Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm solving a linear algebra problem. I have linear transformation $D$:

$D : R_2[t] \rightarrow R_2[t]$

$D(p) = \frac{d}{dt}p$

and bases:

$A = \{1 + t, 1- t, t^2\}$

$B = \{1 + t, 1 - t\}$

Now I need to discover matrix of linear transformation $D$ from $A$ to $B$.

Well, I started up by writing down a typical polynom in canonical base:

$p(t) = a + bt + ct^2$

Then I tried to discover what would be its representation in base $A$, by doing:

$a + bt +ct^2 = x(1 + t) + y(1 - t) +zt^2$

So,

$[p(t)]_A = \left(\frac{a+b}{2}, \frac{a-b}{2}, c\right)$

Good. Now, I know that:

$D(p(t)) = b + 2c$

Then:

$[D(p(t))] = (b, 2c) = \begin{bmatrix} b \\ 2c \\ \end{bmatrix} $

Then

$ [D(p(t))] = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}$

So,

$ [D(p(t))]_A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ \end{bmatrix} \begin{bmatrix} \frac{a+b}{2} \\ \frac{a-b}{2} \\ c \\ \end{bmatrix}$

Ok is clear not the transformation linear from $A$ to $B$, but only this representation from the canonical base to $A$, only? What should I do? Thanks.

share|improve this question
add comment

1 Answer

It looks like you're making it too difficult on yourself. To find the matrix of a linear transformation, it suffices to know where your basis vectors get mapped to, and then find those in your new basis.

So, we know that

$1 + t \mapsto 1$

$1 - t \mapsto -1$

$t^2 \mapsto 2t$.

Now we need to find these represented in the new basis $\beta = \{ 1 + t, 1 - t \}$

It's pretty clear to see that $\frac{1}{2}[(1+t) + (1 - t)] = 1$, so you will have the coordinate vector for $D(1+t)_\beta = (\frac{1}{2},\frac{1}{2})^t$, and similarly for $D(1-t)_\beta = (-\frac{1}{2},-\frac{1}{2})^t$.

So we now must find a linear combo that yields $2t$, which would be $(1+t)-(1-t)$, so our new coordinate vector is $D(t^2)_\beta = (1,-1)^t$.

Putting these three into the columns of a matrix, we get that the matrix of the linear transformation is: $$M = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} & 1 \\ \frac{1}{2} & -\frac{1}{2} & -1 \end{pmatrix}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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