Question

Let Pn be the vector space of polynomials of degree at most n with real coefficients. Let \[V = P _{2} and W = P _{4}\]Let f : V -> W be the function
defined by \[(f(P))(x) = P(x ^{2})-xP(x)\]Find the matrix of f with respect to the bases \[[x^{2}, x,1] \in V\]and\[[x^{4},x^{3},x^{2},x,1] \in W\]

If someone could explain to me the method for doing this, I would be really greatful!

Thankyou :)

Answer 1

It'll take me a couple more minutes to fully re-familiarize myself with the process of this, but the most important part, is to know where the function sends the basis. So you would look at \(f(1)\), \(f(x)\), and \(f(x^2)\) in terms of \(\{1,x,x^2,x^3,x^4\}\).

Answer 2

Okay, I understand that that is what you have to do, but I don't understand actually how to do it!
That's okay, no rush, I'm just learning through it for self study :)

Answer 3

So we have\[f(1)=1-x\]\[f(x)=x^2-x^2=0\]\[f(x^2)=x^4-x^3\]Your matrix will by a 3x5 matrix, and the numbers in it will correspond to different coefficients in that set of equations we have up there.

Answer 4

Sorry, 5x3 matrix.

Answer 5

Apply the matrix to the basis vectors and use these results for your columns.

Answer 6

The first column corresponds to the first equation, and will be [1,-1,0,0,0]. The second column corresponds to the second equation, and since it's 0, everything will be 0.

Making more sense now?

Answer 7

@sarahusher consider that applying a matrix to a vector yields another vector in its *column space*, a linear combination of its columns. Consider, then,$$\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\\j&k&l\\m&n&o\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=x\begin{bmatrix}a\\d\\g\\j\\m\end{bmatrix}+y\begin{bmatrix}b\\e\\h\\k\\n\end{bmatrix}+z\begin{bmatrix}c\\f\\i\\l\\o\end{bmatrix}$$... so clearly applying to a basis vector yields one of our matrix columns; in the case \(x=1,y=z=0\) we get the first column, while for \(y=1,x=z=0\) yields the second, etc. This tells us that our matrix columns are actually the transformed basis vectors of our domain vector space. Here, apply your transformation to those basis vectors of \(\mathbb{P}_2\) to get the columns to use for our matrices.

Answer 8

It is
But, I don't understand where
f(1)=1−x
f(x)=x^2−x^2=0
f(x2)=x^4−x^3

comes from? Sorry if it is really obvious :(

Answer 9

The \(f\) is just the function \(f(P(x))=P(x^2)-xP(x)\). Then I use the set \(\{1,x,x^2\}\) as the different possible polynomials \(P(x)\). So if \(P(x)=x^2\), we get \(f(P(x))=x^4-x^3\).

Answer 10

oh i see!!!
I feel silly now!
Okay that makes sense!
So If I do that, I should end up with a matrix
\[\left[\begin{matrix}1 & 0 &0\\ -1 & 0&0\\0&0&0\\0&0&-1\\0&0&1\end{matrix}\right]\]

Is that right?

Answer 11

That's exactly what I came up with.

Answer 12

nice work @KingGeorge @sarahusher do you understand

Answer 13

Yes I do! Thankyou so much much @KingGeorge @oldrin.bataku , that makes a lot of sense now :)