# Operator

• May 4th 2010, 07:43 AM
gralla55
Operator
Need some help with this... doesn't seem to be a decent example in my book. Appreciate any help!
• May 4th 2010, 08:23 AM
tonio
Quote:

Originally Posted by gralla55
Need some help with this... doesn't seem to be a decent example in my book. Appreciate any help!

1) Apply the transformation T to each and every element of E and write the result as a linear combination of E itself;

2) Take the transpose , if you apply transformations from the left and vectors from the right), or directly (otherwise) the coefficients matrix of the above: this is $\displaystyle [T]_E$.

For example:
$\displaystyle T(1)=1+1 = 2=2\cdot 1+0\cdot x+0\cdot x^2+0\cdot x^3$

$\displaystyle T(x)=x+x-2=-2+2x=(-2)\cdot 1+2\cdot x +0\cdot x^2+0\cdot x^3$ ...

So the first two columns of $\displaystyle [T]_E$ are $\displaystyle \begin{pmatrix}2\\0\\0\\0\end{pmatrix}\,,\,\,\begi n{pmatrix}\!\!-2\\2\\0\\0\end{pmatrix}$ (or the first two rows if you write the map on the right of the vector).

Tonio
• May 6th 2010, 01:03 AM
gralla55

T(x) = -2 + 2x ? Shouldn't that just equal 2?

Thanks again.
• May 6th 2010, 02:26 AM
HallsofIvy
Quote:

Originally Posted by gralla55

T(x) = -2 + 2x ? Shouldn't that just equal 2?

Thanks again.

Yes, that must have been a typo. Since T(f)= f(x)+ f(2- x), T(x)= x+ (2- x)= 2.
• May 6th 2010, 03:08 AM
gralla55
In that case 2 is the answer for all four. And a linear combination would just be every element of E times 1/2 times each column vector? And won't the transpose of this matrix is the same as the matrix itself?
• May 6th 2010, 04:09 AM
tonio
Quote:

Originally Posted by gralla55
In that case 2 is the answer for all four.

Why? $\displaystyle T(x^2)=x^2+(2-x)^2=2x^2-4x+4$ ...

Tonio

And a linear combination would just be every element of E times 1/2 times each column vector? And won't the transpose of this matrix is the same as the matrix itself?

.
• May 6th 2010, 04:18 AM
gralla55
Don't you just substitute the "x" for "x^2" ? I don't see why the ^2 suddenly goes outside the parantheses. But of course, I might be wrong here...
• May 6th 2010, 05:53 AM
gralla55
What confuses me is the notation T(f(x))... In case you're right:

T(1) = 1 + (2-1) = 2
T(x) = x + (2x-2) = 2
T(x^2) = 4 - 4x + 2x^2
T(x^3) = 8 - 12x + 8x^2 - 2x^3

How to proceed?
• May 6th 2010, 06:48 AM
tonio
Quote:

Originally Posted by gralla55
What confuses me is the notation T(f(x))... In case you're right:

T(1) = 1 + (2-1) = 2
T(x) = x + (2x-2) = 2
T(x^2) = 4 - 4x + 2x^2
T(x^3) = 8 - 12x + 8x^2 - 2x^3

How to proceed?

Write each outcome as a linear combination of the given basis and thus the coefficients in each case become the wanted matrix's columns...

Tonio
• May 6th 2010, 06:58 AM
gralla55
So:

2 2 4 8
0 0 -4 -12
0 0 2 8
0 0 0 -2

and then I take the transpose:

2 0 0 0
2 0 0 0
4 -4 2 0
8 -12 8 -2

And that's it?
• May 6th 2010, 08:39 AM
gralla55
Nevermind, I figured it out! And I also did a mistake, the correct answer is:

2 2 4 8
0 0 -4 -12
0 0 2 6
0 0 0 0

Now, how can I find a solution to T(f) = (x-1)^2 ?