3

I'm trying to figure out how to do this, I've checked here and what I got from it is

Suppose you have an operator $D \equiv \frac{d}{dx}$ and some random basis $\vec{e} = \{ cos (ax), sin (bx) \} $

Then the operator $D$ would be represented in matrix notation as

$$D = \begin{pmatrix}0 &b \\ -a & 0 \end{pmatrix}$$

Is this it?

So it is safe to say the operator will always be a $n \times n$ matrix, where $n$ is the vector space's dimension?

Community
  • 1
chili and sea bass
  • 31
  • Also could anyone provide me with a link that would allow a fairly greater insight . – chili and sea bass Jan 11 '16 at 23:33
  • Are you sure that you'd like to consider $\cos(ax)$ and $\sin(bx)$ as a basis? If $a \neq b$, then $D$ doesn't map the (real) span of your basis into itself, i.e. $D:V \rightarrow W$ where $V \neq W$. – Roland Jan 12 '16 at 09:56

1 Answers1

4

Let $V$ be the subspace of smooth functions spanned by \begin{align*} e_1(t) &= \cos(at) & e_2(t)=\sin(bt) \end{align*} Let $W$ be the subspace of smooth functions spanned by \begin{align*} f_1(t) &= \cos(bt) & f_2(t)=\sin(at) \end{align*}

Let $d:V\to W$ be the linear operator defined by $df(t)=f^\prime(t)$. Note that \begin{array}{rcrcrcrc} de_1(t) & = & -a\sin(at) &=& \color{red}{0}\, f_1(t) &+& (\color{red}{-a})\,f_2(t) \\ de_2(t) &=& b\cos(bt) &=& \color{blue}{b}\,f_1(t) &+& \color{blue}{0}\,f_2(t) \end{array} This implies that the matrix representing $d$ in the given bases is $$ D= \begin{bmatrix} \color{red}{0} & \color{blue}{b}\\ \color{red}{-a} & \color{blue}{0} \end{bmatrix} $$ as expected.

Here's the general situation:

Let $T:V\to W$ be a linear map and let \begin{align*} \alpha &= \{v_1,\dotsc,v_n\} & \beta &= \{w_1,\dotsc,w_m\} \end{align*} be bases for $V$ and $W$ respectively. This means there are unique scalars $\lambda_{ij}$ for $1\leq i\leq m$ and $1\leq j\leq n$ such that $$ \begin{array}{ccccccccc} T(v_1) & = & \color{red}{\lambda_{11}}w_1 & + & \color{red}{\lambda_{21}}w_2 & + &\dotsb &+ & \color{red}{\lambda_{m1}}w_m \\ T(v_2) & = & \color{blue}{\lambda_{12}}w_1 & + & \color{blue}{\lambda_{22}}w_2 & + &\dotsb &+ & \color{blue}{\lambda_{m2}}w_m \\ \vdots & \vdots & \vdots &\vdots &\vdots &\vdots &\ddots&\vdots&\vdots \\ T(v_n) & = & \color{green}{\lambda_{1n}}w_1 & + & \color{green}{\lambda_{2n}}w_2 & + &\dotsb &+ & \color{green}{\lambda_{mn}}w_m \\ \end{array} $$ The matrix of $T$ relative to the bases $\alpha$ and $\beta$ is defined as $$ [T]_\alpha^\beta= \begin{bmatrix} \color{red}{\lambda_{11}} & \color{blue}{\lambda_{12}} & \dotsc & \color{green}{\lambda_{1n}} \\ \color{red}{\lambda_{21}} & \color{blue}{\lambda_{22}} & \dotsc & \color{green}{\lambda_{2n}} \\ \vdots & \vdots & \ddots & \vdots \\ \color{red}{\lambda_{m1}} & \color{blue}{\lambda_{m2}} & \dotsc & \color{green}{\lambda_{mn}} \end{bmatrix} $$ Note that this is an $m\times n$ matrix.

Brian Fitzpatrick
  • 26,559
  • Great answer. It took me a bit until I realized why you need to consider different bases for image and pre-image: Because the way OP specified the basis (two separate frequencies for $\sin$ and $\cos$) for $V$, $D$ doesn't map onto itself. – Roland Jan 12 '16 at 10:00