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Suppose $v \in \mathbb{R}^m$ is an $m$-dimensional vector. Then is it true that $v v^T$ is positive semi-definite?

Rodrigo de Azevedo
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MikeL
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4 Answers4

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Yes; it's true since

$$\left\langle vv^Tx,x\right\rangle=\left\langle v^Tx,v^Tx\right\rangle=\left|\left|v^T x\right|\right|^2\ge0 .$$

beep-boop
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Yes, this is true. To verify that this is the case, note that for any complex vector $x$: $$ x^*(vv^*)x = (x^* v)(v^* x) = |\langle x,v \rangle|^2 \geq 0 $$ Or, restrict $x$ to real vectors and note that $vv^T$ is symmetric.

Ben Grossmann
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  • Thanks a lot for the clear answer. Is there also a simple way to prove in this case $A v v^T$ is positive semi-definite if $A$ is positive semi-definite any symmetric? – MikeL Oct 20 '14 at 16:12
  • There isn't because this isn't true. For example, take $$ A = \pmatrix{2&0\0&1}, \quad v = \pmatrix{1\1} $$ the product of two PSD matrices will always, however, have non-negative eigenvalues. – Ben Grossmann Oct 20 '14 at 16:31
  • Oh. Yes. You are absolutely right. I, fortunately, only need the positivity of the eigenvalues. Could you please give me a reference on the proof that product of two PSD matrices has always positive eigenvalues? – MikeL Oct 20 '14 at 16:37
  • See Horn and Johnson (new edition), corollary 7.6.2 page 486. There is an easy proof in the case that $A$ is non-singular, which can be used to give an alternate proof (to that given in the reference) in the general case. – Ben Grossmann Oct 20 '14 at 16:44
  • You may want to ask for a proof to this theorem in a new question-post. – Ben Grossmann Oct 20 '14 at 16:52
  • In my case $A$ is non-singular (I posted a new question as you suggested http://math.stackexchange.com/questions/982797/prove-that-the-product-of-two-positive-semidefinite-and-symmetric-matrices-has-n). Thanks a lot for your helps :-) – MikeL Oct 20 '14 at 17:01
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If $\mathrm v \in \mathbb R^n \setminus \{0_n\}$, then $\mathrm v \mathrm v^T$ is a symmetric, rank-$1$ matrix with a single nonzero eigenvalue

$$\lambda = \mbox{tr} (\mathrm v \mathrm v^T) = \mbox{tr} (\mathrm v^T \mathrm v) = \|\mathrm v\|_2^2 > 0$$

Thus, $\mathrm v \mathrm v^T$ is positive semidefinite.

Rodrigo de Azevedo
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Let $z$ be a vector. Then $z^Tvv^T z = (zv^T)^T(zv^T) = \langle z, v \rangle^2$. This is just a square of a real number. What do you know about squares of real numbers?

rwols
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