I have the following task and just wanted to check weather this is (written) correct(ly).
Let $V$ be the vector space of all polynomials of grade $\le 3$ and $f: V \rightarrow V, p \rightarrow p'$ an $\mathbb{R}-$linear map. Calculate the transformation matrix using the basis $B=(1,x,x^2,x^3)$.
Well first I get the image of the basis-vectors:
$$f(1) = 0, \ f(x) = 1, \ f(x^2) = 2x, \ f(x^3) = 3x^2 $$
The next step is to express those using the canonical basis in $\mathbb{R}^3$, right? So what I did was:
$$f(1) = 0 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \lambda_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + \lambda_2 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \lambda_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \lambda_4 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \Rightarrow \lambda_1 = \lambda_2 = \lambda_3 = \lambda_4 = 0$$
$$f(x) = 1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \lambda_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + \lambda_2 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \lambda_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \lambda_4 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \Rightarrow \lambda_2 = \lambda_3 = \lambda_4 = 0, \ \lambda_1 = 1 $$
$$f(x^2) = 2x = \begin{pmatrix} 0 \\ 2 \\ 0 \\ 0 \end{pmatrix} = \lambda_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + \lambda_2 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \lambda_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \lambda_4 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \Rightarrow \lambda_1 = \lambda_3 = \lambda_4 = 0, \ \lambda_2 = 2 $$
$$f(x^3) = 3x^2 = \begin{pmatrix} 0 \\ 0 \\ 3 \\ 0 \end{pmatrix} = \lambda_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + \lambda_2 \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix} + \lambda_3 \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \lambda_4 \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \Rightarrow \lambda_2 = \lambda_1 = \lambda_4 = 0, \ \lambda_3 = 3 $$
Which results into the following transformation matrix:
$$M_B^B = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0& 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$
My problem is that I am not quite sure whether I've placed the resulting images on the right row. E.g. for $f(x^3) = 3x^2$:
$$ 3x^2 = \begin{pmatrix} 0 \\ 0 \\ 3 \\ 0 \end{pmatrix} \text{ or } 3x^2 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 3 \end{pmatrix}$$
I am quite sure its the first one since one has to order the coefficients in regards to the basis one uses - but I thought it'd be better to check that.
Than you very much for your help.
FunkyPeanut
Let $V$ be a vector space and $T$ an operator on $V$. Since for the basis $(v_1,...,v_n)$ of $V$, $T$ maps each $v_k$ as a linear combination $Tv_k =a_{1,k}v_1 \ + \ ...\ + \ a_{n,k}v_n$ and is represented by the matrix
$$\left[\begin{matrix} a_{1,1}&\cdots& &a_{1,n} \\ \vdots& \ddots & &\vdots\\ a_{n,1}&\cdots & & a_{n,n} \end{matrix}\right]$$
We see that for the your example (extended to $n+1$ dimensions) $f$ maps each $x^k$ to $kx^{k-1}$, which is the linear combination $0 \cdot 1 \ +\ ... \ + \ kx^{k-1} + \ ... \ + \ 0 x^n$
So the matrix representation would be
$$\left[\begin{matrix} 0 & 1& \cdots& &0 \\ \vdots&&\ddots&&\vdots\\ &&&&n\\ 0&&\cdots&&0 \end{matrix}\right]$$
Which is what you have.