I've already asked this question here, but theres a misconception about the phrase "change of basis matrix from B to C", and I think the answers were given in the inverse of what's in my book.

So, in my book, the change of basis matrix from $B$ to $C$ is the matrix $M$ such that

$$[ \ \ \ ]_B = M[\ \ \ ]_C$$

where $M$ has the vectors $c_1,c_2$ of the base $C$ written as a linear combination of the basis $B$.

So, the question is:

The change of basis matrix from $B = \{1+t, 1-t^2\}$ to the base $C = \{c_1, c_2\}$ is:

$$\begin{bmatrix}\color{Red}{1} & \color{Blue}{2}\\\color{Red}{1} & \color{Blue}{-1}\end{bmatrix}$$

Find basis $C$.

So what I did was:

$$c_1 = \color{Red}{1}(1+t) + \color{Red}{1}(1-t^2) = 2 + t -t^2\\c_2 = \color{Blue}{2}(1+t) \color{Blue}{-1}(1-t^2) = 1 + 2t + t^2$$

Am I rigth?

  • $\begingroup$ Not quite; $c_1$ and $c_2$ should be multiplied by $\frac13$, and the end of your computation for $c_2$ is wrong. $\endgroup$ – Bernard Feb 1 '15 at 21:50
  • $\begingroup$ @Bernard why 1/3? $\endgroup$ – Guerlando OCs Feb 1 '15 at 22:28
  • $\begingroup$ When you compute $M^{-1}$, there's a factor $\frac13$. Actually, $M^{-1}=\frac13M$, $\endgroup$ – Bernard Feb 1 '15 at 22:31
  • $\begingroup$ @Bernard why I need to compute $M^{-1}$? I've already found $c_1$ and $c_2$. Are you using the definition I have in the exercise? Thank you! $\endgroup$ – Guerlando OCs Feb 1 '15 at 22:34
  • $\begingroup$ But if $M$ has $c_1$ and $c_2$ expressed with $b_1$ and $b_2$, $M^{-1}$, which is the change of base matrix from $\mathcal C$ to $\mathcal B$ has $b_1$ and $b_2$ expressed with $c_1$ and $c_2$. $\endgroup$ – Bernard Feb 1 '15 at 22:39

This question has been answered in comments. Moving to community wiki so it doesn't hang around forever in an unanswered state.


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