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I have the following matrix which I need to find the determinant of. I am not too sure of how to proceed. Here is my working so far.

\begin{equation} \det(\boldsymbol J (E_i) - \lambda \mathbb I ) = \begin{pmatrix} A_1 -\lambda & \dots & \phi_{1i} & \dots & 0 \\ \vdots & \ddots & \vdots & & \vdots \\ -c_{i1} & \dots & \color{red}{A_i -\lambda} & \dots & -c_{iN}\\ \vdots & & \vdots & \ddots & \vdots \\ 0 & \dots & \phi_{Ni} & \dots & A_N -\lambda \end{pmatrix} \end{equation} The matrix has a diagonal given by $A_j -\lambda$. From the central red element there are vertically and horizontally non-zero elements. All other elements are zero exactly.

I am really not sure how to find the determinant from here and any help or pointers would be greatly appreciated!

Edit

For instance if $N$ where to equal 4 we might have the following case if $i=3$, \begin{equation} \det(\boldsymbol J (E_i) - \lambda \mathbb I ) = \begin{pmatrix} D_1 & 0 & V_1 & 0 \\ 0 & D_2 & V_2&0 \\ H_1 & H_2 & D_3 & H_4\\ 0 & 0 & V_4 & D_4 \\ \end{pmatrix} \end{equation}

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  • $\begingroup$ If from diagonal entries vertical /horizontal elements in that row/column are nonzero, that seems to go against the two $0$ entries upper right and lower left. $\endgroup$
    – coffeemath
    Apr 7, 2017 at 11:28
  • $\begingroup$ @coffeemath Apologies for my poor wording, please see the update! $\endgroup$
    – RedPen
    Apr 7, 2017 at 11:30
  • $\begingroup$ So, wait, your original matrix is like $[[1, 0, 2, 0, 0], [0, 3, 0, 0, 0], [4, 0, 5, 0, 6], [0, 0, 0, 7, 0], [0, 0, 8, 0, 9]]$? $\endgroup$
    – Dan Uznanski
    Apr 7, 2017 at 11:38
  • $\begingroup$ Could the upper right and lower left entries, now labeled as $0,$ also be nonzero in the case $i \neq 1,N$? $\endgroup$
    – coffeemath
    Apr 7, 2017 at 11:42
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    $\begingroup$ Step 1: Change basis from $e_1, e_2, \ldots e_n$ to $e_i, e_2, \ldots, e_1, \ldots e_n$, exchanging rows $1$ and $i$ and columns $1$ and $i$. Then the first row and column become nonzero, and the remainder of the matrix is diagonal. That'll at least simplify the notation without changing the determinant. $\endgroup$
    – John Hughes
    Apr 7, 2017 at 11:44

1 Answer 1

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Edit. Let $B$ the submatrix obtained by deleting the $i$-th row and $i$-th column of the given matrix. Then the required determinant is the product of determinant of the Schur complement of $B$ and $\det B$, i.e. $$ \left(A_i-\lambda + \sum_{k\ne i}\frac{c_k\phi_{ki}}{A_k-\lambda}\right) \prod_{k\ne i}(A_k-\lambda). $$

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  • $\begingroup$ Thank you for your answer. It is a little difficult for me to understand, but does it hold for my edit? I just note that you say that the middle element has to be zero, this might not always be the case though? $\endgroup$
    – RedPen
    Apr 7, 2017 at 15:06

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