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Corrected "exponential"
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John Stillwell
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An example that I like is the proof that $e^{A+B}=e^A e^B$ for commuting matrices $A,B$. Since the matrix exponentionalexponential is defined by the usual exponential series, we have to prove that

$\sum \frac{(A+B)^n}{n!}=\sum\frac{A^n}{n!}\sum\frac{B^n}{n!}$

This follows, without actually computing the two sides, by observing that it is the same computation as for real numbers $A,B$ (because $A$ and $B$ commute). And for real numbers we know the result is correct by first-year calculus.

An example that I like is the proof that $e^{A+B}=e^A e^B$ for commuting matrices $A,B$. Since the matrix exponentional is defined by the usual exponential series, we have to prove that

$\sum \frac{(A+B)^n}{n!}=\sum\frac{A^n}{n!}\sum\frac{B^n}{n!}$

This follows, without actually computing the two sides, by observing that it is the same computation as for real numbers $A,B$ (because $A$ and $B$ commute). And for real numbers we know the result is correct by first-year calculus.

An example that I like is the proof that $e^{A+B}=e^A e^B$ for commuting matrices $A,B$. Since the matrix exponential is defined by the usual exponential series, we have to prove that

$\sum \frac{(A+B)^n}{n!}=\sum\frac{A^n}{n!}\sum\frac{B^n}{n!}$

This follows, without actually computing the two sides, by observing that it is the same computation as for real numbers $A,B$ (because $A$ and $B$ commute). And for real numbers we know the result is correct by first-year calculus.

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Source Link
John Stillwell
  • 12.4k
  • 15
  • 96
  • 118

An example that I like is the proof that $e^{A+B}=e^A e^B$ for commuting matrices $A,B$. Since the matrix exponentional is defined by the usual exponential series, we have to prove that

$\sum \frac{(A+B)^n}{n!}=\sum\frac{A^n}{n!}\sum\frac{B^n}{n!}$

This follows, without actually computing the two sides, by observing that it is the same computation as for real numbers $A,B$ (because $A$ and $B$ commute). And for real numbers we know the result is correct by first-year calculus.