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Problem: Let $D: P_3 (R) \rightarrow P_3(R): p \mapsto p'$ be a linear transformation that maps a polynomial onto its derivative. De matrix representation of this linear map with respect to the standard basis $\beta_S = \left\{1, x, x^2, x^3\right\}$ is given as \begin{align*} A= \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix}. \end{align*} Now find the matrix representation of this linear transformation with respect to the new basis $\beta = \left\{1, x,x(x-1),x(x-1)(x-2)\right\}$.

Attempt at a solution: There's the equation $[T]_{\beta} = Q^{-1} [T]_{\beta_S} Q$. We have $[T]_{\beta_S} = A$, so we need $Q$. I tried to calculate the derivatives of the basis vectors of $\beta$. But then I just got a matrix with one column all zero's, which is not invertible. Anyone got some advice?

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Hint:

Notice that each vector in $\beta$ is a linear combination of vectors in $\beta_s$: this defines what $\beta_s$ is in terms of $\beta$. So, to find $Q$, simply write out the matrix whose columns represent these linear combinations. The rest is as you say.

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  • $\begingroup$ Thanks, but do I always follow this strategy to solve problems like these? For example if I'm given a linear transformation (with domain and codomain equal), then I just express the new basis as a linear combination of the old one? And then find Q from that? $\endgroup$ – Kamil Jan 10 '15 at 10:51
  • $\begingroup$ @Kamil, you don't even need the linear transformation to be invertible; you only need it to map between two vector spaces of the same dimension, and to be able to express one basis in terms of the other. $\endgroup$ – ki3i Jan 10 '15 at 12:08

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