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Suppose $T$ is an element of $L(P_3(\Bbb R), P_2(\Bbb R))$ is the differentiation map defined by $Tp = p'$. Find a basis of $P_3(\Bbb R)$ and a basis of $P_2(\Bbb R)$ such that the matrix of T with respect to these basis is $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}$$
In general, how do I find the bases given a matrix of a linear map? Thanks!
With $e_1={1\over 3}x^3, e_2={1\over 2}x^2, e_3=x, e_4=1$ represented by vectors
$$\begin{pmatrix}1/3 \\ 0 \\0\\0 \end{pmatrix}, \begin{pmatrix}0 \\ 1/2 \\0\\0 \end{pmatrix}, \begin{pmatrix}0 \\ 0 \\1\\0 \end{pmatrix}, \begin{pmatrix}0 \\ 0 \\0\\1 \end{pmatrix}$$
relative to the standard basis, and a basis for the target space as usual just $x^2, x, 1$ represented by the usual basis vectors.
The map in this basis is exactly what you're looking for.
More generally, it's pretty straightforward to write linear transformations in different bases, $\alpha$ of the domain and $\beta$ of the range: start by writing the domain's basis in terms of the standard basis and getting a matrix $T_\alpha$ which changes the standard basis into the basis, $\alpha$. Next find a matrix $T_\beta$ which changes the standard basis on the domain space into the basis you want. Finally find a matrix $T_0$ which does the job on the standard basis.
Then your map is $T=T_\beta T_0 T_\alpha^{-1}$, which you can verify by testing it on a basis, $T_\alpha^{-1}$ takes an alpha basis vector, turns it into a standard one, $T_0$ does the work of the transformation, and $T_\beta$ writes it in the alternate basis.
