1

Could somebody help me to prove that if $B$ is a regular matrix, then $A = B^\intercal B$ is symmetrical and positive definite?

Sebastiano
  • 7,649
batonas su braškėm
  • 15
  • 1
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Oct 20 '19 at 08:36
  • 2
    What do you mean by "regular", in this context? –  Oct 20 '19 at 08:37

1 Answers1

1

For any matrix $C$, we have that $C^TC$ is symmetric. We can check by taking the transpose and verifying we get the same thing: $(C^TC)^T=C^T(C^T)^T=C^TC$ (when applying transpose to a product of matrices you reverse the order of the product).

I'm not sure how you're using the word regular, but I'll consider two possible definitions.

If by regular, you just mean any matrix, then we don't get positive definiteness but rather we only get semi-positive definiteness. If by regular you mean invertible, then yes we do get positive definiteness.

Let's see how we get these results. To check for (semi-)positive definiteness we consider the following product with an arbitrary (but correct dimensionality) vector $x$: $$x^TC^TCx=(Cx)^T(Cx) = ||Cx||^2\geq 0.$$ This show it is at least semi-positive definite.

If $||Cx||^2$ is strictly positive for all $x$ except the zero vector then we have that $C^TC$ is positive definite. But for that to be the case, we would need that $C$'s kernel, or nullspace, is trivial (i.e., only contains the zero vector), which is equivalent to $C$ being invertible. If, however, $C$ is not invertible (for example, the all zeros matrix), then $||Cx||^2 = 0$ for all $x$ and so it is not positive definite.

Math Helper
  • 1,476