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Suppose a linear transformation $T $: $P_3(\Bbb R)$ to $ P_2(\Bbb R))$ has the matrix $$A=\begin{pmatrix} 1 & 2 & 0 & 0 \\ 0 & 1 & 2 & 1 \\ 1& 1 & 1 & 1 \end{pmatrix}$$ relative to the standard bases of $P_3(\Bbb R)$ and $ P_2(\Bbb R))$.

Find bases $\alpha$ of $P_3(\Bbb R)$ and $\beta$ of $ P_2(\Bbb R))$ such that the matrix $T$ relative to $\alpha$ and $\beta$ is the reduced row echelon form of A.

I have been looking everywhere for a similar example. I am struggling with where to begin.

I have calculated that the reduced row echelon form of A is: $$\begin{pmatrix} 1 & 0 & 0 & \frac23 \\ 0 & 1 & 0 & -\frac13 \\ 0& 0 & 1 & \frac23 \end{pmatrix}$$

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    $\begingroup$ Hint: Can you find an invertible matrix $B$ such that $BA$ is the rref of $A$? $\endgroup$
    – amd
    Mar 4 '16 at 19:47
  • $\begingroup$ Thank you! I have figured out B. Is that all that has to be done? $\endgroup$
    – user319635
    Mar 4 '16 at 20:17
  • $\begingroup$ Oops, I still have to find my bases alpha and beta, how do i go about this? $\endgroup$
    – user319635
    Mar 4 '16 at 20:27
  • $\begingroup$ Interpret $B$ as a change-of-basis matrix. Note that the solution isn’t unique. You can start with pretty much any basis for $P_3(\mathbb R)$ and find a corresponding basis for $P_2(\mathbb R)$ so that the matrix of $A$ has the desired form. $\endgroup$
    – amd
    Mar 4 '16 at 20:37
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You have found a product of elementary matrices $B$ such that $\operatorname{rref}(A) = BA$. If we denote the matrix of a transformation $T$ in the bases $B,C$ (for the domain, codomain respectively) by $[T]_B^C$, we have in our situation $$[\mathit{id}]_\mathit{st}^\beta [T]_\mathit{st}^\mathit{st} = [T]_\mathit{st}^\beta = \operatorname{rref}(A) = BA $$

for some unknown basis $\beta$. (And where $\mathit{st}$ represents either of the standard bases, and $\mathit{id}$ is the identity transformation.) This suggests that we take $\alpha = \mathit{st}$, and $\beta$ to be the unique basis such that $$[\mathit{id}]_\mathit{st}^\beta = B,$$ or $$[\mathit{id}]_\beta^\mathit{st} = B^{-1}.$$

This means that the coordinates of the vectors in $\beta$ in the standard basis are simply the columns of $B^{-1}$.

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  • $\begingroup$ So then does this mean $\beta = [1, 0, 1], [2, 1, 1], [0, 2, 1]$ ? I am a bit confused what $\alpha$ is $\endgroup$
    – user319635
    Mar 4 '16 at 23:49
  • $\begingroup$ Wait but this isn't a polynomial of degree 2? $\endgroup$
    – user319635
    Mar 5 '16 at 0:03
  • $\begingroup$ @user319635 $\alpha$ is the standard basis $(1,x,x^2,x^3)$. And not quite; these are the coordinates of the $\beta$-vectors in the standard basis. Thus $\beta = (1 + x^2,2 + x + x^2, 2x + x^2)$. $\endgroup$ Mar 5 '16 at 0:12
  • $\begingroup$ Right! Thank you so much! $\endgroup$
    – user319635
    Mar 5 '16 at 0:29

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