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Let $M_A : C \oplus A \,\tilde{\to} \, C \oplus A$ and $M_B : C \oplus B \,\tilde{\to} \, C \oplus B$ be two symmetric positive definite matrices whose domains share a common subspace $C$.

You can extend them to act on the same space (namely $C \oplus A \oplus B$) by "filling up with zeros" like this (not sure about the notation):

$\tilde{M}_A = M_A \oplus 0_{B \times B}$ and $\tilde{M}_B = M_B \oplus 0_{A \times A}$

Question: Is their sum $\tilde{M}_A + \tilde{M}_B$ also always positive definite?

What I've achieved so far:

  • I know that it's at least positive semi-definite:

    $v^T M_{A/B} \, v \geq 0 \implies v^T \tilde{M}_{A/B} \, v \geq 0$

    and therefore also

    $v^T (\tilde{M}_A + \tilde{M}_B) \, v = v^T M_A \, v + v^T M_B \, v \geq 0$.

  • If it's just about definiteness than in this case it's just about wether or not the sum is invertible. I'm not sure wether this helps.

Please consider in your answer that I'm not a mathematician and only know basic algebra. I hope the question as stated here is well defined (I'm still a bit unfamiliar with (direct) sums). Also feel free to suggest a rephrasing and a better title (this one is horrible).

tobias_t
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  • your title and question body don't match – FShrike Jun 20 '23 at 20:48
  • See https://math.stackexchange.com/questions/544139/sum-of-positive-definite-matrices-still-positive-definite. You need a stronger assumption than just invertibility (so the answer to the question in your title is "no": e.g., take $A = B = 0$ and $M_B = - M_A$). – Rob Arthan Jun 20 '23 at 21:05

1 Answers1

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Yes. Let $v \in A \oplus B \oplus C$. Then, $v = a + b + c$ with $a \in A$, $b \in B$, $c \in C$ (and they are unique). Therefore, if $\tilde{M} = \tilde{M}_A + \tilde{M}_B$, \begin{align*} v^\top\tilde{M}v & = (a + b + c)^\top(\tilde{M}_A + \tilde{M}_B)(a + b + c)\\ & = (a + b + c)^\top\tilde{M}_A(a + b + c) + (a + b + c)^\top\tilde{M}_B(a + b + c)\\ & = (a + c)^\top\tilde{M}_A(a + c) + (b + c)^\top\tilde{M}_B(b + c). \end{align*} Each of the term of this sum is non-negative because each $\tilde{M}_{\cdot}$ is positive semi-definite and this sum is zero if and only if each term is zero, if and only if $a + c = 0$ and $b + c = 0$ because each $\tilde{M}_{\cdot}$ is definite. It implies that $a = b = c = 0$ hence $v = 0$. $\tilde{M}$ is indeed positive definite.

Cactus
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