Using basis $e=[x^3,x^2,x,1]$ instead of $e=[1,x,x^2,x^3]$

Question

So on an exam I've got zero points on the question (and sub-questions) to find matrix of linear operator $L:\Bbb{R}^4[x]\to \Bbb{R}^4[x]$ given by $L(p(x)) = p(x)+xp(2)$ with respect to canonical basis $e$

I've said I'm using notation $(a,b,c,d)$ to mean $(ax^3,bx^2,cx,d)$ I've found the matrix for $L$ lets say $A$ which is $$A=\begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\8 & 4 & 3 & 1\\0 & 0 & 0 & 1\end{bmatrix}$$ Now $$\begin{bmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\8 & 4 & 3 & 1\\0 & 0 & 0 & 1\end{bmatrix}\begin{bmatrix}a \\ b \\ c \\ d\end{bmatrix}=\begin{bmatrix}a \\ b \\ 8a+4b+3c+d \\ d\end{bmatrix}$$ Which is the right result (using my notation), however they've got a different matrix by using $(a,b,c,d) = (a,bx,cx^2,dx^3)$ They found $$A=\begin{bmatrix}1 & 0 & 0 & 0\\1 & 3 & 4 & 8\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{bmatrix}$$ Which is again the right answer (using their notation), so I'm looking for references to use of the first notation or references/reasons to why my notation is wrong.

Answer 1

• First : "I'm using notation $(a,b,c,d)$ to mean $(ax^3,bx^2,cx,d)$" doesn't make any sense. You wanted to say, "I'm using the notation $(a,b,c,d)$ to denote $ax^3+bx^2+cx+d$".

• Secondly : In your notation, does for example $x$ refer to the vector $\begin{pmatrix}0\\1\\0\\ 0\end{pmatrix}$ or to the vector $\begin{pmatrix}0\\0\\1\\0\end{pmatrix}$ ? Because to use the basis $\{x^3,x^2,x,1\}$ for $$\left\{\begin{pmatrix}0\\0\\0\\1\end{pmatrix}, \begin{pmatrix}0\\0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\\0\end{pmatrix}, \begin{pmatrix}1\\0\\0\\0\end{pmatrix}\right\}$$ is correct, but using $\{x^3,x^2,x,1\}$ for $$\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\0\\1\end{pmatrix}\right\}$$ is conventionally not correct but makes sense with your notation (and it's what you seem to have made).

But your work didn't deserve 0 mark. In my opinion, you deserve almost all points for this question (since you precise that $(a,b,c,d)$ refer to $ax^3+bx^2+cx+d$). Strange that you got 0 mark !

Answer 2

It makes no sense to ask what is the matrix of a linear transformation without fixing bases. If the person to whom this question is being asked is free to choose those bases, then both answers are correct.

Answer 3

If $\bf J$ denotes the Exchange matrix $$ {\bf J} = {\bf J}^T = {\bf J}^{\, - 1} = \left( {\matrix{ 0 & 0 & 0 & 1 \cr 0 & 0 & 1 & 0 \cr 0 & 1 & 0 & 0 \cr 1 & 0 & 0 & 0 \cr } } \right) $$ then $$ \left( {\matrix{ {x^{\,3} } \cr {x^{\,2} } \cr {x^{\,1} } \cr {x^{\,0} } \cr } } \right) = {\bf J}\left( {\matrix{ {x^{\,0} } \cr {x^{\,1} } \cr {x^{\,2} } \cr {x^{\,3} } \cr } } \right) $$ and between the matrix you have found and the expected one there is a change of basis relation $$ {\bf A} = {\bf J}\,{\bf A'}\;{\bf J} = {\bf J}\,{\bf A'}\;{\bf J}^{\,{\bf - 1}} = {\bf J}^{\,{\bf - 1}} \,{\bf A'}\;{\bf J} $$

So, if you have specified the base you were considering, your answer is right and shall be accepted.

Answer 4

The other answers so far are missing a key point: the exam question as you’ve presented it here requires you to use the “canonical basis.” That’s a very specific requirement. If the basis that you chose to use was different—remember that these are ordered bases, so the same vectors in a different order is a different basis—then you did not answer the question correctly. Outside references to support your approach are pretty much irrelevant, I think. What matters for this particular exam is the definition of “canonical basis” that was used in that course.

For my part, I would’ve given you partial credit instead of a zero for working through a solution, but you did fail to answer the specific question that you were given.