0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.