$GL(n, \mathbb{C})$ is algebraically closed?

Question

Let $GL(n,\mathbb{C})$ the group of non-singular matrices. Is it algebraically closed? For $GL(1,\mathbb{C})$ is it true; but if I take linear combinations of elements in $GL(n,\mathbb{C})$ with coefficients matrices in $GL(n,\mathbb{C})$ could I find solutions in $GL(n, \mathbb{C})$? We know that diagonalizable applications are dense: maybe it is useful. If $n=2$ can we use the isomorphis of $GL(2, \mathbb{C})$ with quatarnions?

Answer 1

At the moment there is a problem with your question: $\text{GL}_n(\Bbb{C})$ is not a field. The notion of being algebraically closed only applies to fields. In fact $\text{GL}_n(\Bbb{C})$ is not even a ring because the ring axioms mean that we need it to be closed under addition. But then $diag(1,\ldots 1) + diag(-1,\ldots,-1) = 0$ so it is not a ring.

Answer 2

One way (and the only one I know about) to define polynomials over a non-commutative ring $A$ is to consider them as elements of $\bigoplus_{\Bbb N} A$, i.e. as sequences $(a_0,a_1,\dotsc,a_n,\dotsc)$ with only a finite number of $a_i\in A$ not null. Beware, though, that polynomial functions over a non-commutative ring will probably behave in unexpected ways. For example, polynomial functions over $\Bbb H$ can have an infinite number of roots.
Please note that a polynomial is just a formal sequence $f\in A[X]$, while a polynomial function is a map $f(X)\colon A\to A$, defined by substituting an element of $A$ in place of the "variable" $X$. Moreover, in general we can define what a root is only for the latter.

There is a problem, though, in trying to define $GL_n(\Bbb C)[X]$, as $GL_n(\Bbb C)$ is not a ring. In particular, it doesn't have a $0$ element, so we cannot construct sequences as above.

But suppose, for the sake of argument, that we could somehow define polynomial functions with coefficients in $GL_n(\Bbb C)$. In particular, for any $i>0$ we could consider the monomials $X^i=I_nX^i$, where $I_n$ is the identity of $GL_n(\Bbb C)$. Let $A\in M_n(\Bbb C)$ be a matrix which is a root of $X^i=0$ for some $i>0$. Then it would satisfy $A^i=0$, i.e. it would be nilpotent (possibly $0$). Thus $A$ isn't invertible, so $A\notin GL_n(\Bbb C)$.