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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.

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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 $m-n$.

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  • $\begingroup$ Would it be possible to do something similar to find the singular values of the matrix? Usually I would just find the eigenvalues of the matrix multiplied by its' transpose but I'm not sure what matrix this would give me. $\endgroup$
    – amlearn369
    Sep 12, 2021 at 8:09
  • $\begingroup$ It’s a symmetric matrix, so the singular values are just the absolute value of the eigenvalues. $\endgroup$ Sep 12, 2021 at 14:42

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