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)?

  • $\begingroup$ You use the fact that $T$ is linear. By the way, your computations are wrong. $\endgroup$ – José Carlos Santos Dec 15 '18 at 14:41
  • $\begingroup$ So I guess that the values I got after solving the equation was wrong, if that's what you mean. No matter how many times I try to gauss-eliminate, I still get -42, 23. $\endgroup$ – wznd Dec 15 '18 at 14:48
  • $\begingroup$ But you should have got $c_1=-51$ and $c_2=11$. $\endgroup$ – José Carlos Santos Dec 15 '18 at 14:53
  • $\begingroup$ Since $T$ is linear, you might want to understand it as a 2x2 matrix. In this sense, one has $T(1+2x)=T(1)+2T(x)$, where $1$ could be the unit vector in the first direction and $x$ the unit vector perpendicular to it.. You only need to understand $T(1)$ and $T(x)$. If I am not wrong, you should get $T(1)=-40-35x$ and $T(x)=22+19x$. Now you need to figure out, how to combine this to get $T(4-3x)$. $\endgroup$ – Caroline Dec 15 '18 at 14:56
  • $\begingroup$ Which I don't exactly understand, I've done it by hand and tried out several websites aswell and they all get the same result. My matrix: ( 1, 2 ,4 ) ( 5, 9 ,-3). The second paranthesis should be below the first, as a regular matrix, but I don't know how to do that in comments. $\endgroup$ – wznd Dec 15 '18 at 14:56

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 $P_1$.

  • $\begingroup$ Thank you so much for this. I'll definetly start practicing this. You've been a tremendous help! $\endgroup$ – wznd Dec 15 '18 at 16:03

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