The question I'm trying to answer is:

Find an ordered basis $B$ of $\mathbb C_3[x]$, such that $$[p]^B= \begin{pmatrix}1\\0\\0\\i\end{pmatrix}$$ for the polynomial $p=2+2x+2x^2+2x^3$.

The notation I'm struggling with is $[p]^B$. What does that mean? I've also seen notations like $[p]_J^B$. I think it's something to do with basis, but when I read it... I'm blank.

I think in this context, $[p]^B$ is a basis for $p$. Is that correct?

Merry Christmas and thanks for help in advance.

  • $\begingroup$ Is $\mathbb C_3[x]$ all polynomials of degree $\leq 3$ with coefficients in $\mathbb C$? $\endgroup$ Dec 25, 2013 at 13:01
  • $\begingroup$ That's all the information we were given. $\endgroup$ Dec 25, 2013 at 13:03
  • $\begingroup$ I think your question is correct. $\endgroup$ Dec 25, 2013 at 13:05
  • $\begingroup$ The problem with that is that $\mathbb C_3[x]$ has dimension $4$, so the most natural definition of $[p]^B$ would be a column vector with $4$ elements, so it's unclear what the problem actually means by $[p]^B$. $\endgroup$ Dec 25, 2013 at 13:06
  • $\begingroup$ $[p]^B$ is a column vector, I just don't know how to format it. I've put it as a note on the question. $\endgroup$ Dec 25, 2013 at 13:07

1 Answer 1


Here $[p]^B$ means "$p$, written in terms of the basis $B$"; that is, it will be a vector whose entries are the coefficients of the elements of $B$ in the linear combination of elements of $B$ whose sum is $p$.

Almost certainly $[p]^B_j$ would then be the $j$-th entry of that vector, but I can't be sure without further context.


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