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? 

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 $K_{2, 2}$
