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I'm trying to calculate the following determinant: $$\begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & x & a_2 & \dots & a_n \\ a_0 & a_1 & x & \dots & a_n \\ \dots & \dots & \dots & \dots & \dots \\ a_0 & a_1 & a_2 & \dots & x \end{vmatrix} = $$ $$ = \begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & a_1 & a_2 & \dots & a_n \\ \dots & \dots & \dots & \dots & \dots \\ a_0 & a_1 & a_2 & \dots & a_n \end{vmatrix} + \begin{vmatrix} 0 & 0 & 0 & \dots & 0 \\ 0 & x - a_1 & 0 & \dots & 0 \\ 0 & 0 & x - a_2 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & 0 & \dots & x-a_n \end{vmatrix} = 0 + 0 = 0 $$

Still, experimental results contradict, since for one example I get a non-zero determinant.

What am I doing wrong?

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  • $\begingroup$ Did you mean to write and x in the top left element of the very first matrix you've written $\endgroup$ – Triatticus Feb 25 '16 at 15:28
  • $\begingroup$ No, in the top left corner it should be $a_0$ $\endgroup$ – marmistrz Feb 25 '16 at 15:29
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$\text{Det}(AB) = \text{Det}(A)\text{Det}(B)$, but in general, $\text{Det}(A+B)$ is not equal to $\text{Det}(A) +\text{Det}(B)$. It looks like you have used this wrong formula $\text{Det}(A+B) = \text{Det}(A) +\text{Det}(B)$ for your first equality.

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    $\begingroup$ Indeed. Otherwise you'd be able to break down the determinate of a 2x2 into the sum of 4 determinants where the matrices are 0s except for 1 element (which means the determinants are 0) and you could generalize to show that all non 1x1 matrices have a determinant of 0. But that's clearly not true. $\endgroup$ – Dason Feb 26 '16 at 4:54
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    $\begingroup$ In other words, the determinant is an alternating multilinear form :) $\endgroup$ – marmistrz Feb 26 '16 at 10:14
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Determinant is a multilinear map so the linearity works in a different way. Operation that you made corresponds to linearity in a sense $\det(A+B) = \det(A) + \det(B)$. This is true in general only for 1x1 matrices. The correct way to use the linearity of determinant is $$ \det(x_1 + \alpha y_1,x_2,\cdots,x_n) = \det(x_1,x_2,\cdots,x_n) + \alpha \det(y_1,x_2,\cdots,x_n) $$ where $x_1,\cdots,x_n,y_1$ are vectors of rows/columns of the matrix. So in your case the correct manipulation is $$ \begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & x & a_2 & \dots & a_n \\ a_0 & a_1 & x & \dots & a_n \\ \dots & \dots & \dots & \dots & \dots \\ a_0 & a_1 & a_2 & \dots & x \end{vmatrix} = \begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & x - a_1 + a_1 & a_2 & \dots & a_n \\ a_0 & a_1 & x & \dots & a_n \\ \dots & \dots & \dots & \dots & \dots \\ a_0 & a_1 & a_2 & \dots & x \end{vmatrix} = \begin{vmatrix} a_0 & 0 & a_2 & \dots & a_n \\ a_0 & x - a_1 & a_2 & \dots & a_n \\ a_0 & 0 & x & \dots & a_n \\ \dots & \dots & \dots & \dots & \dots \\ a_0 & 0 & a_2 & \dots & x \end{vmatrix} + \begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & a_1 & x & \dots & a_n \\ \dots & \dots & \dots & \dots & \dots \\ a_0 & a_1 & a_2 & \dots & x \end{vmatrix} $$

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I assume you actually want to know what the determinant is. For a 2 by 2 we have, $$(1) \quad \begin{vmatrix} a_0 & a_1 \\ a_0 & x \\ \end{vmatrix} =a_0 \cdot (x-a_1)$$ For a 3 by 3, we have, $$ \begin{vmatrix} a_0 & a_1 & a_2 \\ a_0 & x & a_2 \\ a_0 & a_1 & x \\ \end{vmatrix} $$ Since the upper right 2 by 2 is already known, we can evaluate along the bottom row. If we also note that the expansion along the bottom $a_1$ is zero, we get, $$(2) \quad \begin{vmatrix} a_0 & a_1 & a_2 \\ a_0 & x & a_2 \\ a_0 & a_1 & x \\ \end{vmatrix} =a_0 \cdot (x-a_1) \cdot (x-a_2)$$

In fact, we now know that the determinant $D_n$ for a n by n matrix of this form obeys,

$$(3) \quad D_{n+1}=D_n \cdot (x-a_{n-1})$$

Which implies using $D_1=a_0$, that

$$(4) \quad D_{n}=a_0 \cdot \prod_{k=1}^{n-1} (x-a_k)$$

I'm more physically oriented, so I'd just take this result and calculate the corresponding path integral. However, if you wish, I leave it as an exercise to rigorously prove this using induction.

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