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So I need to find the matrix representation of the function f.

f = [.]β given by [$\vec v$]β = $\begin{pmatrix}a_1\\a_2\\a_3\\a_4\end{pmatrix}$ where $\vec v = a_1\vec v_1 + a_2\vec v_2 + a_3\vec v_3 + a_4\vec v_4$.

The basis β = {$\begin{pmatrix}1\\1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\\1\end{pmatrix}, \begin{pmatrix}0\\0\\0\\1\end{pmatrix}$} = {$\vec v_1, \vec v_2, \vec v_3, \vec v_4$}.

The problem is that I don't know what f = [.]β means. I was told that its the function that takes as input a vector $\vec v$, and outputs the column vector $a_1 a_2 a_3 a_4$ otherwise known as the "coordinates with respect to the basis beta" function but I'm still not sure what I need to do.

david mah
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  • Well firstly, you need some prescription for the function $f$. –  Nov 22 '16 at 21:56
  • Then you can look at my answers to these questions to try to understand what matrix representation of a function means: 1, 2, 3, 4. –  Nov 22 '16 at 22:03
  • I do know what the matrix representation of a function is, I just don't know how to find the specific function. – david mah Nov 22 '16 at 23:09
  • That conflicts with the first sentence of your question. But OK. You still need some prescription for $f$ -- or its action on a basis or something tangible about it. Is it supposed to be the change of basis matrix from the standard basis to $\beta$? –  Nov 22 '16 at 23:12
  • It is a change of basis matrix from the standard basis to β. I know how to find the change of basis but what I have no idea how to do is with that dot. All it says is that f is some dot with respect to the basis but I have no idea how to do that. – david mah Nov 22 '16 at 23:18
  • The notation $[\cdot]_\beta$ means a matrix with stuff inside (rows of numbers in this case) wrt the basis $\beta$. That's all the dot means: "stuff goes here". –  Nov 22 '16 at 23:19
  • He also said something about how it can also be written as f(b) = [v] with the subscript beta. He also said that the function takes the input of v and outputs the column vector a1 a2 a3 a4. So if that's the case then would f just be the v function that I listed above? Then would that make [f] just an identity matrix? That last paragraph in the question is how the problem was described to me and I don't really know what that description means. – david mah Nov 22 '16 at 23:39
  • Without knowing more about what $\vec v$ is, this is a ambiguous question. My guess would be that $\vec v$ is supposed to be a vector in $\Bbb R^4$ and $f$ is the change of basis function from the vector $\vec v$ to its representation wrt the basis $\beta$, $[\vec v]\beta$. In that case, yes: $f$ is an identity function, but its matrix representation won't be the identity matrix $I = \pmatrix{1 & 0 & \cdots \ 0 & 1 & \cdots \ \vdots & \vdots &\ddots}$ -- it'll be the change of basis matrix $[I]{\beta \leftarrow \mathcal{E}}$ where $\mathcal E$ is the standard basis. –  Nov 22 '16 at 23:44
  • I said that $\vec v = a_1\vec v_1 + a_2\vec v_2 + a_3\vec v_3 + a_4\vec v_4$ and that each $\vec v_1, \vec v_2$ ect is a column in the basis vector. So $\vec v_1$ would be the first column in the basis vector and so on. – david mah Nov 22 '16 at 23:51
  • OK, then $\vec v$ is some vector in $\Bbb R^4$ so everything starting from "$f$ is the change of basis function" in my previous comment is true. Find the change of basis matrix. –  Nov 23 '16 at 00:13

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