I'm given the exercise:

Given the subspace: $$U =\left\{ \textit{x} = \left(\begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right) \in \mathbb{F} | x_1=x_2+x_3 \right\} \subset \mathbb{F}^3$$ and the function $f: U \rightarrow \mathbb{F}^2$ defined by: $$f \left(\begin{matrix} x_1 \\ x_2 \\ x_3 \end{matrix} \right) = \left(\begin{matrix} x_1 \\ x_3 \end{matrix} \right) $$ a) Find a basis for $U$
b) Find the matrix $A$ that represents the linear transformation with respect to the found basis of $U$ and the standard basis $(e_1,e_2)$ for $\mathbb{F}^2$
c) Show that $f: U \rightarrow \mathbb(F)^2$ is isomorphic

I have found the basis of $U$ to be $\left(\left(\begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right) , \left(\begin{matrix} 1 \\ 0 \\ 1 \end{matrix} \right)\right)$. However I have no idea how I should interpret or solve question b. Question c shouldn't be a problem when I've solved b. Any help is much appreciated.

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    $\begingroup$ Do you know the definition of the matrix $A$ in (b)? $\endgroup$ – Jack Jan 2 '17 at 16:03
  • $\begingroup$ Just find a matrix $A$ that sends vectors from the basis in $U$ to the basis in $\mathbb{F}^2$ $\endgroup$ – Jesús Ros Jan 2 '17 at 16:06
  • $\begingroup$ No, I have no more information, besides what is written in the OP. $\endgroup$ – limeeattack Jan 2 '17 at 16:06
  • $\begingroup$ (b) is a very standard question in linear algebra. One must have seen the definition of the matrix $A$ before doing such an exercise. $\endgroup$ – Jack Jan 2 '17 at 16:09

Here is the more general setting.

Let $V$ (resp. $W$) be an $n$ (resp. $m$) dimensional vector space over $\mathbb{F}$. Let $\alpha=(v_1,\cdots,v_n)$ be an ordered basis in $V$ and $\beta=(w_1,\cdots,w_m)$ an ordered basis in $W$.

In your question, $V=U$ and $W=\mathbb{F}^3$. You have found $\alpha$. Note that $\beta$ is given. Also, $n=2$ and $m=2$.

For any vector $x\in V$, denote its coordinate w.r.t. the basis $\alpha$ as $ [x]_\alpha=(x_1,\cdots,x_n)^T $ and for any vector $y\in W$, denote its coordinate w.r.t. the basis $\beta$ as $ [y]_\beta=(y_1,\cdots,y_m)^T. $

Let $T:V\to W$ be a linear transformation. Let $[T]_\alpha^\beta$ denotes the matrix for $T$ w.r.t. the bases $\alpha$ and $\beta,$ i.e., $$ [T]^\alpha_\beta=\bigr[[Tv_1]_\beta,\cdots,[Tv_n]_\beta\bigr]. $$ Note in particular that $[T]^\alpha_\beta$ is an $m\times n$ matrix.

So what are the steps to find $[T]^\alpha_\beta$?

  • Note that you are given $\alpha$ and $\beta$, and $T$. Identify what are $m$ and $n$ for your problem;
  • Find $Tv_j$ for each $j$;
  • and then find $[Tv_j]_\beta$.

For the very last step, you need to know how to find $[z]_{\beta}$ given $z\in W$. Suppose $[z]_\beta=(z_1,\cdots,z_m)^T$ Then by the definition of coordinates: $$ z=\sum_1^m z_iw_i $$ which essentially gives you a linear equation about the $z_i$'s.

  • $\begingroup$ So my $[T]^\alpha_\beta$ matrix should be a 2 x 2 matrix? $\endgroup$ – limeeattack Jan 2 '17 at 16:23
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Jack Jan 2 '17 at 16:24
  • $\begingroup$ So the result would be given by: $$\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right) \left(\begin{matrix} 1 & 1 \\ 1 & 0 \\ 0 & 1 \end{matrix} \right) $$ But how can that be when $f$ takes a 3-d vector? $\endgroup$ – limeeattack Jan 2 '17 at 16:27
  • $\begingroup$ $f$ is a map on $U$. An element of $U$ is a 3D vector but the coordinate of the element is a 2D vector since $U$ is of dimension $2$. $\endgroup$ – Jack Jan 2 '17 at 16:38
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    $\begingroup$ I did, however with a rep below 15 it's not shown $\endgroup$ – limeeattack Jan 2 '17 at 17:00

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