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Ok so we know that it is an n order determinant but I do not know how to calculate it $\begin{vmatrix} 0 & a & a^2 & ... &a^{n-1} \\ \frac{1}{a} &0 & a & ... &a^{n-2} \\ \frac{1}{a^2}&\frac{1}{a} & 0 & ... &a^{n-3} \\ ...& ... & ... &... &... \\ \frac{1}{a^{n-1}} & \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & ... & 0 \end{vmatrix}$

Ghost
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2 Answers2

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Denote the matrix in question as $A$. The matrix $$A + I = \begin{bmatrix} 1 \\ \frac{1}{a} \\ \cdots \\ \frac{1}{a^{n-1}}\end{bmatrix} \cdot \begin{bmatrix} 1, a, \cdots, a^{n-1}\end{bmatrix}$$ is of rank $1$. Therefore, its eigenvalues are $0$ with multiplicity $n - 1$ and $n$ with multiplicity $1$. See for example this question.

Note that, if $\lambda$ is an eigenvalue for $A + I$, then $\lambda - 1$ is an eigenvalue for $A$. Therefore, the eigenvalues of $A$ are $-1$ with multiplicity $n - 1$ and $n - 1$ with multiplicity $1$. Thus $$ |A| = (-1)^{n-1}(n-1) $$ is the final answer.

Another method that is via basic row operations is as follows. I show the details so that you can follow it.

  • First, multiplying the second row of $A$ by $a$ and subtracting it from the first row, we get $$\begin{vmatrix} -1 & a & 0 & ... & 0 & 0 \\ \frac{1}{a} &0 & a & ... &a^{n-3} & a^{n-2} \\ \frac{1}{a^2}&\frac{1}{a} & 0 & ... & a^{n-4} & a^{n-3} \\ ...& ... & ... &... &... & ... \\ \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & \frac{1}{a^{n-4}} & ... & 0 & a \\ \frac{1}{a^{n-1}} & \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & ... & \frac{1}{a} & 0 \end{vmatrix}$$

  • Second, multiplying the $3$rd row of $A$ by $a$ and subtracting it from the $2$nd row, we get $$\begin{vmatrix} -1 & a & 0 & ... & 0 & 0 \\ 0 & -1 & a & ... & 0 & 0 \\ \frac{1}{a^2}&\frac{1}{a} & 0 & ... & a^{n-4} & a^{n-3} \\ ...& ... & ... &... &... & ... \\ \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & \frac{1}{a^{n-4}} & ... & 0 & a \\ \frac{1}{a^{n-1}} & \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & ... & \frac{1}{a} & 0 \end{vmatrix}$$

  • You can continue this routine by multiplying the $i+1$th row by $a$ and substracting it from the $i$th row, for $3 < i \leq n - 1$ and we finally obtain $$\begin{vmatrix} -1 & a & 0 & ... & 0 & 0 \\ 0 & -1 & a & ... & 0 & 0 \\ 0 & 0 & -1 & ... & 0 & 0 \\ ...& ... & ... &... &... & ... \\ 0 & 0 & 0 & ... & -1 & a \\ \frac{1}{a^{n-1}} & \frac{1}{a^{n-2}} & \frac{1}{a^{n-3}} & ... & \frac{1}{a} & 0 \end{vmatrix}$$

  • We deal with the last row as follows: For $1 \leq i \leq n - 1$ in order, repeatedly multiply the $i$th row by $\frac{i}{a^{n-i}}$ and add it to the last row, leading to an upper triangle matrix $$\begin{vmatrix} -1 & a & 0 & ... & 0 & 0 \\ 0 & -1 & a & ... & 0 & 0 \\ 0 & 0 & -1 & ... & 0 & 0 \\ ...& ... & ... &... &... & ... \\ 0 & 0 & 0 & ... & -1 & a \\ 0 & 0 & 0 & ... & 0 & n-1 \end{vmatrix}$$ Therefore, the determinant is $(-1)^{n-1}(n-1)$, which is the product of the elements on the diagonal of the triangle matrix.

Community
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PSPACEhard
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  • So... what is the final answer? – Ghost Nov 13 '16 at 19:28
  • @OvyOvy $(-1)^{n-1}(n-1)$ – PSPACEhard Nov 13 '16 at 23:54
  • I have not learned these things so far. I should be able to do it with a classic method like making 0 on a row. Do you have any idea how to do it like that? – Ghost Nov 14 '16 at 05:32
  • @OvyOvy Please see whether it is clear now. – PSPACEhard Nov 14 '16 at 06:17
  • Ok i think i got it. Thank you very much. – Ghost Nov 14 '16 at 12:33
  • How did you get on the last row 0? Cause if you multiply the first row with 1/a^(n-1) and add it to the last you get 0 1/a^(n-2) + a^(n-2) and the others remain the same. After you multiply the second row with 2/a^(n-2) you get on the last row 0 a^(n-2) - 1/a^(n-2) (cause you had that from adding thr first row) and so on...? – Ghost Nov 14 '16 at 12:56
  • @OvyOvy After you multiply the first row by $\frac{1}{a^{n-1}}$ and add it to the last row, the last row becomes $0, \frac{2}{a^{n-2}}, \frac{1}{a^{n-3}}, \cdots, 0$. – PSPACEhard Nov 14 '16 at 13:00
  • Ohhhh yeah sorry I failed at calculating a*1/a^(n-1) XD it is a^(1-n+1)=a^(2-n)=1/a^(n-2) and when you add the last row number it is 2/a^(n-2). Thank you for help. – Ghost Nov 14 '16 at 13:04
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Scalar multiplication on second row with an 'a', third row with a squared, and so on until the last row with an 'a' to the power of n-1 on the last row. You will end up seeing something interesting. Do some row operation to clean out such as using the first row to subtract the last row. Using the property of determinant from 3 types of row operated matrix. You'll find the answer. I'm new to this site, so I didn't learn how to type in this format very well. sorry bro

robster
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