Why is every representation of $\textrm{GL}_n(\Bbb{C})$ completely determined by its character? I know that every (Lie group) representation of $\textrm{GL}_n(\Bbb{C})$ is completely reducible; this I believe comes from the fact that every representation of the maximal compact subgroup $\textrm{U}(n)$ is completely reducible. More explicitly, suppose $V$ is a representation of $\textrm{GL}_n(\Bbb{C})$. Then $V$ is also a representation of $\textrm{U}(n)$, by complete reducibility of the unitary group we know that there is a $\textrm{U}(n)$ invariant inner product such that if $U$ is any $\textrm{GL}_n$ - invariant subspace of $V$ (and hence $\textrm{U}(n)$ invariant), there is an orthogonal complement $W$ such that 
$$V = U \oplus W$$
with $W$ invariant under $\textrm{U}(n)$. Now $W$ as a representation of the real Lie algebra $\mathfrak{u}(n)$ is invariant and hence under the complexified Lie algebra
$$\mathfrak{gl}_n = \mathfrak{u}_n \oplus i \hspace{1mm} \mathfrak{u}(n).$$
Since $\textrm{GL}_n(\Bbb{C})$ is connected $W$ is also invariant under $\textrm{GL}_n$ showing that every representation of it is completely reducible.
Now I have read several textbooks on representation theory (e.g. Bump's Lie Groups, Procesi's book of the same name) and they all seem to tacitly assume that every representation of $\textrm{GL}_n$ is completely determined by its character; i.e. if two representations have the same character then they are isomorphic.
Now in the finite groups case, we concluded this fact based on 1) Maschke's Theorem and 2) Linear independence of characters.
We do not necessarily have 2) so how can we conclude the fact I said about about $\textrm{GL}_n$?
Thanks.
 A: The character of a representation $V$ tells you how a maximal torus $T$ acts on $V$.  Moreover, you can then figure out how any maximal torus acts on $V$ because any other maximal torus is of the form $g T g^{-1}$ for $g \in G$. (Here, $G = GL_n(\mathbb{C})$.)  That is, if $v_1, \dots, v_n$ are the $T$-eigenvectors of $V$, then $g \cdot v_1, \dots, g \cdot v_n$ are the $gTg^{-1}$-eigenvectors of $V$.
In the case of $GL_n(\mathbb{C})$, this says that the character determines how every diagonalizable element $g \in GL_n(\mathbb{C})$ acts on $V$.  But the diagonalizable matrices are dense in $GL_n(\mathbb{C})$, and thus representations are uniquely determined by their characters.
A: This boils down to facts about the representation theory of compact groups:  there every complex representation is determined by its character.  Now a representation of $GL(n,\mathbb C)$ is determines by a representation of its Lie algebra.  But this is the complexification of $u(n)$ and complex representations of a Lie algebra are in one to one correspondence with complex representations of its complexification.
A: One does have linear independence of characters.  The characters of the various irreps. form a basis for the space of conjugation invariant algebraic functions on $GL_n(\mathbb C)$.  
Added some time later: The ring of conjugation invariant functions on $GL_n(\mathbb C)$ is precisely the ring of symmetric functions in the eigenvalues (with the inverse of the determinant, i.e. the inverse of 
the product of all the eigenavlues, adjoined), 
which is generated as a polynomial ring by the elementary symmetric polynomials in the eigenvalues (with the inverse of the product $\lambda_1\cdots\lambda_n$
adjointed).
Thus this ring is spanned as a $\mathbb C$-vector space by the monomials in the elementary symmetric polynomials.  
The Grothendieck  semiring of isomorphisms classes of finite dim'l reps. of $GL_n(\mathbb C)$ is generated by the fundamental representations $\wedge^i ($standard rep$)$, for $i = 1,\ldots,n$, together with the inverse of the 
last of these (the $n$th exterior power of the standard rep. is just $\det$,
and so I am just saying that we throw in $\det^{-1}$ as well as $\det$).
The character of $\wedge^i$ of the standard rep. is the $i$th symmetric polynomial in the eigenvalues, and now by inductively writing "monomials" in
the $\wedge^i$ (i.e. tensor products of the various $\wedge^i$) in terms 
of irreps., it is easy to verify my claim that the characters of the irreps.
span the conjugation invariant functions.  
