0

\begin{vmatrix} 3& 2& 0& 0& 0& \cdots&\\ 1& 3& 1& 0& 0& \cdots&\\ 0& 2& 3& 2& 0& \cdots&\\ 0& 0& 1& 3& 1& \cdots&\\ \vdots& \vdots& \vdots& \vdots& \vdots& \ddots&\\ \end{vmatrix}

How to compute this determinant?

I tried to multiply the frist column by $-2/3$ and then adding it to the second column. The second column would thus be $0$ instead of $2$ in the first row. However I do not know how to continue this procedure, or if it even is correct from the start.

VLC
  • 2,527
  • Cols and rows starting with $0$ can be eliminated. A simple $2x2$ determinant is left. – Ripi2 Apr 06 '21 at 19:11
  • 2
    @Ripi2 Are you sure? By your logic, the matrix $2I$ has determinant 2 for every dimension, because every row/column other than the first can be eliminated. – Calvin Lin Apr 06 '21 at 19:11
  • @bili 1) Is the matrix finite dimensional? 2) If yes, for the first column, is the last entry 1? (Likewise, for the first row, is the last entry 2). – Calvin Lin Apr 06 '21 at 19:13
  • I think this is useful for you https://math.stackexchange.com/questions/148532/general-expression-for-determinant-of-a-block-diagonal-matrix – pawel Apr 06 '21 at 19:13
  • @CalvinLin Well, no. Better I' d say that minors related to those "$0$-cols" can be eliminated on the calculation proccess. – Ripi2 Apr 06 '21 at 19:13
  • @Ripi2 Right, the issue is that it's tricky to account for all the permutations. In cases like this, the solution is unlikely to arise from a pure expansion. There might be an approach by expanding the first (and second?) column, with induction. – Calvin Lin Apr 06 '21 at 19:15
  • @CalvinLin It doesn't say if it is finite or not. What I tried now is, to multiply by $-2/3$ for odd and add one to the right column, and for even columns I would $-1/3$ and add one to the right. What I would get is $(\frac{1}{3})^{n-2}$, because the first $3$ in the diagonal and one $1/3$ cancel out. But it looks like it is not correct, since the solution to the problem should be : $2^{n+1}-1$ – VLC Apr 06 '21 at 19:17
  • 2
    @CalvinLin there is an easy recurrence indexed by the dimension. Requires proof by induction.... – Will Jagy Apr 06 '21 at 19:32

1 Answers1

2

Denote $D_n$ the determinat of an $n\times n$ matrix, which follows the pattern.
Compute the first determinants: $$D_1=3,\quad D_2=7, \quad D_3=15$$ With the use of the last row, compute $$D_4=\left[\begin{array}[cccc][3&2&0&0\\1&3&1&0\\0&2&3&2\\0&0&1&3\end{array}\right]=-1\cdot \left[\begin{array}[ccc][3&2&0\\1&3&0\\0&2&2\end{array}\right]+3\cdot D_3$$ or equivalently $$D_4=-1\cdot 2 \cdot D_2+3\cdot D_3$$

If we did it manually, it is then easy to observe that for any odd $n=2k-1$ $$D_{2k}=-1\cdot 2 \cdot D_{2k-2}+3\cdot D_{2k-1}$$ and for even $n=2k$ $$D_{2k+1}=-2\cdot 1 \cdot D_{2k-1}+3\cdot D_{2k},$$ which gives a common reccurence relation.

The determinants satisfy $$\forall n\in \mathbb{N}:\;\;D_{n+2}=-2D_{n}+3D_{n+1}$$

user376343
  • 8,311
  • 1
    The recurrence gives a nice closed form for $D_n$ as the characteristic roots are pleasant – Will Jagy Apr 06 '21 at 20:17
  • The matrix is actually similar to a symmetric tridiagonal Toeplitz matrix. Call it $A_n$ and let $D=\operatorname{diag}(\sqrt{2},1,\sqrt{2},1,\ldots)$. Then $$ B_n=D^{-1}A_nD=\pmatrix{3&\sqrt{2}\ \sqrt{2}&3&\sqrt{2}\ &\sqrt{2}&\ddots&\ddots\ &&\ddots&\ddots&\sqrt{2}\ &&&\sqrt{2}&3}. $$ For such a $B_n$, there is no need to consider even values and odd values of $n$ separately. – user1551 Apr 06 '21 at 20:46
  • @WillJagy I agree, I found that form a bit later, while preparing my classes. It is now too late (midnight) to edit carefully :) – user376343 Apr 06 '21 at 21:53
  • @user1551 thanks for your remark. I agree that it would not be necessary to consider even/odd cases separately. It was even not necessary in my approach, but I wanted to write the solution as it appeared for me. – user376343 Apr 06 '21 at 21:59