[Math] Linear transformations and polynomials

linear algebra

I've been having a really hard time recently understanding linear transformations. I would really like to know conceptually what exactly is going on.

Let $T: P_3 \rightarrow P_2$ be defined by

$$T(a + bx + cx^2 +dx^3) = ax^2 + b$$

$$B = \{1,x,x^2,x^3\} $$

$$B' = \{1-x+x^2, 1-x, -1\} $$

Find the matrix $[T]_{B',B}$ of the linear transformation $T$ with respect to the bases $B$, $B'$.

  • Now, I know that $T$ is an operator and that the question is asking for you to find the transformation matrix from $B'$ to $B$

  • So to start, if I'm going from from $B'$ to $B$, I should first find these three transformations and then, once I have that, I can put them into a matrix and that should be the matrix $[T]_{B',B}$.
    $$T(1-x+x^2)$$
    $$T(1-x)$$
    $$T(-1)$$

  • Is this correct? What's confusing me is that in order to find these transformations, I need to know how the transformation is acting on the input.

  • I know that if I was to put the polynomials into a matrix, this transformation is essentially:

$$A \vec{P} = \vec{p}$$

where $A$ is the transformation in matrix form, $\vec{P}$ is the polynomial input, and $\vec{p}$ is the output polynomial.

  • If this is the case then I can solve for $A$ and then apply that matrix to the other polynomials, but I'm not exactly sure how to set that up. Can I get some pointers? Thank you very much.

Best Answer

I think there should be a $x$ in front of $b$ in your first equation defining the action of $T$.

To get the matrix form of $T$ with respect to the bases $B$ and $B'$, we must evaluate $T$ on each element of $B$ and write the result in term of elements of $B'$. To be more precise, let $\mathrm{e}_1:=1-x+x^2$, $\mathrm{e}_2:=1-x$ and $\mathrm{e}_3:=-1$ ; we compute : $$T(1)=x^2=\mathrm{e}_1-\mathrm{e}_2,$$ $$T(x)=x=-\mathrm{e}_2-\mathrm{e}_3,$$ $$T(x^2)=T(x^3)=0.$$ Thus the representative matrix of $T$ relatively to the bases $B$ and $B'$ is given by \begin{align} \begin{pmatrix}1&0&0&0 \\-1&-1&0&0\\ 0&-1&0&0 \end{pmatrix}. \end{align} We can see that the first column is the coordinates vector of $T(1)$ in the basis $B'$, that the second column is the coordinates vector of $T(x)$ in this same basis, and so on.

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