Let $f:U \rightarrow V$ be a linear map where $U$ and $V$ are finite-dimensional vector spaces. Let $U$ be the vector space of polynomials degree $3$ in variable $t$. $F:U \rightarrow \mathbb{R}$ be a linear map defined by $f(u)=u'(1)$. Find the matrix of $f$ with respect to bases $\{1,t,t^2,t^3\}$ and $\{1\}$.
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$\begingroup$ I don't know if you have to do f(1) and then f(t) etc for elements of the basis? $\endgroup$– HannahMay 6, 2016 at 9:29
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$\begingroup$ Yes, and then these are the coefficients of your matrix (and this works for any map between two finite-dimensional vector spaces). Check out the answer below. $\endgroup$– B. PasternakMay 6, 2016 at 9:30
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$\begingroup$ Thankyou that makes sense! $\endgroup$– HannahMay 6, 2016 at 9:32
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1 Answer
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The matrix will have the shape $$\begin{bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \end{bmatrix}$$ Where $a_i=f(t^i)$.
For example $a_0=f(1)=(t\mapsto 1)'(1)=(t \mapsto 0)(1)=0$ and $a_1=f(t)=(t\mapsto t)'(1)=(t \mapsto 1)(1)=1$ and $a_2=f(t^2)=(t\mapsto t^2)'(1)=(t \mapsto 2t)(1)=2$.
So the matrix is : $$\begin{bmatrix} 0 \\ 1 \\ 3 \\ a_3 \end{bmatrix}$$
I will let you calculate $a_3$ in order to complete the matrix.