# Proof with symmetric positive definite matrices

Prove that if $K_1$ and $K_2$ are positive definite $n × n$ matrices, then $$K = \begin{pmatrix}K_1& 0\\ 0 &K2\end{pmatrix}$$ is a positive definite $2n × 2n$ matrix. Is the converse true?

I know that since $K_1, K_2$ are both SPD, then they're non-singular and regular, so $K$ should be composed of linearly independent columns which implies that it is SPD.

I'm having trouble writing it formally.

• What do you mean by "regular?" Nov 21, 2014 at 3:27
• Just work it out by the definition of SPD $x^TKx > 0, \forall x \in \mathbb R^n$. Nov 21, 2014 at 3:30

It's not true that any matrix composed of linearly independent columns is symmetric and positive definite. Just consider $\begin{pmatrix} 1&0\\1&1\end{pmatrix}$. Instead, observe that $K$ is symmetric because $K^T=\begin{pmatrix} K^T_1&0\\0&K^T_2\end{pmatrix}=K$. And for positivity, if $x=(x_1,x_2)$ where the $x_i$ are of length $n$, then $x^TKx=x^T\begin{pmatrix} K_1x_1\\K_2x_2\end{pmatrix}=x_1^TK_1x_1+x_2^TK_2x_2>0$.
Hint: Use the definition that $A$ is (symmetric) positive definite iff it is symmetric and $$x^TAx > 0 \quad \forall x \in \Bbb R^n$$ then note that $x$ can be written as a "block vector" $\pmatrix{x_1\\x_2}$. The converse is indeed true: the matrix $$\pmatrix{K_1\\&K_2}$$ is symmetric positive-definite if and only if each $K_i$ is symmetric positive definite.