Question

Suppose that A is an m x m skew-symmetric matrix. Show that -A^2 is a nonnegative definite matrix.

Answer 1

Really, all I know is that since A is skew symmetric, this is equivalent to wanting to prove that \(AA^{T}\) is nonnegative definite matrix. But after that, I really don't know what to do with it. I'm still trying to figure out how to visualize what is happening. Taking a look at \(x^{T}(-A^{2})x\) might look like seems pretty disgusting, so not really sure on the correct manipulation of definition or thought process I should have.

Answer 2

hint:
we can write these equalities:
\[ {x^T}{A^2}x = {x^T}A \cdot Ax = - \left( {{x^T}{A^T}} \right) \cdot \left( {Ax} \right) = - \left\langle {Ax,Ax} \right\rangle = - {\left\| {Ax} \right\|^2} < 0\]

Answer 3

Surely it would be a nonnegative matrix...because as u know that the transpose of the matrix is -ve of the matrix in case of skew symmetric matrix n then when u multiply the matrix there will be positive terns remaining....

Answer 4

so we get:
\[\Large {x^T} \cdot \left( { - {A^2}x} \right) = {\left\| {Ax} \right\|^2} > 0\]

Answer 5

Im just bad with my matrix properties I guess, lol. I'm here confused trying to look at fancy summation definitions they give in the textbook and I guess that's making me not think the right way about these.

But okay, what you typed up made sense @Michele_Laino , but we would also need for some \(x \neq 0\), \(x^{T}(-A^{2})x =0\)

Answer 6

yes! of course!

Answer 7

by definition it is \(x \neq 0\)

Answer 8

RIght, right. But don't we need to find some \(x\) that will give us 0? Or is it something automatic from what we have?

Answer 9

I think that there is some vector \(x\) such that \(Ax=0\), if \(A\) is a singular matrix

Answer 10

Hmm...because if it were nonsingular then you'd only have the trivial solution. So if we can say or prove that \(-A^2\) is singular we have it.

Answer 11

yes! I think so!

Answer 12

Alright, I'll take it from there, thank you! ^_^

Answer 13

:)