Is the commutator of two complex matrices nilpotent?

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  • Thread starter Ackbach
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    2017
In summary, the commutator of two matrices is defined as [A,B] = AB - BA and measures how much the matrices fail to commute. To determine if the commutator is nilpotent, it must be true that (AB)^n = 0 for some positive integer n. A matrix is nilpotent if repeated multiplication by itself results in a zero matrix. The commutator of two complex matrices can be non-nilpotent, meaning (AB)^n ≠ 0 for any positive integer n. Commutativity of matrices does not necessarily guarantee nilpotency of the commutator, but if two matrices do commute, their commutator is always nilpotent.
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Ackbach
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Here is this week's POTW:

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If the commutator, $Z$, of two complex $n\times n$ matrices commutes with one of those matrices, must $Z$ be nilpotent?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered this week's POTW. Source credit: this problem, and quite possibly a few more in the coming weeks, is due to Euge. Here is the solution:

[sp]Yes. Let $Z = [X,Y]$ and suppose $ZX = XZ$. Inductively, $Z^kX = XZ^k$ for all natural numbers $k$. For every $k \ge 1$, $$\operatorname{trace}(Z^k) = \operatorname{trace}(Z^{k-1}XY) - \operatorname{trace}(Z^{k-1}YX) = \operatorname{trace}(XZ^{k-1}Y) - \operatorname{trace}(XZ^{k-1}Y) = 0$$ using the properties $\operatorname{trace}(MN) = \operatorname{trace}(NM)$ and $Z^{k-1}X = XZ^{k-1}$. This implies $Z$ is nilpotent. Indeed, since $Z$ is complex, it has $n$ eigenvalues. If its eigenvalues are not identical, suppose $\lambda_1,\ldots, \lambda_d$ are the distinct eigenvalues of $Z$ with multiplicities $m_1,\ldots, m_d$, respectively. Since $\operatorname{trace}(Z^k) = 0$ for all $k$, then $\sum_{j = 1}^d m_j \lambda_j^k = 0$ for $k = 1,2,\ldots, d$. The Vandermonde determinant of $\lambda_1,\ldots, \lambda_d$ is nonzero since the $\lambda_i$ are distinct, whence $m_1 = \cdots = m_d = 0$. This contradicts the equation $\sum_{j = 1}^d m_j = n$. So let $\lambda$ be the unique eigenvalue of $Z$. The condition $\operatorname{trace}(Z) = 0$ forces $\lambda = 0$. Finally, the characteristic polynomial of $Z$ is $p(t) = t^n$, implying the $Z$ is nilpotent.[/sp]
 

FAQ: Is the commutator of two complex matrices nilpotent?

What is a commutator of two matrices?

The commutator of two matrices A and B is defined as the matrix [A,B] = AB - BA. It measures the extent to which the two matrices fail to commute.

How do you determine if the commutator of two matrices is nilpotent?

The commutator [A,B] is nilpotent if and only if (AB)^n = 0 for some positive integer n.

What does it mean for a matrix to be nilpotent?

A matrix A is nilpotent if and only if A^n = 0 for some positive integer n. In other words, repeated multiplication of the matrix by itself eventually results in a zero matrix.

Can the commutator of two complex matrices be non-nilpotent?

Yes, it is possible for the commutator of two complex matrices to be non-nilpotent. This means that (AB)^n ≠ 0 for any positive integer n.

How is the nilpotency of the commutator related to the commutativity of the matrices?

If two matrices A and B commute (i.e. AB = BA), then their commutator [A,B] is always nilpotent. However, the converse is not necessarily true. The commutator can be nilpotent even if the matrices do not commute.

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