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A is $n \times n$ symmetric and positive semidefinite. So we can use $A^T = A$ and $x^TAx \geq 0$ for all x $\in \mathbb{R}^n$.

Proof by contradiction comes to mind, or $x^TAx = 0 \; \land \; Ax \neq 0$. This might seem like a given at first but I'm thinking what if there is some linear dependence between $x^T$ and the $Ax$ vector?

Edit: Ideally, I'm supposed to prove this using the rank of a matrix.

If I were to prove it by contradiction, $x^TAx = 0 \; \land \; Ax \neq 0$, I'd start off by saying that:

  • $x$ cannot be a zero vector in this case. The original statement $x^TAx = 0 \Rightarrow Ax = 0$ is trivial for a zero $x$ vector and needs no proof.
  • from $Ax \neq 0$ we have $rank(Ax) = 1$ since $Ax$ is also a non-zero vector.
  • problem: how to show that the $x^T$ vector and the $Ax$ vector cannot be orthogonal? (this would be the point of the contradiction proof)
  • problem 2: the only idea I have for a proof barely utilizes the rank of a matrix, which is my assignment.
supinf
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Milly
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  • I have seen it and I didn't really like any answers in that thread except for the one using the rank of the matrix but that one seemed to be done as if X was a matrix not a vector. – Milly Nov 22 '20 at 12:07
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    Please don't vandalize your question. Please restore it. – Gerry Myerson Nov 23 '20 at 12:33
  • @GerryMyerson I rolled back the question to an unvandalized version – supinf Nov 23 '20 at 12:36
  • @supinf Excuse me but by what authority did you do that? It's not your question. – Milly Nov 23 '20 at 13:11
  • @Milly no special authority. Afaik, bandalizing questions is not welcome on this site, even if it is your own question. Plus its unfair to the people who answered in my opinion. You can ask a mod about this if you want. – supinf Nov 23 '20 at 15:20

2 Answers2

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I think you have the best hint from Gerry Myerson in the link from @projectilemotion.

Symmetric $A \Rightarrow$ we can choose orthogonal eigenvectors $v$ for A. These form a basis for $\mathcal{R}^n$, so you can write $x=\sum b_jv_j$.

Positive semi-definite $A \Rightarrow$ its eigenvalues are non-negative. Express $x^TAx$ in terms of the eigenvector basis and use these two facts, you should be home.

Gerry Myerson
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balaji
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Consider the EVD of the matrix $A=UDU^T$.

Then we can rewrite $x^TAx=x^TUD^\frac12D^\frac12U^Tx=\|D^\frac12U^Tx\|^2$

Hence if $x^TAx=0$, we must have $D^\frac12Ux=0$ and hence $$UD^\frac12(D^\frac12U^Tx)=(UD^\frac12D^\frac12U^T)x=Ax=0$$

Siong Thye Goh
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