If T: P1 -> P1 is a linear transformation such that T(1 + 2x) = 4 + 3x and T(5 + 9 x) = -2 - 4x, then T(4 - 3 x) =?
I started off with expressing (4-3x) as a linear combination of the two other polynomials:
c1(1+2x) + c2(5+9x) = 4-3x.
I then solved the equation with gauss, which gave me: c1 = -42 and c2= 23. To solve the equation i continued with: T(4-3x) = { T(-42)(1+2x) + 23(5+9x)}.
This is where I am stuck though. How do I proceed to figure out what's on the right hand side of T(4-3x)?
The linear map $T$ can be understood as 2x2 matrix in the following sense. Take the canonical basis in $P_1$, that is: $B=\{1,x\}$ (why is this a basis?) Then the linear map $T$ is fully defined by defining the image of $T$ of the elements of $B$. That is, we need to understand $T(1)$ and $T(x)$. Let me try to clearify this. Since any element of $P_1$ can be written as linear combination of $1$ and $x$ in a unique way, we may identify \begin{align} p=a+bx = a (1) + b(x) \longleftrightarrow p = \begin{pmatrix} a\\b \end{pmatrix}. \end{align} Now we can write \begin{align} T(4-3x) = \begin{pmatrix} T_{11} & T_{12}\\T_{21}&T_{22}\end{pmatrix} \begin{pmatrix} 4\\-3 \end{pmatrix}. \end{align} Now we need to compute all the $T_{ij}$s. Lets start! We are given $T(1+2x)=4+3x$ and $T(5+9x)=-2-4x$. In our notation, this means: \begin{align} \begin{pmatrix} T_{11} & T_{12}\\T_{21}&T_{22}\end{pmatrix} \begin{pmatrix} 1\\2 \end{pmatrix} = \begin{pmatrix} 4\\3 \end{pmatrix} \text{ and }\begin{pmatrix} T_{11} & T_{12}\\T_{21}&T_{22}\end{pmatrix} \begin{pmatrix} 5\\9 \end{pmatrix} = \begin{pmatrix} -2\\-4 \end{pmatrix} \qquad \qquad (\ast) \end{align} The upper two lines of these equations read \begin{align} T_{11}+2T_{12}=4 \text{ and }5T_{11}+9T_{12}=-2. \end{align} Do Gauß here, or invert the matrix (whatever you like more)! You should find $T_{11}=-40$ and $T_{12}=22$. Analogously, you proceed to find $T_{21}=-35$ and $T_{22}=19$ (using the lower two equations from ($\ast)$. If you have done this, you have full knoledge of $T$. The answer to you question then simply reads \begin{align} T(4-3x) = \begin{pmatrix} T_{11} & T_{12}\\T_{21} & T_{22}\end{pmatrix}\begin{pmatrix}4\\-3 \end{pmatrix} = (4T_{11} -3T_{12}) + (4T_{21}-3T_{22})x = -226-197x. \end{align} I left out the Gauß part, so you can practice (I hope that is ok). Note, that one should be more careful with equality signs, but I chose to ignore this ('canonical' isomorphism between $\mathbb{R}^2$ and $P_1$.