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The problem I'm solving involves a proof using the operator $ad A$, which is defined as follows:

$ad$ $A $ $\cdot =[A,\cdot]$

What does this notation mean?

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    It is the [adjoint action](http://en.wikipedia.org/wiki/Adjoint_endomorphism). Namely, if $A\in\mathrm{End}(V)$, then $\mathrm{ad}_A:X\mapsto [A,X]:=A\circ X-X\circ A$ is, for all $A$, a linear map on $\mathrm{End}(V)$. When working with the endomorphism ring as something called a *Lie algebra*, the bracket notation by convention refers to the commutator bracket I just wrote. We can also define this Lie bracket in abstract settings where the bracket satisfies the axioms of bilinearity, skew-symmetry, and the Jacobi identity, but you don't need to know these things for just linear algebra.2012-11-15
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    I'm not quite understanding. I'm given a matrix A, and the question involves e^(ad A). What does this mean?2012-11-15
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    $\mathrm{ad}_A$ is an operator on the vector space $\mathrm{End}(V)$. That is, it is an endomorphism of the space of endomorphisms. Thus it makes sense to talk about powers of $\mathrm{ad}_A$ e.g. $$\mathrm{ad}_A^2:X\to [A,[A,X]]=A^2X-2AXA+XA^2.$$ Then, when the ground field has char zero, we define the maps $\mathrm{End}(V)\to\mathrm{End}(V)$ $$\exp(t~\mathrm{ad}_A):X\mapsto \mathrm{Id}+\sum_{n\ge1}\frac{t^n}{n!}[\underbrace{A,[A,\cdots,[A}_n,X]\cdots]]$$ when $t$ is a scalar. Are you sure the context of the question is *just* linear algebra, and not more informatively *Lie theory*?2012-11-15
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    Since mathjax is buggy: $$\exp(t~\mathrm{ad}_A)=\sum_{n=0}^\infty\frac{(t\,\mathrm{ad}_A)^n}{n!}:~~ X\mapsto \mathrm{Id}+\sum_{n=1}^\infty\frac{t^n}{n!}[\underbrace{A,[A,\cdots,[A}_n,X] \cdots]].$$2012-11-15
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    Thanks, that makes a lot more sense now.2012-11-15

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