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Let the matrix $T_n\in M(n\times n,\mathbb{F})$, where $\mathbb{F}$ denotes a field, be defined by $T_n=(t_{ij})$ with

$$t_{ij}= \begin{cases} \alpha\beta & 1\leq i\leq n-1,\;j=i+1 \\ \alpha+\beta & 1\leq i\leq n,\; j=i \\ 1 & 2\leq i\leq n,\; j=i-1 \\ 0 &\textrm{otherwise} \end{cases} $$

Show that $$\det(T_n)=\sum_{k=0}^{n}\alpha^{n-k}\beta^k$$

My approach: I tried a proof via induction and while the basis step is trivial, I can't seem to solve the induction step since the matrix is never in upper or lower triangular form but always a block matrix, which makes this seemingly difficult as when calculating the determinant of block matrices, one usually calculates the product of all the "diagonal blocks".

I would very much appreciate help, thank you very much.

  • I think you get a recurrence relation by expanding by first column, and then expanding the second term by the first row. – ancient mathematician Jan 18 '20 at 10:01
  • Also: https://math.stackexchange.com/q/383735/42969, https://math.stackexchange.com/q/2216162/42969, https://math.stackexchange.com/q/2262488/42969. – Martin R Jan 18 '20 at 10:02
  • Have you seen your determinant: $$\begin{vmatrix}\alpha+\beta&\alpha\beta&0&0&\ldots&0&0\1&\alpha+\beta&\alpha\beta&0&\ldots&0&0\0&1&\alpha+\beta&\alpha\beta&\ldots&0&0\0&0&1&\alpha+\beta&\ldots&0&0\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\0&0&0&0&\ldots&\alpha+\beta&\alpha\beta\0&0&0&0&\ldots&1&\alpha\beta\end{vmatrix}$$? – PinkyWay Jan 18 '20 at 10:06
  • Thanks that really is the same matrix but the identity they have to prove looks different. –  Jan 18 '20 at 10:09
  • @user: The lower right entry in your matrix is $\alpha + \beta$, isn't it? So that is not quite the same matrix. – But note that $\sum_{k=0}^{n}\alpha^{n-k}\beta^k = \frac{\alpha^{n+1} - \beta^{n+1}}{\alpha - \beta}$ for $\alpha \ne \beta$, so the solutions in the above referenced Q&As are exactly what you are looking for. – Martin R Jan 18 '20 at 10:16
  • @MartinR Yes, that's true thanks, but where did you get that identity? This doesn't look like we can simply apply the geometric series formula. –  Jan 18 '20 at 11:06
  • @user: What happens if you multiply $\sum_{k=0}^{n}\alpha^{n-k}\beta^k$ with $(\alpha - \beta)$? Can you see how all terms cancel, with the exception of $\alpha^{n+1} - \beta^{n+1}$? – See also https://math.stackexchange.com/q/2476730/42969. – Martin R Jan 18 '20 at 11:34
  • @MartinR All right thank you very much, now I see it. –  Jan 18 '20 at 11:41

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