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For the question below, I am not sure how to find the desired matrix relative to the standard basis. $P_3$ refers to polynomials with up to degree $3$

Thanks in advance

for the linear map

$$T:P_3(\mathbb{R})\rightarrow\mathbb{R^4}, p\mapsto \big(p(0), p(1), p\prime(0), p\prime(1) \big)$$

a) Find the matrix $[T]_\beta^\gamma$ of $T$ relative to the standard bases of $P_3(\mathbb{R})$ and of $\mathbb{R^4}$

b) Show that $T$ is an isomorphism

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  • $\begingroup$ Do you know the definition of a linear transformation's matrix with respect to certain basis? If so, what have you tried so far? $\endgroup$
    – qualcuno
    Nov 4 '17 at 2:25
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Let $B := \{1, X, X^2, X^3\}$ be the standard basis of $\mathbb{R}_{\leq3}[X]$, and $B' = \{e_1, e_2, e_3, e_4\}$ the standard basis of $\mathbb{R}^4$. Now, by definition, $[T]_{B}^{B'}$ has on column $i$ the coordinates in $B'$ of the image of the $i$-th vector of $B$. So, since $B'$ is the canonical basis, taking coordinates is redundant: it suffices to place $T(X^{i-1})$ in each column $i$. So,

$$ T(1) = (1,1,0,0)\\ T(X) = (0,1,1,1)\\ T(X^2) = (0,1,0,2)\\ T(X^3) = (0,1,0,3) $$

and therefore,

$$[T]_{B}^{B'}= \begin{bmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 1 & 1\\ 0 & 1 & 0 & 0\\ 0 & 1 & 2 & 3 \end{bmatrix} $$

To finish we must conclude that $T$ is an isomorphism. This follows from the fact that $[T]_{B}^{B'}$ is invertible, since

$$\det \begin{bmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 1 & 1\\ 0 & 1 & 0 & 0\\ 0 & 1 & 2 & 3 \end{bmatrix} = \det \begin{bmatrix} 1 & 1 & 1\\ 1 & 0 & 0\\ 1 & 2 & 3 \end{bmatrix} = \det \begin{bmatrix} 1 & 1\\ 2 & 3 \end{bmatrix} = 1 \neq 0 $$

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