# Showing a matrix is positive semi definite if certain principal submatrix

In my matrix theory studies I recently came across the following problem:

We have the following real symmetric $n \times n$ matrix $A=\begin{pmatrix} d_1 & u^T E^\dagger v & u^T \\ v^T E^\dagger u & d_2 & v^T \\ u & v & E \end{pmatrix}$ where $d_1,d_2$ are real numbers and $E$ is a square $n-2\times n-2$ matrix, $u,v$ are column vectors of length $n-2$, T denotes transpose as usual and $\dagger$ denotes generalized inverse (or classic inverse if it is easier to solve) and we know that any principal submatrix of $A$ that does not include entries $a_{1,2},a_{2,1}$ is positive semidefinite, and we are asked to show $A$ is positive semidefinite.

I thought a direct approach might be difficult, so I though about using the characterization theorem of positive semidefinite matrices to show an easier equivalent condition. I would certainly appreciate all help.

## 1 Answer

Well, according to what you say if you choose
$$n = 4$$ $$E = I$$ $$u = v = [1 \quad 1]^T$$ $$d_1 = d_2 = 1$$ You can show that matrix that the principle submatrices by deleting the second row and column are positive definite, however $A$ is not positive definite. The eigenvalues, in that case, are given as $\lambda = [-1,\ -0.2361,\ 1, \ 4.2361]$.

I hope i understood your question correctly.

• I mean principal submatrices that don't include index 2 – kroner Oct 17 '16 at 23:09
• Can you elaborate on what you mean ? – Ahmad Bazzi Oct 17 '16 at 23:10
• I mean any principal submatrix of A that does not include entries $a_{1,2}, a_{2,1}$ – kroner Oct 17 '16 at 23:11
• Aha that is different now.. – Ahmad Bazzi Oct 17 '16 at 23:12
• sorry about that – kroner Oct 17 '16 at 23:12