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Let $$ M = \begin{bmatrix} d_{1} - \lambda & 0 & \cdots & 0 & \zeta_{1} \\ 0& d_{2} - \lambda & \cdots & 0 & \zeta_{2} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & d_{n-1} - \lambda & \zeta_{n-1} \\ \zeta_{1} & \zeta_{2} & \cdots & \zeta_{n-1} & \alpha - \lambda \end{bmatrix}. $$ Compute determinant of $M$.

So, my guess was to solve this using Laplace expansion, but I got stuck. Is there any faster and easier way to solve this, since $M$ is symmetric? Any hint helps!

user560461
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1 Answers1

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Let M be the matrix $\text {diag}(d_1-\lambda_1,...,d_{n-1}-\lambda_{n-1})$ Use what I call the 'bordered determinant' formula $i.e.$ your determinant = $$(\alpha-\lambda)\det M$$ $$-\begin{bmatrix}\zeta_1&...&\zeta_{n-1}\end{bmatrix}\text {cofactor}(M)\begin{bmatrix}\zeta_1\\...\\\zeta_{n-1}\end{bmatrix}$$ $$=(\alpha-\lambda)\prod_{i=1}^{n-1}(d_i-\lambda_i)-\sum_{i=1}^{n-1}(\prod_{j \ne i}(d_j-\lambda_j))\zeta_i^2$$

P. Lawrence
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