Let $x$ be a real number such that $|x|<1$. Let the matrix to be checked be
$$\rho = \frac{1}{2}\begin{bmatrix} 1 & 0 & 0 & (1-x) \\ 0& 0& 0& 0 \\ 0& 0& 0& 0 \\ (1-x)& 0 & 0 & 1 \end{bmatrix}$$
I want to check that this matrix $\rho$ is positive semi - definite.
To do so I compute the eigenvalues and check to see if they are non - negative.
However, this is a sparse matrix and I am running into some trouble.
Here is my attempt:
$\rho v = \lambda v \\ \rightarrow det(\rho - \lambda I) = 0 \\ \rightarrow \frac{1}{2}(1-\lambda)(-1)^{2} \begin{vmatrix} -\lambda & 0& 0 \\ 0& -\lambda & 0&\\0& 0& (1-\lambda)& \end{vmatrix} -\frac{1}{2}\begin{vmatrix} 0& -\lambda& 0& \\ 0& 0& -\lambda& \\ (1-x) &0 &0 \end{vmatrix} =0 $
As can be observed, the 3x3 determinant can be computed again but the zeros at the top rows for each determinant are causing issues.
Any help is appreciated.
This symmetric matrix has a $2$-dimensional kernel. To check for positive semi-definiteness, it suffices to restrict to the orthogonal complement of the kernel, which is also $2$-dimensional. Finding the eigenvalues of the restriction is relatively straight forward.