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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.

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

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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 $x^TKx=x^T\begin{pmatrix} K_1x_1\\K_2x_2\end{pmatrix}=x_1^TK_1x_1+x_2^TK_2x_2>0$.

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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.

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