# How to prove a matrix is positive semidefinite?

Let $$X\in S^3_+$$ be a semidefinite cone. Show the explicit conditions on the components of $$X$$.

I wanted to show for a positive semidefenite matrix $$X$$ we have $$z^T Xz\geq0\forall z$$:

$$\begin{bmatrix} z_1& z_2& z_3 \end{bmatrix}\begin{bmatrix} x_1& x_2& x_3\\ x_2& x_4& x_5\\ x_3& x_5& x_6 \end{bmatrix}\begin{bmatrix} z_1\\ z_2\\ z_3 \end{bmatrix}=z_1^2x_1+2z_1z_2x_2+2z_1z_3x_3+z_2^2x_4+z_3z_2x_5+z_3^2x_6\geq 0 \forall z$$

This is the point where I am lost. I have seen people continue by assuming $$x_1=0$$ and deducing $$x_2=x_3=0$$ so that $$X\succeq0$$ iff $$\begin{bmatrix} x_4& x_5\\ x_5& x_6\end{bmatrix}\succeq0$$.

I am trying to understand the path I have started. Specifically, why is for the $$x_1=0$$ case we must have $$x_2=x_3=0$$? also, what about the $$x_1\neq0$$ case?

Referance to the source, First problem

First $$X$$ needs to be symmetric, that is: $$x_{i,j} = x_{j,i}$$. Then its eigenvalues need to be $$\geq 0$$. Express the eigenvalues through the elements and set the conditions.
Edit: To see why this is so, do an eigendecomposition of $$X = Q\Lambda Q^T$$, we know that it exists, since the matrix is symmetric so all its eigenvalues are real numbers. Now: $$v^TXv= (Q^Tv)^T\Lambda Q^Tv= \sum_{i=1}^{n}{\lambda_iu_i^2} \geq 0$$ Where $$u = Q^Tv$$. This implies that $$\lambda_i \geq 0$$ for every $$i$$, since we can always pick a vector $$v$$ such that $$u_i = 1, u_j = 0, \forall j \neq i$$.
To find the eigenvalues simply express the roots of $$det(X-\lambda I)= 0$$ through the elements.
Edit2: Consider $$x_{1,1} = \lambda_1 q_{1,1}^2 + \lambda_2 q_{1,2}^2 + \lambda_3 q_{1,3}^3 = 0$$, I do not believe that it implies $$x_{1,2} = x_{1,3} = 0$$.