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?
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6It's not clear what you are asking; being algebraically closed is a property of fields. I think you are asking whether every polynomial equation with $n \times n$ complex matrices as coefficients has a root? But you might want to specify what you mean by a polynomial in this case, since matrix multiplication is not commutative. (In particular, the expressions $AX^2$, $XAX$ and $X^2A$ are all different in the noncommutative world.) – Michael Joyce May 11 '13 at 15:24
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4Also, under any sensible interpretation of your question, $GL(1,\mathbb{C})$ will not be "algebraically closed" because it has no solution to the simple equation $X = 0$. Your attempted question (as I understand it) has nothing to do with invertible matrices and must be interpreted as a statement about the noncommutative ring of $n \times n$ matrices (with complex entries). – Michael Joyce May 11 '13 at 15:26
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@MichaelJoyce Let $n$ be a natural number. We can take $n$ elements in $GL$ and with them we can build a polinomial $p$ with $deg(p)=n-1$. Is it true that this polynomial has roots in GL? – ArthurStuart May 11 '13 at 15:49
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2You really should say what you mean exactly by algebraically closed here. (And add it to your question!) – tomasz May 11 '13 at 16:03
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I think that the answer is yes if we formulate the question as "Let $L$ be the noncommutative polynomial ring over $\operatorname{GL}_n(\mathbb{C})$. Does every nonconstant element in $L$ completely factor into linear factors?" But I'm not sure how to prove it. Let me see if I can find a reference. (Obviously, I vote not to close.) – Alexander Gruber May 11 '13 at 16:56
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@YACP I'm still doing my research on this topic. My opinion is that the topic should remain open because, while OP is being unresponsive, it is an interesting question, and I am not as convinced as others seems to be that it is ill-founded. – Alexander Gruber May 11 '13 at 18:07
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@AlexanderGruber, what is ¸the «noncommutive polynomial ring over $GL_n(C)$»? – Mariano Suárez-Álvarez May 14 '13 at 17:48
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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.
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1The notion of "algebraically closed" under consideration here is (sort of) defined after "but if I take linear combinations..." – Martin May 11 '13 at 15:26
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@BenjaLim Let $n$ be a natural number. We can take $n$ elements in $GL$ and with them we can build a polinomial $p$ with $deg(p)=n−1$. Is it true that this polynomial has roots in $GL$? – ArthurStuart May 11 '13 at 15:58
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2@ArthurStuart How do I add $AX + XA$? Please define what it means to have a polynomial with coefficients in $\text{GL}_n(\Bbb{C})$. – May 11 '13 at 15:59
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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 $A\notin GL_n(\Bbb C)$.
