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The matrix $A_n\in\mathbb{R}^{n\times n}$ is given by

$$\left[a_{i,j}\right] = \left\lbrace\begin{array}{cc} 1 & i=j \\ -j & i = j+1\\ i & i = j-1 \\ 0 & \text{other cases} \end{array} \right.$$

I already showed that it holds

$$\det{A_n}= \det{A_{n-1}}+\left(n-1\right)^2\cdot\det{A_{n-2}}$$

However, can we find an explicit expression for the determinant of $A_n$?

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    $\begingroup$ A simple way is to prove that the eigenvalues of $A_n$ are $1,\ldots,n$, (by making some Gauss moves that put $A_n$ in upper triangular form, for example) from which $A_n=n!$ immediately follows. $\endgroup$ – Jack D'Aurizio Jan 15 '14 at 0:29
  • $\begingroup$ @JackD'Aurizio No, the spectrum of $A_n$ is not $\{1,2,\ldots,n\}$. To find the determinant (instead of eigenvalues), your idea works, though. $\endgroup$ – user1551 Jan 15 '14 at 22:19
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I had an idea by myself right now...

It holds

  • $\det(A_1) = 1$
  • $\det(A_2) = 2$
  • $\det(A_3) = 6$
  • $\det(A_4) = 24$
  • $\dots$

Thus I assume the determinant of $A_n$ is given by $n!$


Proof: (by induction over $n$)

  • The statement is true for $n=1\dots 4$
  • Now let it be true for $n-1$ and $n-2$ then:

$$\begin{array}{rcl} \det(A_n) &=& \det(A_{n-1}) + \left(n-1\right)^2\det(A_{n-2})\\ &=& (n-1)! + (n-1)^2\cdot(n-2)! \\ &=& (n-1)! + (n-1)\cdot(n-1)! \\ &=& (1+(n-1))\cdot(n-1)! \\ &=& n! \end{array}$$ Hence $\det(A_n) = n!$ is true for every $n\geq 1$

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Put $a_n=\det A_n$ and $$ f(x)=\sum_{n=1}^{+\infty}a_n \frac{x^n}{n!},\tag{1}$$ in order that: $$f'(x) = \sum_{n=0}^{+\infty}a_{n+1}\frac{x^{n}}{n!},\tag{2}$$ $$x\,f(x) = \sum_{n=2}^{+\infty}n\, a_{n-1} \frac{x^{n}}{n!},$$ $$f(x)+x\,f'(x) = \sum_{n=2}^{+\infty}n^2\, a_{n-1} \frac{x^{n-1}}{n!},$$ $$x\,f(x)+x^2\,f'(x) = \sum_{n=2}^{+\infty}n^2\, a_{n-1} \frac{x^{n}}{n!}.\tag{3}$$ Now $a_{n+1}=a_n+n^2\,a_{n-1}$ gives: $$f'(x) = f(x) + x\, f(x) + x^2\,f'(x),$$ or: $$ (1+x)\,f(x) = (1-x^2)\, f'(x)$$ $$ f(x) = (1-x)\, f'(x)\tag{4}. $$ The solutions of this differential equation are $f(x)=\frac{K}{1-x}$, from which we get that $a_n = K\,n!$. Since $a_1=1$, $a_n=n!$.

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    $\begingroup$ Very elegant re-framing $\endgroup$ – nbubis Jan 15 '14 at 0:23

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