*symmetry*. =D (*)

But because we launch into that, some of you might find this blog entry from a previous year useful. Those who are in my tutes will have seen the very first thing in this blog post, but those who're only in my lab might find it interesting. I also go through a line-by-line analysis of the code at the end of last week's lab - if you totes understood everything, then great! If not, this might be a useful reference (plus, I point out a few things along the way - who knows, maybe you missed something?).

Okay, on to symmetry?

Now, we want to prove that \begin{align}\left| \begin{array}{ccc} 1 & 1 & 1 \\a & b & c \\ a^3 & b^3 & c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c).\end{align} And the first thing that I want to do is to make the following observations:

- The determinant is going to be a degree 4 polynomial in \(a,b,c\). This is because each (monomial) term in the determinant is a product of matrix entries taken from different rows and different columns (have a gooooood think about how cofactor expansion works).
- If any two of the three variables are equivalent, e.g.: \(a=b\), then the matrix has repeated columns and hence will have determinant equal to \(0\) (**). This tells us that \((a-b),(a-c)\) and \((b-c)\) must all be factors of the determinant.
- Switch any two of the variables in the matrix has the same effect as switching the two corresponding rows! This means that the determinant changes sign every time you switch two variables

Aaaaand, that should be enough to get us started.

Now, we know that the determinant is some polynomial \(d(a,b,c)\). And observation 2 tells us that \begin{align} d(a,b,c)=(a-b)(b-c)(c-a)\times p(a,b,c),\end{align} where \(p(a,b,c)\) is another polynomial in \(a,b,c\). Add on the fact that observation 1 tells us that every single term in \(d(a,b,c)\) has to be degree 4, this tells us that the polynomial \(p(a,b,c,)\) actually has to be linear: \begin{align} p(a,b,c)=\alpha a+\beta b+\gamma c\text{, for some }\alpha,\beta,\gamma\in\mathbb{R}.\end{align}

Okay, so it's lookin' preeeetty good...we're getting a good handle-hold on what \(d(a,b,c)\) looks like.

Thanks to observation 3, we know that \(d(a,b,c)=-d(b,a,c)\) and so: \begin{align} (a-b)(b-c)(c-a)p(a,b,c)&=-(b-a)(a-c)(c-b)p(b,a,c)\\&=(a-b)(b-c)(c-a)p(a,b,c)\\ \Rightarrow p(a,b,c)&=p(b,a,c)\\ \Rightarrow \alpha a +\beta b+\gamma c &= \alpha b+\beta a+\gamma c.\end{align} Since \(a\) and \(b\) are independent variables, this means that \(\alpha=\beta\). By the same logic applied to \(a\) and \(c\), we get that: \begin{align} \alpha=\beta=\gamma.\end{align} Soooo, we've just shown that the determinant may be written as \begin{align} d(a,b,c)=\alpha (a-b)(b-c)(c-a)(a+b+c),\end{align} and all we need to do now, is to show that \(\alpha=1\). One way to do this is to substitute nice numbers into \(a,b,c\) and to work it out. In fact, if we just set \(a=0\) we see that \begin{align}d(0,b,c)=\left| \begin{array}{ccc} 1 & 1 & 1 \\ 0 & b & c \\ 0 & b^3 & c^3\end{array}\right|=\left|\begin{array}{cc}b & c\\ b^3 & c^3\end{array}\right|=bc(c-b)(b+c).\end{align} Since we know that \begin{align}d(0,b,c)=\alpha bc(c-b)(b+c),\end{align} this gives us the desired conclusion that \(\alpha=1\).

AND WE'RE DONE! Now, this technique can also be used to do the other determinant in this question. It's a little bit trickier though...and you'll need to show, using symmetry arguments, that at some point you can get the determinant \(d(x,y,z)\) into the following form: \begin{align} d(x,y,z)=(x-y)(y-z)(z-x)[\alpha (x^2+y^2+z^2)+\beta (xy+yz+zx)],\end{align} and then show that \(\alpha=0\) and \(\beta=1\). This is a healthy exercise and you should all totes give it a try!

*: Note to self: smileys look vaguely wrong in a serif font... =/

**: if you didn't know this, combine the fact that determinants aren't changed by transposition and the method of computing determinants using row reduction

**: if you didn't know this, combine the fact that determinants aren't changed by transposition and the method of computing determinants using row reduction

## No comments:

## Post a Comment