Why do matrices act on vectors from one side but on other matrices from both sides? I have a very silly linear algebra question. If we have a matrix $\hat{\mathbb{M}}$ and we want to act on an arbitrary vector $\mathbf{V}$ with the matrix, we do it by simply calculating $\hat{\mathbb{M}}{\mathbf{V}}\equiv\mathbf{V}'$. This makes a good deal of sense.
What doesn't make so much sense is that if we wish to transform another matrix, call it $\hat{\mathbb{T}}$ with that same matrix $\hat{\mathbb{M}}$, we have to do it by "sandwiching" $\hat{\mathbb{T}}$ between $\hat{\mathbb{M}}$ and its adjoint such that $\hat{\mathbb{T}}'\equiv\hat{\mathbb{M}}^{\dagger}\hat{\mathbb{T}}\hat{\mathbb{M}}$.
Is there an intuitive explanation for why (insofar as mathematics has "why" answers) matrices acting on other matrices have to act this way? What transformation would result if we simply did the "natural" thing of assuming (wrongly) that $\hat{\mathbb{T}}'$ is given by $\hat{\mathbb{T}}'\equiv\hat{\mathbb{M}}\hat{\mathbb{T}}$ and why is it in some sense the "wrong" transformation?
 A: In both cases, you are using matrix multiplication (a vector is just an $n\times 1$ matrix), but you are doing quite different things.
Multiplying a matrix $M$ with a vector $v$ is computing what the image of $v$ is under the linear transformation associated to $M$. If you compute $M^{-1}TM$ for some matrix $T$, you are changing the basis.
This latter operation only works if $M$ is invertible, of course. In that case, the columns of $M$, let's call them $b_{1},...,b_{n}$, form a basis. If you have the coordinate vector $v$ of a point with respect to this basis, then $Mv$ gives the coordinates of $v$ with respect to the standard basis $e_{1},e_{2},...,e_{n}$. Then you apply the linear transformation associated to $T$, i.e. you compute $T(Mv)$. To change the coordinates back from the standard basis to the basis $b_{1},...,b_{n}$, you now multiply this with $M^{-1}$. In total, this yields $M^{-1}TMv$.
Certainly, something like $MT$ still exists (provided $M$ and $T$ are square matrices of the same size). Both this corresponds to the composition of two linear transformation. In this case you first apply $T$, then $M$.
In short: you have different operations because you are doing different things.
