Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
    
Did you mean to write and x in the top left element of the very first matrix you've written – Triatticus Feb 25 at 15:28
    
No, in the top left corner it should be $a_0$ – marmistrz Feb 25 at 15:29

$\text{Det}(AB) = \text{Det}(A)\text{Det}(B)$, 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.

share|cite|improve this answer
1  
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. – Dason Feb 26 at 4:54
2  
In other words, the determinant is an alternating multilinear form :) – marmistrz Feb 26 at 10:14

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 $$ \begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & x & a_2 & \dots & a_n \\ a_0 & a_1 & x & \dots & a_n \\ \dots & \dots & \dots & \dots & \dots \\ a_0 & a_1 & a_2 & \dots & x \end{vmatrix} = \begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & x - a_1 + a_1 & a_2 & \dots & a_n \\ a_0 & a_1 & x & \dots & a_n \\ \dots & \dots & \dots & \dots & \dots \\ a_0 & a_1 & a_2 & \dots & x \end{vmatrix} = \begin{vmatrix} a_0 & 0 & a_2 & \dots & a_n \\ a_0 & x - a_1 & a_2 & \dots & a_n \\ a_0 & 0 & x & \dots & a_n \\ \dots & \dots & \dots & \dots & \dots \\ a_0 & 0 & a_2 & \dots & x \end{vmatrix} + \begin{vmatrix} a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & a_1 & a_2 & \dots & a_n \\ a_0 & a_1 & x & \dots & a_n \\ \dots & \dots & \dots & \dots & \dots \\ a_0 & a_1 & a_2 & \dots & x \end{vmatrix} $$

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.