I am confused by the definition of the general linear group of a vector space. Let $V$ be a vector space of dimension $3$ and $W$ be a subspace of $V$ with dimension $2$. (For example, $V$ can be the three dimensional Euclidean space, and $W$ can be a plane in $R^3$.) We can observe that elements of $V$ and $W$ are going to be $3$-tuples.

When the general linear group of a vector space is defined, the size of matrix elements is given by the dimension of the vector space. So $GL(V,R)$ will have invertible real matrices of size $3\times3$ as elements. However, I am really confused when I am looking at $GL(W,R)$. Using the definition it seems its elements are invertible $2\times2$ real matrices. However, these matrices should be able to act on elements of $W$, so if $A$ is in $GL(W,R)$ and $w$ is in $W$ (note that $w$ is a vector in $R^3$). Then $A*w$ should be well defined, but it's not because $A$ is $2\times2$ and $w$ is $3\times1$.

I am pretty sure the elements of $GL(W,R)$ should be $3\times3$ matrices. However, how does this fulfill the definition of the general linear group of the finite dimensional vector space $W$? The dimension of $W$ is certainly not $3$. There is something I am misinterpreting, please any help would be appreciated. Thanks!

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    $\begingroup$ $GL(W)$ is the set of linear mappings from $W$ to $W$, not a set of matrices. Once a particular basis for $W$ has been selected, then you can of course write all those linear mappings in that basis, and you'll get the set of all invertible $2 \times 2$ matrices. When $V$ is a general three-dimensional vector space, the same problem arises: you need to select a basis in order to identify elements of $GL(V)$ with matrices. When $V = R^3$, you have a ready-made choice: the standard basis. When $W$ is a plane in $R^3$, there's no basis that is really singled out that way. $\endgroup$ – user49640 Jun 2 '17 at 3:46
  • $\begingroup$ The set of linear maps is a set of matrices.... but... After re-reading your original post. $W$ has a two dimensional basis, and so a $2\times 2$ matrix $A$ can act on vectors in the vector space $W$ even though $W$ is represented by 3-tuples in the basis of $V$ $\endgroup$ – Doug M Jun 2 '17 at 3:50
  • $\begingroup$ Sorry, I should have said the set of bijective linear mappings. That was a mistake. $\endgroup$ – user49640 Jun 2 '17 at 7:24

As others have pointed out in the comments, $\DeclareMathOperator{\GL}{GL} \GL(V)$ is the set of invertible linear maps $V \to V$; only by choosing a basis do we get an isomorphism $\GL(V) \cong \GL_3(\mathbb{R})$. Nonetheless, I think there is a way in which your question does make sense. Given $\varphi \in \GL(W)$, $\varphi: W \to W$ is only defined on $W$, not on all of $V$, but there is a way to extend $\varphi$ to all of $V$.

Choose a basis $\mathcal{B} = \{e_1, e_2\}$ for $W$ and let $[\varphi]_\mathcal{B}$ denote the matrix representation of $\varphi$ with respect to $\mathcal{B}$. By the basis extension theorem, we can extend the basis $\mathcal{B} = \{e_1, e_2\}$ of $W$ to a basis $\mathcal{C} = \{e_1, e_2, e_3\}$ of $V$. Define $\widetilde{\varphi} : V \to V$ by $\widetilde{\varphi}(e_i) = \varphi(e_i)$ for $i = 1,2$ and $\widetilde{\varphi}(e_3) = e_3$, i.e., $\widetilde{\varphi}$ acts as $\varphi$ on $W$ and $\widetilde{\varphi}$ fixes $e_3$. Then the matrix representation of $\widetilde{\varphi}$ with respect to $\mathcal{C}$ is block-diagonal: $$ [\widetilde{\varphi}]_\mathcal{C} = \left( \begin{array}{c|c} [\varphi]_\mathcal{B} & 0 \\ \hline 0 & 1 \end{array} \right) \, . $$

So in some sense we have have embedded $\GL(W) \subseteq \GL(V)$, which in coordinates amounts to embedding $2 \times 2$ matrices in $3 \times 3$ matrices. (More generally, this sort of block diagonal representation will arise whenever there is a $\varphi$-invariant subspace.)

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    $\begingroup$ I was actually reading about invariant subspaces and that's where I came up with this doubt, this helped me tie up many concepts together, thanks you! $\endgroup$ – Cami77 Jun 2 '17 at 4:46
  • $\begingroup$ Your $\widetilde\varphi$ is not linear! In fact, you should never define a linear map piecewise (but rather by choosing complements or bases). $\endgroup$ – Claudius Jun 2 '17 at 7:10
  • $\begingroup$ @user218931 Yes, you're right, my attempt to write $\widetilde{\varphi}$ without coordinates was incorrect. The definition I gave in coordinates was what I intended, so I've kept that. $\endgroup$ – Richard D. James Jun 2 '17 at 8:01

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