Determinant factorisation (1 Viewer)

[martin]

Hi,
to anyone that knows tricks of determinant evaluation how would you do these questions?

i) factorise
| 1 1 1 |
| x y z |
|x^3 y^3 z^3 |

ii) deduce the factors of

| 1 1 1 1 |
| x y z w |
|x^2 y^2 z^2 w^2 |
|x^4 y^4 z^4 w^4 |

I'm actually tutoring this course at Uni of Queensland and the lecturer's solutions are less than illuminating so I'd like to have something helpful to show the kids on monday.

If anybody needs reminding of properties of determinants, reply and I'll put up some quick notes.

thanks,
Martin

 

[Euler]

there doesn't seem to be a nice way to put matrices into txt format, so each line will have 3 entries, giving a 3 by 3 matrix.

recall that operations like (row2)-k(row1) do not affect the determinant. Thus, the determinant of

1, 1, 1,
x, y, z,
x^2, y^2, z^2,

is the same as

1, 1, 1,
0, y-x, z-x,
0, y^2-x^2, z^2-x^2,

This comes from (row2)-x(row1) and (row3)-x^2(row1).

Calculating the determinant of this matrix is alot easier (note the zeros), and upon factorising gives
(z-x)(z-y)(y-x).

As for the 4 by 4 matrix, without explicitly doing the calculation, I would guess the answer to be
(w-x)(w-y)(w-z)(z-x)(z-y)(y-x).

when I have verified it, I'll tell you.

You can sort of guess it by looking at the 2 by 2 case first.

hope this helps.

 

[Euler]

Just as I put pen to paper to do the 4 by 4 case, I remembered that matrices of the given form are called Vandermonde matrices, and their determinants are called Vandermonde determinants.

what I wrote down for the 4 by 4 case is correct (note that with the order x,y,z,w in the matrix, the order of subtraction is important).

To prove it in general is a tedious application of induction involving clever matrix manipulations, and knowing what you are doing.

 

[martin]

What you wrote seems to be correct but the question actually had x^3 not x^2 etc.

I think this makes it a bit trickier and maybe the general results don't apply?

I've been working on it and I think something like

1,1,1
x,y,z
x^3,y^3,z^3

doing column 2 - column 1 and column 3 - column 2 gives

1,0,0
x,y-x,z-y
x^3,y^3-x^3,z^3-y^3

expanding by top row and factorising cubics equals

y-x,z-y
(y-x)(y^2+xy+x^2), (z-y)(z^2+zy+y^2)

then taking out factors from columns equals
(y-x)(z-y)*
|1,1|
|y^2+xy+y^2,z^2+zy+y^2|

expanding 2*2 determinant

(y-x)(z-y)*[z^2+zy+y^2-(y^2+xy+x^2)]

= (y-x)(z-y)[(z-x)(z+x)+y(z-x)]
= (y-x)(z-y)(z-x)(z+x+y)

so if I didn't make a mistake we get those three factors and a linear addition.

Now I don't want to do something similar for the second determinant so wondering if anyone knows a trick.

thanks,
Martin

 

[maniacguy]

If you take the generalization to the n x n case:
1 1 1 1 ... 1
x_1 x_2 ... x_n
.
.
.
(x_1)^(n-2) ... (x_n)^(n-2)
(x_1)^n ... (x_n)^n

Notice that the degree of every term in this polynomial is:
0+1+2+...+(n-2)+n
= 1 + 2 + ... + (n-2) + (n-1) + 1
= n(n-1)/2 + 1
= nC2 + 1

Now, also observe that setting x_i = x_j gives us a determinant of 0 (two columns identical), so that (x_i-x_j) is a factor of the determinant (i not equal to j).

But there are nC2 ways to choose distinct i,j from 1,2,...,n

Thus we have identified all but one of the factors in the product.
It will turn out this factor is:
x_1+x_2+... + x_n

Unfortunately I can't remember the exact proof of the very last part, although I am confident it is right, if that's any encouragement...

 

[KeypadSDM]

Matricies.

Can't wait till I tackle them.

 

[martin]

Alright! Thanks, this is what I was looking for.
I still don't quite understand the connection between the determinant and the polynomial you get when you expand it.

You get the polynomial degree by adding degrees of the terms on diagonal right?

Then determinant is 0 when x=y so by factor theorem (from polynomials) (x-y) is a factor.

I think I'm starting to understand (still a bit hazy though).

So the answer to part 2 should be
(y-x)(z-x)(w-x)(z-y)(w-y)(w-z)(w+x+y+z).

Originally posted by maniacguy

Notice that the degree of every term in this polynomial is:
0+1+2+...+(n-2)+n
= 1 + 2 + ... + (n-2) + (n-1) + 1
= n(n-1)/2 + 1
= nC2 + 1

Now, also observe that setting x_i = x_j gives us a determinant of 0 (two columns identical), so that (x_i-x_j) is a factor of the determinant (i not equal to j).

 

[Euler]

I tricked myself into thinking they were Vandermonde matrices.

Very sorry!

 

[Euler]

when setting x=y, you get determinant zero. (x-y) is then a factor, but so is (y-x). Which one do you take? when n is even, it shouldn't matter, but if n is odd, then you still have a negative sign to deal with.

I'm not sure what the matrix in the general case looks like. are the powers increased one at a time until you reach the bottom row in which the degree now jumps two?

In the case set by maniac guy, the last factor has to be symmetric in all the x_i's. how many degree one polynomials are symmetric in n variables? I can only think of two: the sum of them all, and the negative of it.

My guess is as good as yours, martin, for the four by four case.

 

[maniacguy]

Thanks for that, Euler. That's what I was missing!

As for whether the sign should be positive or negative, the result should be:
determinant
= PROD[(x_j-x_i) for i=1,2,..,n-1, j=i+1,i+2,..,n]*SUM[x_i,i=1..n]

(you can see this by taking the term down the main diagonal, that is 1*x_2*x_3^2*...*x_(n-1)^(n-2)*x_n^n. Work out what each term in the product contributes to this, so for example from the term (x_2-x_1) we obviously must have taken the x_2. Then from (x_3-x_2) and (x_3-x_1) we must have taken the x_3 in each case, and so forth...)

Sorry if that isn't very clear, I really need a bunch of matrix diagrams to explain clearly and those are a pain to type. Hopefully the meaning comes across...