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Assuming that $A$ is a real symmetric matrix and $(\forall x: \ x^tAx=0)$, is it possible to prove that $A$ is the zero matrix?

I am trying to prove example 2.15 from the Book "Convex Optimization" by Stephen Boyd. The example states that $S^n_+$ is proper cone, when $S^n_+$ is the group containing all symmetric and positive semi-definite (PSD) matrices of size nxn. According to the book, a proper cone has to be pointed, when the term "pointed" is defined by:

$K$ is pointed iff $(x ∈ K\text{ and } −x ∈ K\Longrightarrow x=0)$

I deduced that if $A ∈ S^n_+$ and $-A ∈ S^n_+$ then $(\forall x: \ x^tAx=0)$, but I wasn't sure if it is possible to continue and deduce that $A$ is the zero matrix from that information.

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  • $\begingroup$ Hi, and welcome to StackExchange! You could diagonalize the matrix $A$ and prove that all of its eigenvalues are zeroes, for example. $\endgroup$ – Thomas Bakx Jun 8 '19 at 15:29
  • $\begingroup$ Thank you Thomas! $\endgroup$ – Neta S Jun 8 '19 at 15:30
  • $\begingroup$ Welcome to MSE! It is customary here to include your attempts when proving/disproving things. For example, where did you get the problem from? What have you tried? What examples have you examined? Cheers! $\endgroup$ – ε--δ Jun 8 '19 at 15:32
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    $\begingroup$ Pick $x$ to have one or two nonzero entries. $\endgroup$ – Lord Shark the Unknown Jun 8 '19 at 15:41
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first let $x=e_i,$ the column vector with all zeroes except a single $1$ at position $i.$ Multiply it out, it tells you something very specific.

Then, for each pair $i \neq j,$ let $x = e_i + e_j$

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I'll offer an alternative answer: since $A$ is real and symmetric, we know that it can be diagonalized, and in fact can be diagonalized by a matrix $S = [ \vec{u}_1 \text{ } \vec{u}_2 \cdots \vec{u}_n]$ where the $u_i$'s form an orthonormal basis for $\mathbb{R}^n$. (This is the Spectral Theorem for Real Symmetric Matrices.) Letting $A = SDS^{-1} =SDS^t$, we update our equation to be $\vec{x}^t S D S^t \vec{x} = 0$ for diagonal matrix $D$. Letting $\vec{y} = S^t\vec{x}$, our equation becomes $\vec{y}^t D \vec{y} = 0$, for all $\vec{y} \in \mathbb{R}^n$. Why can I say this? In fact, the matrix $S$ represents a bijective map (it is an orthogonal matrix), so there is exactly one vector $\vec{x}$, namely $\vec{x} = S\vec{y}$, such that $\vec{y} = S^t \vec{x}$.

Now if we let $D$'s diagonal entries be $\{ \lambda_1, \lambda_2, \cdots, \lambda_n \}$, and $\vec{y} = (y_1, y_2, ... y_n)$, then $\vec{y}^t D \vec{y} = 0 = \sum\lambda_i y_i^2$, which has only solution $\lambda_i = 0 ,\text{ } \forall i$. Therefore $A = SDS^t = S(0)S^t = 0$.

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For all its eigenvalues $\lambda$ and corresponding eigenvector x, we have $x^tAx = x^t\lambda x = \lambda x^tx = 0$. As x is a non zero vector, $\lambda$ must be zero (i.e. all $\lambda = 0$). Then as A is real symmetric, expressing it as $A = SDS^{-1}$, we see D is the diagonalized zero matrix, so A is zero matrix.

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