# Finding bases for a Linear Transformation of a Matrix

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: $$\begin{pmatrix} 1 & 0 & 0 & \frac23 \\ 0 & 1 & 0 & -\frac13 \\ 0& 0 & 1 & \frac23 \end{pmatrix}$$

• Hint: Can you find an invertible matrix $B$ such that $BA$ is the rref of $A$?
– amd
Mar 4, 2016 at 19:47
• Thank you! I have figured out B. Is that all that has to be done?
– user319635
Mar 4, 2016 at 20:17
• Oops, I still have to find my bases alpha and beta, how do i go about this?
– user319635
Mar 4, 2016 at 20:27
• 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.
– amd
Mar 4, 2016 at 20:37

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 $B^{-1}$.

• So then does this mean $\beta = [1, 0, 1], [2, 1, 1], [0, 2, 1]$ ? I am a bit confused what $\alpha$ is
– user319635
Mar 4, 2016 at 23:49
• Wait but this isn't a polynomial of degree 2?
– user319635
Mar 5, 2016 at 0:03
• @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 $\beta = (1 + x^2,2 + x + x^2, 2x + x^2)$. Mar 5, 2016 at 0:12
• Right! Thank you so much!
– user319635
Mar 5, 2016 at 0:29