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I have a couple questions actually, couldn't get both of them in the title.

First suppose I have a positive definite matrix H and a positive semidefinite matrix D. I believe that H*D would be a positive semidefinite matrix. Is this correct?

Second suppose I have a vector $g$. If I were to multiply $g$ by its transpose so that I get a matrix and not a single value, would the resultant matrix be positive semidefinite? I believe that it will be.

I'm not looking for a proof (though it wouldn't be out of place for others looking for answers to similar questions) just yes or no answers and a brief explanation of why I am wrong, if am wrong. I am working on a proof and want to make sure that my intuition is correct.

HXSP1947
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    For the second question, use the fact that a real symmetric matrix $S$ is PSD iff $x^TSx\geq 0$ for all appropriately sized real vectors $x$. If $S=gg^T$, then $S$ is symmetric since $(gg^T)^T=(g^T)^Tg^T=gg^T=S$ and for any vector $x$, $x^Tgg^Tx=(g^Tx)^T(g^Tx)=(g^Tx)^2\geq 0$. Hence $S$ is PSD. – Casteels Apr 03 '14 at 08:46

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The matrix $uu^T$ is a special case of $\mathsf{PSD}$, (Positive semidefinite), matrices called a symmetric dyad.

Indeed, if $A = uu^T$ for some vector $u \in \mathbf{R}^n$, then for every $x$:

$$q_A (x) = x^T uu^T x = (u^Tx)^2 \ge 0. $$

Another way to look at this is thus: $$\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \cdot \begin{bmatrix} x_1 & x_2 & \ldots & x_n \end{bmatrix} = \begin{bmatrix}x_1^2 & x_1x_2 & \ldots & x_1x_n \\ x_2x_1 & x_2^2 & \ldots & x_2x_n \\ \vdots & \vdots & \ddots & \vdots \\ x_nx_1 & x_nx_2 &\ldots &x_n^2 \end{bmatrix}$$ What can you say about the above matrix? In particular, it's symmetric, and what property do all of its leading principal minors have in common?


To answer your first question, I would direct you to an answer of Plainview's in
the following question: Is the product of symmetric positive semidefinite matrices positive definite?

The product of two symmetric $\mathsf{PSD}$ matrices is $\mathsf{PSD}$, iff the product is also symmetric. More generally, if $A$ and $B$ are $\mathsf{PSD}$, $AB$ is $\mathsf{PSD}$ iff AB is normal, ie, $(AB)^T AB = AB(AB)^T$.



Since your matrix $H$ is positive definite, it is also positive semidefinite. Therefore, the product $HD$ is positive semidefinite iff $$ (HD)^T HD = HD(HD)^T $$

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Rustyn
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    I read the wiki article as well. That is for pos def matrices. I am not making any conclusions with pos def, only pos semi def. What would be a counter example? – HXSP1947 Apr 03 '14 at 08:01
  • @user2736423 For example, $H = 2I$, and $D$ contains only ones (both of the same arbitrary order). – Vedran Šego Sep 07 '14 at 00:19