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In spite of the fact that the matrix ring $\mathbb{C}^{n \times n}$ is not a field, is it still possible to talk about it being 'algebraically closed' in the sense that $\forall f \in \mathbb{C}^{n \times n}[x]$ does $\exists A \in \mathbb{C}^{n \times n}$ such that $f(A) = 0$? If so, then is it 'algebraically closed'?

Are there any other non-field sets that this idea can be extended to?

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    $\begingroup$ The real question is here what $\mathbb{C}^{n\times n}\left[x\right]$ means. With $\mathbb{C}^{n\times n}$ being non-commutative, it is not clear what kind of polynomials you wish to allow. For instance, the Amitsur-Levitzki theorem ( gilkalai.wordpress.com/2009/05/12/… ) gives a nontrivial polynomial relation between any $2n$ matrices in $\mathbb{C}^{n\times n}$; if you replace the right hand side by $1$ rather than $0$, you will get a polynomial that never attains zero (though it is hardly the zero polynomial.) $\endgroup$
    – darij grinberg
    Feb 1, 2010 at 10:29
  • $\begingroup$ On the other hand, Ore polynomials (coefficients on the left, powers of $x$ on the right) may have a chance of making the assertion correct. $\endgroup$
    – darij grinberg
    Feb 1, 2010 at 10:30
  • $\begingroup$ Ok, suppose we define $f \in \mathbb{C}^{n \times n}[x]$ by $f(x) = \sum_{i=0}^n A_i x^i$ where $A_i \in \mathbb{C}^{n \times n}$, so we do in fact have coefficients on the left, powers of x on the right (this way infact what I was thinking of). Does this help or is there a better definition to use? I would imagine the simplest way to find a counter example would be finding a matrix with no 'square root', i.e. an $A \in \mathbb{C}^{n \times n}$ such that $\forall B \in \mathbb{C}^{n \times n}$, $B^2 \neq A$. $\endgroup$
    – Mark Bell
    Feb 1, 2010 at 10:57
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    $\begingroup$ Yuck. People actually call the ring of matrices $\mathbb C^{n\times n}$? When I see that, I assume you mean something like $\prod^{n^2} \mathbb C$, the commutative product of rings. Much better you be something like $\text{End}(\mathbb C^n)$. $\endgroup$
    – Theo Johnson-Freyd
    Feb 1, 2010 at 16:22
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    $\begingroup$ There's a model-theoretic generalization of "algebraically closed", namely "existentially closed": see Wikipedia. It basically says that if some system of equations has a solution in a larger algebra, then it has a solution within the algebra. Douglas' comment is that some nilpotent matrix in $M_2(C)$ has no square root. Embedding $M_2(C)$ into $M_4(C)$ diagonally, it has a square root in the larger algebra. So $M_2(C)$ is not existentially closed. $\endgroup$
    – YCor
    Jan 30, 2020 at 12:33

2 Answers 2

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The matrix $\left( \begin{array}{cc} 0 & 1 \\\\ 0 & 0 \end{array} \right)$ has no square root.

Polynomials make sense for continuous complex functions on a space. If that space is $\mathbb R$, then polynomial equations with complex coefficients are solvable. If that space is $\mathbb C$ or $S^1$ then $g^2 = f$ may not be solvable.

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    $\begingroup$ And so $f(x) = \left(\begin{array}{cc}1 & 0 \\ 0 & 1 \\ \end{array}\right) x^2 - \left(\begin{array}{cc}0 & 1 \\ 0 & 0 \\ \end{array}\right) = 0$ has no root over $\mathbb{C}^{n \times n}$. Hence $\mathbb{C}^{n \times n}[x]$ is not 'algebraically closed'. $\endgroup$
    – Mark Bell
    Feb 1, 2010 at 11:52
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    $\begingroup$ What if you consider H = n x n Hermitian matrices? Does the spectral theorem provide the necessary push in this context? $\endgroup$
    – Tom LaGatta
    Feb 1, 2010 at 16:26
  • $\begingroup$ @Tom: do you mean allowing the polynomial to have coefficients in H? $\endgroup$
    – Yemon Choi
    Feb 1, 2010 at 19:48
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I'd like to add that a nice theory of roots of polynomials over noncommutative rings was developed by I. Gelfand, V. Retakh, and R. Wilson, see the paper arXiv:math/0208146 and references therein (in particular, the earlier paper by Gelfand and Retakh on the noncommutatoive Vieta theorem).

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