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I having trouble solving this problem:

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For the change of basis matrix of this transformation, I plugged in each entry of the basis $\{t_2, t, 1\}$ in for $T(p(t))$ like so:

$T(t^2)$ = some $2\times 2$ matrix... find the weights $c_1 \ldots c_n$ using the basis for $M(R)$ and put the weights in a vector

$T(t) =$ same

$T(1) =$ same

Then I am not sure on how to continue. I know that $T(t^2)$ will output some $2\times 2$ matrix, but how do I actually solve something like $T(t^2)$?

My attempt:
$T(t^2) = p(t) = t^2$, and $p(0) = 0, p(1) = 1, 0 = 0$, and $p'(0) = 0$ so the $2\times 2$ would be:

$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$

For the size, I am guessing it wants the $m \times n$ size, I know for sure this change of basis matrix will be a $4 \times 3$ matrix.

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1 Answer 1

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The $i$th column of $[T]_{\beta,\gamma}$ (the matrix of $T$ with respect to $\beta$ and $\gamma$) is given by $[T(b_i)]_\gamma$, the coordinate vector of $T(b_i)$ with respect to $\gamma$, where $b_i$ is the $i$th vector of $\beta$. So for instance $$ T(b_1)=T(t^2)=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ and so $[T(b_1)]_\gamma=[0,1,0,0]^\top$.

So do the same for $[T(t)]_\gamma$ and $[T(1)]_\gamma$ to find the other columns of $[T]_{\beta,\gamma}$, and hence determine its size at the same time.

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For T(b2) = T(t), I am not sure for p'(0). I derive t, which gives me 1, so p'(t) = 1, and plugging in 0 doesn't do anyting, and so it is 1? –  Dog Dec 18 '12 at 5:51
1  
@Dog, yes, $p'(t)=1$ means $p'(t)$ is $1$ for any value of $t$, including $t=0$. –  yunone Dec 18 '12 at 5:59
    
thanks yunone!! –  Dog Dec 18 '12 at 6:07
    
@Dog Glad to help. –  yunone Dec 18 '12 at 6:20

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