0
$\begingroup$

I have a matrix of: $$ A = \begin{bmatrix} a & b & b & \cdots & b \cr -1 & 1 & 0 & \ddots & \vdots \cr 0 & -1 & 1 & \ddots & \vdots \cr \vdots & \vdots & \ddots & \ddots & 0 \cr 0 & 0 & \cdots & -1 & 1 \cr \end{bmatrix}_{N \times N} $$ To calculate the determinant I followed the most general approach:

$$ \begin{aligned} \Delta_A &= a \cdot \text{det} \begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \cr -1 & 1 & 0 & \cdots & \vdots \cr 0 & -1 & 1 & \cdots & \vdots \cr 0 & 0 & -1 & 1 & \vdots \cr \vdots & \ddots & \ddots &\ddots & 1 \cr \end{bmatrix} - b \cdot \text{det} \begin{bmatrix} -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 1 & \vdots \\ \vdots & 0 & 0 & 0 & \ddots & 1 \\ 0 & 0 & 0 & 0 & 0 & -1 \end{bmatrix} + b \cdot \text{det} (A_{13}) + \cdots \end{aligned} $$ So the good news is the second matrix that I faced is a diagonal matrix consisting of either 1 or -1 on the diagonal. Actually, it's not a diagonal matrix but it's a diagonal matrix with non-zero off-diagonal values. I numerically calculate the determinant of such a matrix (up to size 100) and it seems that the determinant is no different from the determinant of a simple diagonal matrix:

$\text{det} A = \prod_{i=1}^{N} A_{ii}$

I wonder if there's any proof for that.

$\endgroup$
1
  • 1
    $\begingroup$ For any upper triangular (or lower triangular) matrix the determinant is the product of the diagonal elements. $\endgroup$ Commented Apr 21, 2022 at 8:40

1 Answer 1

1
$\begingroup$

Seems to me that the matrix is a particular upper Hessenberg so the closed form of its determinant in this case should be $$\sum_{k=0}^{n-1} \left(a_k\prod_{i=1}^{k} y_i\prod_{i=k+1}^{n-1} x_i\right)$$

where $a_k = \begin{cases}a & k = 0 \\ b & k \geq 1\end{cases}$ and $x_i = y_i = 1$ for each $i$. It follows that the formulas simplifies to $$\sum_{k=0}^{n-1} a_k = a + \sum_{k=1}^{n-1} a_k = a + \sum_{k=1}^{n-1} b = a+(n-1)b$$

$\endgroup$
2
  • $\begingroup$ @SassatelliGiulio Seems really nice this way. I don't see the companion matrix according to this though. Aren't the $b$'s in wrong place? Pheraphs they are similar? $\endgroup$ Commented Apr 21, 2022 at 9:23
  • $\begingroup$ The form of $I-A$ is the form a companion matrix? @SassatelliGiulio $\endgroup$ Commented Apr 21, 2022 at 9:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .