# Tricks to find determinant of large symmetric matrix

Are there any tricks to find the determinant of a symmetric matrix in the form:

$$\begin{bmatrix} \mathbf{I} & \mathbf{B} \\ \mathbf{B}^T & \mathbf{0} \end{bmatrix}$$ where $$\mathbf{I}$$ is the identity matrix of size $$m \times m$$ matrix and $$\mathbf{B}$$ is an $$m \times n$$ matrix.

Suppose $$\mathbf{B}$$ is full column rank. Would this imply that the determinant is non-zero? I intuitively think this is but cannot prove this. I also want to know how to find the determinant to find eigenvalues of such a matrix.

So are there any tricks to do this quickly? I can't seem to put this matrix in a special category other than it is symmetric and can be used in the Schur complement.

Using the Schur complement, we know that the determinant of this matrix can be expressed as $$\det \pmatrix{I & B\\ B^T & 0} = \det\pmatrix{I & 0\\0 & -B^TB} = \det(-B^TB).$$ We therefore see that it is indeed the case that this matrix has non-zero determinant if (and only if) $$B$$ has full column rank.
Regarding the eigenvalues of this matrix, the singular value decomposition (SVD) of $$B$$ can be used as a helpful intermediate step. In particular, if $$B = U \Sigma V^T$$ is a singular value decomposition, then the matrix of interest is similar to $$\pmatrix{U & 0\\0 & V}^T \pmatrix{I & B\\B^T & 0}\pmatrix{U & 0\\0 & V} = \pmatrix{I & \Sigma\\ \Sigma^T & 0}.$$ If $$m \leq n$$, then there exists a permtuation matrix $$P$$ such that $$P\pmatrix{I & \Sigma\\ \Sigma^T & 0}P^T = \pmatrix{A_1 \\ & \ddots \\ && A_m\\ &&& 0},$$ where each block $$A_k$$ has the form $$A_k = \pmatrix{1 & \sigma_k\\ \sigma_k & 0 },$$ where $$\sigma_k$$ denotes the $$k$$th singular value of $$B$$. It follows that the eigenvalues of the matrix will be equal to $$0$$ and the solutions to $$\lambda^2 - \lambda - \sigma_k^2 = 0, \quad k = 1,\dots,n \implies\\ \lambda = \frac{1 \pm \sqrt{1 + 4\sigma_k^2}}{2}, \quad k = 1,\dots,n.$$ In the case that $$m>n$$, we have essentially the same thing except that there is now an identity matrix of size $$m-n$$ added to the diagonal. Consequently, the matrix will have (in addition to the eigenvalues noted in the previous case) $$1$$ as an eigenvalue of multiplicity at least $$m-n$$.