Is X Always Commutative with Y in Matrix Invariance?

  • Context: Undergrad 
  • Thread starter Thread starter Euge
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    2015
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SUMMARY

The discussion centers on the problem of proving that if $X$ is a $Y$-invariant matrix, defined by the condition $e^{zY}Xe^{-zY} = X$ for all complex numbers $z$, then $X$ and $Y$ must commute. This conclusion is derived from the properties of matrix exponentiation and invariance under transformations. The problem highlights the relationship between matrix invariance and commutativity, which is crucial in linear algebra and matrix theory.

PREREQUISITES
  • Understanding of matrix exponentiation
  • Familiarity with complex matrices
  • Knowledge of linear algebra concepts, specifically matrix invariance
  • Basic principles of commutative properties in algebra
NEXT STEPS
  • Study the properties of matrix exponentiation in detail
  • Explore the implications of matrix invariance in quantum mechanics
  • Learn about the commutation relations in linear algebra
  • Investigate applications of invariant matrices in differential equations
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Mathematicians, students of linear algebra, and anyone interested in the theoretical aspects of matrix theory and its applications in various fields such as physics and engineering.

Euge
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Since many schools have just started the school year, I'd like to begin September's POTW with an important, yet simple problem.

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Suppose $X$ and $Y$ are $n \times n$ complex matrices such that $X$ is $Y$-invariant, i.e., $e^{zY}Xe^{-zY} = X$ for all $z\in \Bbb C$. Prove that $X$ and $Y$ commute.
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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!
 
Last edited:
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No one solved this week's problem correctly. You can read my solution below.
Differentiate the equation $e^{zY}Xe^{-zY} = X$ with respect to $z$ to obtain

$$Ye^{zY}Xe^{-zY} - e^{zY}XYe^{-zY} = 0,$$

then set $z = 0$ to get $YX - XY = 0$, or $YX = XY$.
 

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