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Linear Algebra,regarding commutator

My brother asked me for help on this but I quite embarrassingly can't figure it out. I vaguely recall solving a similar problem in school, but with the added condition that B also commutes with (AB-BA).

Any help would be appreciated.

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marked as duplicate by Matt N., Gerry Myerson, BenjaLim, Did, DonAntonio Nov 3 '12 at 10:55

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This is a result called Jacobson's Lemma, first proven in the paper "Rational Methods in the Theory of Lie Algebras" by Nathan Jacobson.

Let me essentially reproduce his very ingenious proof here.

First let us notice that the commutator satisfies properties very similar to a derivative. Fix some $M$ and consider $[M,\ A] = A'$ as an operator on $A$. Notice that $$(A+B)' = A' + B',\ \ (cA)' = cA'$$ for matrices $A,\ B$ and scalar $c$ so that the operator is linear. It also satisfies the product rule $$(AB)' = MAB - ABM = (MA - AM)B + A(MB - BM) = A'B + AB'$$ If $A$ and $A'$ commute, then we recover the power rule in the form of $$(A^k)' = kA^{k-1}A'$$ to see this, we proceed by induction with the base case already covered. Then $$(A^{k+1})' = A'A^k + A(A^k)' = A'A^k + A(kA^{k-1})A' = (k+1)A^{k}A'$$ As a consequence, if we have $p(A)$ for some polynomial $p$, then we have the identity $$[p(A)]' = p'(A)A'$$ where $p'$ is the (ordinary) derivative of the polynomial $p$. Now take some annihilating polynomial of $A$ such that $p(A) = 0$. Then consider repeatedly "differentiating" the polynomial $$[p(A)]' = p'(A)A' = 0$$ $$[p(A)]'' = [p'(A)A']' = p''(A)A'A' + p'(A)A'' = 0$$ Now notice that $A'A = AA'$ will imply $A''A = AA''$ so let us post multiply by $A'$ to have $$p''(A)(A')^3 + p'(A)A'A'' = p''(A)(A')^3 = 0$$ If we carry on in this fashion, we will find that $$[p(A)]^{(k)} = p^{(k)}(A)(A')^{2k-1} = 0$$ where the superscript $(k)$ indicates the $k$th derivative. In particular, if $p$ is $n$th degree, then $$p^{(n)}(A)(A')^{2n-1} = n!(A')^{2n-1} = 0$$ which shows that $A'$ is nilpotent.

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Wow, that is very clever. –  wj32 Nov 3 '12 at 8:46
    
I think you didn't want to say $[p(A)]^{(k)} = p^{(k)}(A)(A')^{2k-1}$ (which is false). –  user10676 Nov 3 '12 at 16:55
    
@user10676 Not sure what you mean, the statement seems to hold. –  EuYu Nov 3 '12 at 19:03
    
I think what user10676 meant is that, while $[p(A)]^{(k)} = p^{(k)}(A)(A')^{2k-1}$ is true (because both terms are equal to zero), the expression $p^{(k)}(A)(A')^{2k-1}$ is actually derived from $[p(A)]^{(k)} (A')^{k-1}$. So, writing $[p(A)]^{(k)} (A')^{k-1} = p^{(k)}(A)(A')^{2k-1} = 0$ is clearer than writing $[p(A)]^{(k)} = p^{(k)}(A)(A')^{2k-1} = 0$. Nice answer, anyway. –  user1551 Jan 24 '13 at 10:10

Hint: Notice $A$ is nilpotent iff $Tr(A^j) = 0$ for all $j >0$ (Proof?)

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It's not immediately clear how this leads to a proof. Would you care to elaborate a little bit? –  EuYu Nov 3 '12 at 7:34

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