Sums of positive definite matrices are still positive definite

For a matrix \(A\), the notation \(A\succeq0\) and \(A\succ0\) when \(A\) is often used to denote positive semi-definite (PSD) or positive definite (PD) matrices, respectively. Using the definition of a PD matrix, we can prove that the sum of two PD matrices is also PD. A very similar approach can be used to prove the sum of two PSD matrices is also PSD.

\(\textbf{Proof:}\) Let both \(A\) and \(B\) be \(d \times d\) matrices. Also, let \(A \succ 0\) and \(B \succ 0\). Then by the definition of positive definite matrices, \(\textbf{x}^T A \textbf{x} > 0\) and \(\textbf{x}^T B \textbf{x} > 0\) for all non-zero vectors \(\textbf{x}\).

Now, notice that \[\textbf{x}^T (A + B) \textbf{x} = \textbf{x}^T A \textbf{x} + \textbf{x}^T B \textbf{x}\] by the matrix distributive property. Since we know that both \(\textbf{x}^T A \textbf{x} > 0\) and \(\textbf{x}^T B \textbf{x} > 0\), and that the sum of two positive numbers is also positive. Then, \[\textbf{x}^T A \textbf{x} + \textbf{x}^T B \textbf{x} > 0.\] This implies that \[\textbf{x}^T (A + B) \textbf{x} > 0.\] Thus, the sum of two positive definite matrices \(A\) and \(B\) is positive definite. \(\blacksquare\)

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Eric McKinney
PhD Student in Statistics
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