# Discover matrix of linear transformation between non-canonical bases

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.

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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: $$M = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} & 1 \\ \frac{1}{2} & -\frac{1}{2} & -1 \end{pmatrix}$$

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