7
$\begingroup$

Hello everybody

There is a nice classical result in linear algebra: if $A, B$ are two matrices in $M_n(k),$ where $k$ is a field, and $B$ commutes with every element of $M_n(k)$ which commutes with $A$, then $B = f(A)$ for some polynomial $f(x)$ in $k[x].$

I was wondering if anybody knows any (important) theorem which is proved using this result. Thank you.

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ This kind of idea occurs in noncommutative algebra under the label "double centralizer theorem". For instance, you can find some applications in Lam's book on noncommutative rings. $\endgroup$
    – KConrad
    Commented Aug 27, 2010 at 16:16
  • 3
    $\begingroup$ But... is this assertion always true? Suppose your field k is the complex number field $\mathbb{C}$. Doesn't exp(A) commute with all the matrices with which A commutes? $\endgroup$
    – Qfwfq
    Commented Aug 27, 2010 at 17:07
  • 3
    $\begingroup$ That means, that $exp(A)$ can be written a polynomial in $A$. The point is that this polynomial may (and will) depend on $A$ itself. $\endgroup$
    – Johannes Hahn
    Commented Aug 27, 2010 at 17:37
  • 5
    $\begingroup$ @unkown(google). Despite its appearance, exp(A) is a polynomial on A. That is, for every A, you can find a polynomial r(t) such that exp(A)= r(A). It is the Lagrange-Sylvester interpolation polynomial: see the book of Gantmacher, The theory of matrices, books.google.es/… . $\endgroup$
    – Agustí Roig
    Commented Aug 27, 2010 at 17:49
  • 6
    $\begingroup$ Just to make it a little clearer, the reason that $exp(A)$ is a polynomial in $A$ is that $A$ satisfies its own minimal polynomial, so all terms of degree at least that of the minimal polynomial can be expressed as terms of lower degree, etc. $\endgroup$
    – MTS
    Commented Aug 27, 2010 at 19:00

3 Answers 3

Reset to default
4
$\begingroup$

Tate's famous "Endomorphisms of Abelian Varieties over Finite Fields," which proves the Tate conjecture in the finite field case, uses the full force of the theorem of bicommutation in a reduction lemma. As KConrad mentions in the comments, the result you've cited is the special case of this theorem where one works with the subalgebra generated by one element.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Nice! I had no idea, thanks. That's the paper Tate published in 1966. I think that's kind of applications I'm looking for. Thank you again. $\endgroup$
    – iravan
    Commented Aug 27, 2010 at 17:02
  • $\begingroup$ +1 for "Tate´s famous..." ;) $\endgroup$
    – M.G.
    Commented Aug 27, 2010 at 17:22
2
$\begingroup$

That result sits inside a wider set of results. Search for spectral theorem, functional calculus of linear operators.

Books could be Halmos, A Hilbert Space problem book if you also need to read more about linear operators in general I think in Conway's Functional Analysis there is also stuff about these results, together with an introduction to functional analysis.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Unfortunately I don't know that much about functional analysis but I appreciate your comment very much. Thank you. $\endgroup$
    – iravan
    Commented Aug 27, 2010 at 17:06
  • 2
    $\begingroup$ It is interesting to see an answer from the point of functional analysis after an answer involving abelian varieties over finite fields! $\endgroup$
    – M.G.
    Commented Aug 27, 2010 at 17:25
1
$\begingroup$

This probably doesn't qualify as "important", but you put that in parentheses so I'll mention it anyway.

I used that result when figuring out some basic facts about polynomial loops in a compact, connected Lie group which I needed for my paper the co-Riemannian structure of smooth loop spaces.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .