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.
2 Answers
$\begingroup$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.
$\endgroup$ 4 $\begingroup$When you are trying to understand what conceptually is going on you need to know one thing and that would be it: All about the matrix of transformation.
I believe the most important thing you are asking is "What exactly is going on?". Well, the best way of representing linear transformations which are in short black boxes that rotate, elongate or shorten a vector(and so the word transformations!) "linearly", which essentially means that the grid of the plane is not squished or curved, nor is any vector that lies on it. The most a transformations does is magnify(+/-) or rotate the plane.
Well what this essentially means is that any basis vector in the domain plane, when transformed linearly, retains the "coordinates". Coordinates of a vector are nothing but a vector's position on the plane with respect to the basis of the plane, technically those constants(to not sound bombastic if I say 'scalars') which when multiplied to the basis vectors give you the the vector in hand.
The 'linearity' of the transformations guarantee you this one amazing thing: that these coordinates of any vector remain same. This means for any vector in the Domain space, its coordinates are going to be safely preserved even after it's transformation, which means that the image of the vector will have the same coordinates with respect to the images of the basis vectors. This is undoubtedly the best part because what it also means is that any vector's image can be written as the linear combination of the images of the domain's basis vectors,where in the linear combination the scalars are the coordinates of the vector with respect to the basis vectors.
Now what you see in the matrix that @Nicolas has provided you with is that the columns represent the coordinates of the images of domain's basis vectors which is what you get when you write the image of basis vectors as a scalar combination of vectors of $B′=\{1−x+x^2,1−x,−1\}$.
When you multiply the matrices, all that you essentially do is 'multiply the "first coordinate" of every basis vector's image with the invariant coordinate of the "input vector" (as discussed above due to 'linearity'). You will have as many columns as you have basis vectors of domain space. Now for every k'th row of coordinates in the matrix, it will have 'j' coordinates. This means that there are k basis vectors of image space and j basis vectors of domain space. When the j coordinates of the domain's basis is multiplied to the j coordinates of the input vector, we will get the output vector.
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