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

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- May 4th 2010, 07:43 AMgralla55Operator
Need some help with this... doesn't seem to be a decent example in my book. Appreciate any help!

- May 4th 2010, 08:23 AMtonio

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 .

For example:

...

So the first two columns of are (or the first two rows if you write the map on the right of the vector).

Tonio - May 6th 2010, 01:03 AMgralla55
Thanks for your reply! But how do you get:

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

Thanks again. - May 6th 2010, 02:26 AMHallsofIvy
- May 6th 2010, 03:08 AMgralla55
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 AMtonio
- May 6th 2010, 04:18 AMgralla55
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 AMgralla55
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 AMtonio
- May 6th 2010, 06:58 AMgralla55
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 AMgralla55
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 ?