# What is the difference between a “change of basis” and a “similarity transformation”?

In crystallography we define a "misorientation", $M_{AB/A}$, as the rotation required to bring crystal A into coincidence with crystal B, as measured with respect to the reference frame of crystal A. Assume that by "rotation" we are referring to members of the group SO(3), and that we are using the active rotation convention, whereby we interpret a rotation as a rotation of space with respect to a fixed coordinate system. In this case we have $M_{AB/A}=R_{A/0}^{-1}R_{B/0}$, where $R_{i/j}$ is the orientation of crystal $i$ (itself a rotation) with respect to reference frame $j$, and $0$ is the reference frame of the fixed coordinate system. What I am confused about is the difference between a similarity transformation and a change of basis.

For example, according to the above definitions, $R_{B/0}=R_{A/0}M_{AB/A}$. However, if I form the product in the opposite order I get $M_{AB/A}R_{A/0}=R_{A/0}^{-1}R_{B/0}R_{A/0}$, which is clearly just a similarity transformation of $R_{B/0}$, and I believe it gives $R_{B/A}$. But it was my understanding that changing the reference frame from which we measure a rotation was equivalent to simply a change of basis from one set of coordinate axes (defined by the first orthonormal basis) to another (defined by the second orthonormal basis). But that would mean that my similarity transformation was equivalent to a change of basis, and I thought the following relationships (which are not equivalent) defined a similarity transformation and change of basis respectively:

Similarity: $A'=B^{-1}AB$

Change of Basis: $A_{[u]}=PA_{[v]}$

Could someone please shed some light on this issue? Thank you.

-
The "change of basis transformation" takes a vector in one reference frame, and gives you the coordinates of that same vector in a different reference frame. By contrast, the "similarity transformation" takes a linear transformation written in terms of one reference frame, and outputs the same linear transformation, but in terms of the other reference frame. That is, "change of basis" takes coordinate vectors to coordinate vectors; "similarity transformations" take operators to operators. They are closely related, though, since they both involve a change of reference frame. – Arturo Magidin Jun 2 '11 at 19:51

Let $\mathcal{V}$ be a n-dimensional vector space (over a field), and let $\alpha$ and $\beta$ be two bases for $\mathcal{V}$. Let $\boldsymbol \tau$ be a linear operator on $\mathcal{V}$, and $\gamma$ any vector in $\mathcal{V}$.
So then $\gamma_{[\alpha]}=P\gamma_{[\beta]}$, where $P$ is the non-singular $\beta$-to-$\alpha$ change of basis matrix. Also $\boldsymbol \tau(\gamma)_{[\alpha]}=A\gamma_{[\alpha]}$ where $A$ is the matrix representative of $\boldsymbol \tau$ with respect to the $\alpha$ basis. Similarly we have $\boldsymbol \tau(\gamma)_{[\beta]}=B\gamma_{[\beta]}$ where $B$ is the matrix representative of $\boldsymbol \tau$ with respect to the $\beta$ basis.
So now making the necessary substitions from the paragraph above: $$P\boldsymbol \tau(\gamma)_{[\beta]}=\boldsymbol \tau(\gamma)_{[\alpha]}=A\gamma_{[\alpha]}=AP\gamma_{[\beta]},$$ and since $P$ is non-singular we then have: $$\boldsymbol \tau(\gamma)_{[\beta]}=P^{-1}AP\gamma_{[\beta]},$$ and so we must have $B=P^{-1}AP$, that is $A$ and $B$ are similar.