# Is it true that if $A$ is real and symmetric and $(\forall x: \ x^tAx=0)$, then $A$ is the zero matrix?

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.

• Hi, and welcome to StackExchange! You could diagonalize the matrix $A$ and prove that all of its eigenvalues are zeroes, for example. – Thomas Bakx Jun 8 '19 at 15:29
• Thank you Thomas! – Neta S Jun 8 '19 at 15:30
• 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! – ε--δ Jun 8 '19 at 15:32
• Pick $x$ to have one or two nonzero entries. – Lord Shark the Unknown Jun 8 '19 at 15:41

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 $$x = e_i + e_j$$
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 $$A = SDS^t = S(0)S^t = 0$$.
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.