Determinant

**dwsmith** · September 26th 2011, 01:51 PM

Let $a_1,\cdots , a_n$ be given numbers. Compute the determinant of the nxn matrix $A=(a_{ij})$ , where $a_{ij}=a^{i-1}_j$ .

I don't understand what is meant by this: $a_{ij}=a^{i-1}_j$

**Deveno** · September 26th 2011, 01:57 PM

without being given any more context, i would assume it means $a_{ij}$ is the (i-1)-th power of $a_j$ . so the first row should all be 1's. this kind of matrix is called a Vandermonde matrix.

you'll probably arrive at a better understanding if you try it first for n = 2, and n = 3. try to factor the result into binomials.

if you are clever, and figure out (guess) the general form, you can actually prove it by induction on n.

**dwsmith** · September 26th 2011, 02:02 PM

Originally Posted by Deveno

without being given any more context, i would assume it means $a_{ij}$ is the (i-1)-th power of $a_j$ . so the first row should all be 1's. this kind of matrix is called a Vandermonde matrix.

you'll probably arrive at a better understanding if you try it first for n = 2, and n = 3. try to factor the result into binomials.

That is the question 100% verbatim.

**pickslides** · September 26th 2011, 02:08 PM

I see it as $a^{i-1}_j \in \{a_1,\cdots , a_n\}$

**Deveno** · September 26th 2011, 02:08 PM

well it's an important matrix, both for the study of polynomials, alternating bilinear forms, and permutation groups. it crops up in a lot of different places. time to roll up your sleeves and make some messy calcs...lol

@pickslides: i don't think so....there's nothing in the wording to indicate membership. and the calculation of such a determinant (a Vandermonde matrix) is the sort of thing you might encounter in a variety of different courses, inclusing differential geometry, linear algebra and group theory. i think the first time i saw it was in a physics class.

**dwsmith** · September 26th 2011, 02:12 PM

Originally Posted by Deveno

well it's an important matrix, both for the study of polynomials, alternating bilinear forms, and permutation groups. it crops up in a lot of different places. time to roll up your sleeves and make some messy calcs...lol

I have used it before in my undergraduate linear alg. class. I am familiar with it. I just don't know it when I am presented with it.

**Deveno** · September 26th 2011, 02:22 PM

$V\ =\ \begin{bmatrix}1&1&\cdots&1\\a_1&a_2&\cdots&a_n\\a _1^2&a_2^2&\cdots&a_n^2\\ \vdots&\vdots&\ddots&\vdots\\a_1^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1} \end{bmatrix}$

does it look better written out like this?

**dwsmith** · September 26th 2011, 02:28 PM

Originally Posted by Deveno

$V\ =\ \begin{bmatrix}1&1&\cdots&1\\a_1&a_2&\cdots&a_n\\a _1^2&a_2^2&\cdots&a_n^2\\ \vdots&\vdots&\ddots&\vdots\\a_1^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1} \end{bmatrix}$

does it look better written out like this?

I understood when you set it was a Vandermonde matrix. That wasn't needed.

**Deveno** · September 26th 2011, 02:48 PM

fair enough. but perhaps if someone else peruses this thread one day, it will be helpful to them. no slight intended.

Thread: Determinant

LinkBack

Thread Tools

Display

Determinant

Re: Determinant

Re: Determinant

Re: Determinant

Re: Determinant

Re: Determinant

Re: Determinant

Re: Determinant

Re: Determinant

Similar Math Help Forum Discussions

Determinant

Determinant help

Determinant

Determinant 3

Determinant 2