} \\
  rK_{2,1} & r^2K_{2,2} \\
 \end{pmatrix} 
+
\begin{pmatrix}
  0 & 0 \\
  0 & qK_{2,2} \\
 \end{pmatrix}, \text{ where $q = 1 - r^2 > 0$} \\
&= 
\begin{pmatrix}
  I & 0 \\
  0 & rI \\
 \end{pmatrix}^T
\begin{pmatrix}
  K_{1,1} & K_{1,2} \\
  K_{2,1} & K_{2,2} \\
 \end{pmatrix} 
\begin{pmatrix}
  I & 0 \\
  0 & rI \\
 \end{pmatrix} 
+
q\begin{pmatrix}
  0 & 0 \\
  0 & K_{2,2} \\
 \end{pmatrix}
\end{align*}
Matrix $K$ is PSD by definition and so is its submatrix $K_{2, 2}$
A: This is a nice opportunity to apply the definitions: no advanced theorems are needed.
To simplify the notation, for any number $\rho$ let $$\mathbb{A}(\rho)=\pmatrix{A&\rho B\\\rho B^\prime&D}$$ be a symmetric block matrix. 
 (If working with block matrices is unfamiliar to you, just assume at first that $A$, $B$, $D$, $x$, and $y$ are numbers. You will get the general idea from this case.)
For $\mathbb{A}(\rho)$ to be positive semidefinite (PSD) merely means that for all vectors $x$ and $y$ of suitable dimensions
$$\eqalign{
0 &\le \pmatrix{x^\prime&y^\prime} \mathbb{A}(\rho) \pmatrix{x\\y} \\
&= \pmatrix{x^\prime&y^\prime} \pmatrix{A&\rho B\\\rho B^\prime&D}\pmatrix{x\\y} \\
&=x^\prime A x + 2\rho y^\prime B^\prime x + y^\prime D y.\tag{1}
}$$
This is what we have to prove when $|\rho|\le 1$.
We are told that $\mathbb{A}(1)$ is PSD.  I claim that $\mathbb{A}(-1)$ also is PSD.  This follows by negating $y$ in expression $(1)$: as $\pmatrix{x\\y}$ ranges through all possible vectors, $\pmatrix{x\\-y}$ also ranges through all possible vectors, producing
$$\eqalign{
0 &\le \pmatrix{x^\prime&-y^\prime}\mathbb{A}(1)\pmatrix{x\\-y} \\
&= x^\prime A x + 2(-y)^\prime B^\prime x + (-y)^\prime D (-y) \\
&= x^\prime A x + 2(-1)y^\prime B^\prime x + y^\prime D y \\
&= \pmatrix{x^\prime&y^\prime}\mathbb{A}(-1)\pmatrix{x\\y},
}$$
showing that $(1)$ holds with $\rho=-1.$
Notice that $\mathbb{A}(\rho)$ can be expressed as a linear interpolant of the extremes $\mathbb{A}(-1)$ and $\mathbb{A}(1)$:
$$\mathbb{A}(\rho) = \frac{1-\rho}{2}\mathbb{A}(-1) + \frac{1+\rho}{2}\mathbb{A}(1).\tag{2}$$
When $|\rho|\le 1$, both coefficients $\color{blue}{(1-\rho)/2}$ and $\color{blue}{(1+\rho)/2}$ are non-negative.  Therefore, since both ${\pmatrix{x^\prime&y^\prime}\mathbb{A}(1)\pmatrix{x\\y}}$ and $\pmatrix{x^\prime&y^\prime}\mathbb{A}(-1)\pmatrix{x\\y}$ are nonnegative, so is the right hand side of
$$\eqalign{
&\pmatrix{x^\prime&y^\prime}\mathbb{A}(\rho)\pmatrix{x\\y} \\ 
&= \color{blue}{\left(\frac{1-\rho}{2}\right)}\pmatrix{x^\prime&y^\prime}\mathbb{A}(-1)\pmatrix{x\\y} + \color{blue}{\left(\frac{1+\rho}{2}\right)}\pmatrix{x^\prime&y^\prime}\mathbb{A}(1)\pmatrix{x\\y} \\
&\ge \color{blue}{0}(0) + \color{blue}{0}(0) = 0.
}$$
(I use colors to help you see the four separate non-negative terms that are involved.)
Because $x$ and $y$ are arbitrary, we have proven $(1)$ for all $\rho$ with $|\rho|\le 1$.
