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Given $ A_n = \left[ {\begin{array}{cc} 2 & 1 & \\ 1 & 2 & 1\\ & 1 & 2 \\ & & & ... \\ & & & & 2 & 1 \\ & & & & 1 & 2 \end{array} } \right] \in M_n(\mathbb{R})$. What I have to do is calculate the determinant with respect to $n\in \mathbb{N}$. I already found out $\det(A_n) = n+1$. I tried to show this with induction. I get confronted by the following problem:

I showed the statement for $n=1$. Lets consider the statement is true for one $n \in \mathbb{N}$. Now show that the statement is also true for $n+1$ using laplace expansion for the first row.

We get $\det(A_{n+1}) = (-1)^2\cdot 2\cdot \det(A_n)+(-1)^3\cdot 1\cdot \begin{vmatrix} 1 & 1 & \\ 0 & 2 & 1\\ & 1 & 2 \\ & & & ... \\ & & & & 2 & 1 \\ & & & & 1 & 2 \end{vmatrix} $. The second determinant should be $1\cdot \det(A_{n-1}) = n$ but we only considered the statement tu be true for $n$ and not for $n-1$. How do I continue from here?

David C. Ullrich
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Arji
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  • Use "complete induction". Assume the hypothesis, not just for $n$, but for all of $1,2,\ldots,n$, and then deduce the $n+1$ case. – Angina Seng Apr 29 '18 at 16:06
  • I've already thought of that. I was not sure if this is legit to do. – Arji Apr 29 '18 at 16:08
  • I know the matrix in the question I marked this one as duplicate of has sub- and sup-diagonal entries $-1$ rather than $1$, but the difference is just a base change where basis vectors $e_i$ are replaced by $(-1)^ie_i$ (conjugation by a diagonal matrix with diagonal entries $1,-1,1,-1,1,-1,\ldots$). – Marc van Leeuwen Apr 29 '18 at 17:14
  • As for the question of strong induction, you can use as induction hypothesis that the statement holds for all $i\leq n$. Since you explicitly need the statement for the last two of these values, you need to prove two initial case separately from the induction argument, so that validity for two previous values is assured even for the first application of the induction step. – Marc van Leeuwen Apr 29 '18 at 17:17

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