linear maps

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March 9th 2008, 06:16 AM
Deadstar

linear maps

Let T be the linear map on $P_2$ defined by
$T(p)(t) = (1+t)p^{'}(t)$
Determine the matrix of T with respect to the basis $(1, t, t^2)$ of $P_2$ .
What are the eigenvalues of T? Is T diagonalisable?
March 9th 2008, 09:23 AM
Plato

I will help with the basic matrix.
From $T(a + bt + ct^2 ) = (1 + t)(b + 2ct) = \left( {b + (b + 2c)t + 2ct^2 } \right)$ we get the matrix
$\left[ {\begin{array}{ccc}<br /> 0 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \\ \end{array}} \right]$
March 9th 2008, 10:50 AM
CaptainBlack

Quote:

Originally Posted by Deadstar

Let T be the linear map on $P_2$ defined by
$T(p)(t) = (1+t)p^{'}(t)$
Determine the matrix of T with respect to the basis $(1, t, t^2)$ of $P_2$ .
What are the eigenvalues of T? Is T diagonalisable?

The columns of the matrix are the images of each of the basis vectors under
the transformation (in component form with respect to the basis)

RonL
March 9th 2008, 11:11 AM
Deadstar

cheers guys. One more question, what does the symbol that looks like a "U to the power 'an upside down T'" mean? I dont know how to notate that with latex.
March 9th 2008, 11:58 AM
Chloroform

lol, you miss those lectures too? I'm doing the same course.

The 'U to the power 'an upside down T' thing you speak of is the orthogonal complement of U. The orthogonal compliment of a subspace U of an inner product V is the set of all vectors in V that are orthogonal to every vector in U.

Know Q3 yet? :p
March 9th 2008, 12:09 PM
Deadstar

no ive not really looked at it yet, Q.1s sorted. Trying to do the Stats assignment as well, bloody hard...
March 9th 2008, 06:39 PM
lllll

For $T(p)(t) = (1+t)p^{'}(t)$

is $T$ invertible?

& would $T^{-1} = -b - (b+2c)t - (2c)t^2$ ?
March 9th 2008, 09:01 PM
CaptainBlack

Quote:

Originally Posted by lllll

For $T(p)(t) = (1+t)p^{'}(t)$

is $T$ invertible?

& would $T^{-1} = -b - (b+2c)t - (2c)t^2$ ?

No $T$ is not invertible, the matrix of $T$ has zero determinant.

Another way to look at it is that if it were invertibable then for every quadratic function $f(x)$ (quadratic function here means a polynomial of degree not more than 2):

$p(t)=\int_0^t \frac{f(\zeta)}{1+\zeta} ~d\zeta + p_0$

would be a unique quadratic, but it is neither unique nor always a quadratic
(put $f(x)=1$ to see that it is not always a quadratic)

RonL