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Suppose $V$ is a two-dimensional $K$-vector space. Does an embedding $\varphi: GL(V) \hookrightarrow GL(K^n)$ exist?

I tried multiplying $A \in GL(V)$ with the basis vectors of $K^n \cong \operatorname{Sym}^n(V)$ via $A(v\cdot w) = Av\cdot Aw$ and take the resulting coordinates as colums for $\varphi (A)$, but I don't see how this is well-defined (i.e. yields linearly independent colums and thus an invertible matrix) let alone a group homomorphism.

Edit: I forgot to mention that this embedding is in particular required to preserve the set of power tensors $v^n \in Sym^n(V)$ in a way that $A(v)^n = \varphi (A)(v^n)$ - So it is better to think of it as an embedding $\varphi: GL(V) \hookrightarrow GL(Sym^n(V))$.

I computed the above construction for $n = 3$ and it seemed to work, but I don't know why $\varphi$ needs to be a well-defined homomorphism in the general case.. Any hints would be greatly appreciated

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    Thank you, I'll look into it (funny coincidence by the way, the reason I asked this question is a certain passage in "Algebraic Geometry - A First Course" by Harris).2011-11-10

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What about a map like $g\mapsto\begin{pmatrix}g&0\\0&\mathbf{1}\end{pmatrix}$?$\qquad$ (assuming that $V$ sits inside $K^n$ as $K^2\times 0^{n-2}$)

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    You are absolutely right. I forgot to add my additional condition, this is edited now - Thanks though, this was a silly mistake of me.2011-11-10
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Yes, your formula gives a homomorphism in general. The point is that the formation of $Sym^n V$ is functorial in $V$. Since $GL(V)$ acts on $V$, it thus acts on each $Sym^n V$ as well.

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    whoops, $A \cdot (v_1 \cdot v_2 ..) = (A \cdot v_1) \cdot (A \cdot v_1) ..$ is of course supposed to be $A \cdot (v_1 \cdot v_2 ..) = (A \cdot v_1) \cdot (A \cdot v_2) ..$2011-11-10