Matrix Problem From Hell - Inverse of Linear Transformation

[blashmet]

Homework Statement

http://img847.imageshack.us/img847/2783/77597781.jpg

Homework Equations

See below.

The Attempt at a Solution

I get all the way to the last step, but I'm not sure how to perform it.I get to the point where I have:

[(T^-1)(ax^2+bx+c)]\alpha= [T^-1]\alpha\beta*[v]\beta=

[ a/4 + b/8 - (3c/8)]
[ 3(b+c) - (a/2) ]
[(a/2) - ((b+c)/4) ]

Now, at this point, I need to get those 3 elements into a 2x2 symmetric matrix. At first I just made the top element the repeated element along the diagonal, but apparently this is wrong.

At this point on my homework my teacher wrote, "Now use \alpha={basis v1, basis v2, basis v3} (the given basis in the problem statement).

I'm still not sure exactly what to do at this point. I don't know how to get those three elements into a 2x2 symmetric matrix "under" the basis alpha.

Can someone please show me how to perform this last step? Thanks! :)

 

[Deveno]

i've seen this problem before...

it's a lot of work to verify that those coordinates are right (and i don't feel like spending the hour or so it would take), but if they are, then you just need to form the matrix:

c₁α₁ + c₂α₂+ c₃α₃

where c₁ = a/4 + b/8 - (3c/8), etc.

your resulting matrix will be god-awful ugly, but it WILL be symmetric.

 

[blashmet]

Ahh Deveno! lol...

I didn't know you were on this forum too. I was going to have another look at our convo about this problem on MHF, but it's down apparently.


Anyways,

why do you think the matrix will be ugly?

The solution is posted as:

http://img408.imageshack.us/img408/7732/la2c.jpg



It doesn't look that bad...