Conjugation in $GL(V)$? Given $T,P\in GL(V)$, what does it mean to conjugate $T^{-1}PT$? I understand that if we choose a basis for $V$, then conjugation on $GL_n(F)$ just becomes representing the same linear transformation under different choices of basis. However, it seems to me that conjugation in $GL(V)$ is different because if it were the same, we'd be saying $T^{-1}PT=P$ which is definitely false. However, $GL_n(F)$ and $GL(V)$ are isomorphic as groups which further confuses me. Could anyone shed some insight? 
 A: The word conjugate can have several meanings depending on the context, even restricted to the scope of your question.


*

*First, in abstract algebra, an entity such as $$T^{-1}PT$$
is called the conjugate of $P$ by $T$. This terminology holds in any group. One will say that a group (here $GL(V)$) acts on itself by conjugation.

*Completely unrelated, some people use the word conjugate to refer to the matrix $M^*$ of the adjoint operator, when the vector space $V$ is endowed with an inner product.

*Finally, when the base field is $\Bbb C$, the conjugate of a matrix can refer to the matrix $\bar M$ obtained by applying complex conjugation to every entry.

Now I believe that your question is about the first definition. It is true that $T^{-1}PT$ represents the same operator (call it $L$) as $P$ but in a different basis. That does not make $T^{-1}PT$ equal to $P$ as a matrix.  For the same input vector $\vec v$ would have different coordinates and correspond to a different column-vector depending on whether you want to use the matrix $P$ or the matrix $T^{-1}PT$ to compute the output $L(\vec v)$.
