Determinant of sort of companion matrix

Question

Note: I'm trying no EROs and no induction.

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This is the matrix I'm given:

$$A = \begin{bmatrix} b & 0_F & 0_F & \cdots & 0_F & 0_F & a_1\\ -1_F & b & 0_F & \cdots & 0_F & 0_F & a_2\\ 0 & -1_F & b & \cdots & 0_F & 0_F & a_3\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\ 0_F & 0_F & \cdots & -1_F & b & 0_F & a_{n-2}\\ 0_F & 0_F & \cdots & 0_F & -1_F & b & a_{n-1}\\ 0_F & 0_F & \cdots & 0_F & 0_F & -1_F & b+a_n \end{bmatrix}$$

This is for $b, a_1, ..., a_n \in$ a field $F$ with $b \ne 0_F$ s.t. $A \in F^{n \times n}$. (There's no mention of $n$, but I guess $n \ge 1$.)

1. Looks like this: Determinant of a companion matrix but my method is different because I plan to not use EROs.

2. Also looks like this: Finding determinant for characteristic polynomial and this: The characteristic and minimal polynomial of a companion matrix but I'm expanding over the last column...but these don't seem to do that.

3. Also looks like transpose of this, but my method is still different. Maybe I'm doing something wrong if my methods are different from the ones on maths se... Determinant of companion matrix - Specifically here, someone says in a comment that we're indeed supposed to use induction. I tried but it was very messy with the indices and stuff. Like the minors of $n=k+1$ don't quite match up with the induction hypothesis re $n=k$

Luckily, I'm already given the determinant and just have to prove it. It's not like I have to compute the determinant from scratch. The determinant is given as $$\sum_{j=1}^{n-1} a_jb^{j-1} +(b+a_n)b^{n-1}$$ Here $b^0$ makes sense as $b^0=1_F$ because $b \ne 0_F$.

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Question 1: I think I get it for $n \ge 3$, but for $n \le 2$, is the given matrix, and maybe even the given determinant, perhaps kind of ambiguous/not well-defined or something? My concern is that for $n \le 2$, we don't quite have any $0_F$'s.

Based on the given determinant I just guess what the matrix looks like and proceed to compute the determinant.

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Question 2: Is this correct for $n \le 2$?

Case 1: $n=1$. Then $A = [b+a_1]$ and $\det(A)=b+a_1=b^{n}+a_1$.

Case 2: $n=2$. Then $A = \begin{bmatrix} b & a_1\\ -1_F & b+a_2 \end{bmatrix}$. Do cofactor expansion about the 1st row. Then $$\det(A)$$

$$=b(b+a_2)-(a_1)(-1_F)$$ $$=b^2+ba_2+a_1$$ $$=b^n+b^{n-1}a_{n}+a_1$$

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Question 3: Is this correct for $n \ge 3$?

Case 3: $n \ge 3$

Step 1: Do cofactor expansion about the last (nth) column. Then

$$\det(A) = \sum_{j=1}^{n-1}(-1_F)^{j+n}a_j\det(M_{(j,n)})+(-1_F)^{n+n}(b+a_n)\det(M_{(n,n)})$$

Step 2: Use [the rule that determinant of upper or lower triangular equals product of diagonal entries] to get, for $j=1,...n-1, n$ (I'm including $n$ this time for the index $j$ unlike in Step 1), that $\det(M_{(j,n)}) = b^{j-1}(-1_F)^{n-j}$ because $M_{(j,n)}$ is upper triangular or lower triangular with diagonal entries starting with $j-1$ of $b$'s and then $n-j$ of $-1_F$'s.

Step 3: Therefore, by Steps 1 and 2, we have

$$\det(A) = \sum_{j=1}^{n-1}(-1_F)^{j+n}a_j\det(M_{(j,n)}) +(-1_F)^{n+n}(b+a_n)\det(M_{(n,n)})$$

$$ = \sum_{j=1}^{n-1}(-1_F)^{j+n}a_jb^{j-1}(-1_F)^{n-j} +(-1_F)^{n+n}(b+a_n)b^{n-1}(-1_F)^{n-n}$$

$$ = \sum_{j=1}^{n-1}(-1_F)^{j+n+n-j}a_jb^{j-1} +(-1_F)^{n+n+n-n}(b+a_n)b^{n-1}$$

$$ = \sum_{j=1}^{n-1}(-1_F)^{2n}a_jb^{j-1} +(-1_F)^{2n}(b+a_n)b^{n-1}$$

$$ = \sum_{j=1}^{n-1} a_jb^{j-1} +(b+a_n)b^{n-1}$$

Answer 1

If $R_i$ is the ith row of the determinant, add $R_1/b$ to $R_2$, then $R_2/b$ to $R_3$ etc and finally $R_{n-1}/b$ to $R_n$. It remains a diagonal determinant equal to \begin{equation} b^{n-1}\left(b + \sum_{k=0}^{n-1}a_{n-k} b^{-k}\right) =b^n + \sum_{k=1}^{n}a_k b^{k-1} \end{equation}