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?
Best Answer
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 $x_{1,2} = x_{1,3} = 0$.