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

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

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- Sep 26th 2011, 01:51 PMdwsmithDeterminant
Let $\displaystyle a_1,\cdots , a_n$ be given numbers. Compute the determinant of the nxn matrix $\displaystyle A=(a_{ij})$, where $\displaystyle a_{ij}=a^{i-1}_j$.

I don't understand what is meant by this: $\displaystyle a_{ij}=a^{i-1}_j$ - Sep 26th 2011, 01:57 PMDevenoRe: Determinant
without being given any more context, i would assume it means $\displaystyle a_{ij}$ is the (i-1)-th power of $\displaystyle a_j$. 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. - Sep 26th 2011, 02:02 PMdwsmithRe: Determinant
- Sep 26th 2011, 02:08 PMpickslidesRe: Determinant
I see it as $\displaystyle a^{i-1}_j \in \{a_1,\cdots , a_n\}$

- Sep 26th 2011, 02:08 PMDevenoRe: Determinant
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. - Sep 26th 2011, 02:12 PMdwsmithRe: Determinant
- Sep 26th 2011, 02:22 PMDevenoRe: Determinant
$\displaystyle 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? - Sep 26th 2011, 02:28 PMdwsmithRe: Determinant
- Sep 26th 2011, 02:48 PMDevenoRe: Determinant
fair enough. but perhaps if someone else peruses this thread one day, it will be helpful to them. no slight intended.