Matrix change of basis, versus vector I am under the impression, that changing the basis of a vector involves multiplication by a matrix:
$$
\mathbf{x}'=\mathbf{A}\mathbf{x}
$$
where $\mathbf{x}',\mathbf{x}\in \mathbb{R}^n$ and $\mathbf{A}\in\mathbb{R}^{n\times n}$.
But the same concept of a change of basis over a matrix involves the multiplication be a similarity transformation;
$$
\mathbf{A}'=P\mathbf{A}P^{-1}
$$
Is this correct? Why not $\mathbf{A}'=\mathbf{B}\mathbf{A}$?
The reason I am asking about the difference, is when it comes to integrating the determinant of a matrix, say
$$
1=\int \det \mathbf{M}(x,y)dxdy
$$
To perform a change of coordinates, such that the integral remains invariant, must one write it as such:
$$
1=\int \det \mathbf{M}(x,y)\sqrt{|g|}dxdy
$$
How then is the transformation done with both a similarity transformation and a matrix-multiplication vector transformation? Must I do both $M'=PMP^{-1}$ and a coordinate transformation $(x,y)\to (f(x,y),g(x,y))$ for instance:
$$
M'=PM(f(x,y),g(x,y)) P^{-1}
$$
Is the function
$$
1=\int \det \mathbf{M}(x,y)\sqrt{|g|}dxdy
$$
invariant with respect to arbitrary coordinate changes?
 A: To answer your first question:
This 'similarity' transformation for a matrix arises because (!) of the way a vector is transformed. Consider a vector $x \in \mathbb{R}^n$ and a matrix $A \in \mathbb{R}^{n\times n}$.
Now, consider a basis transformation, such that $x' = Bx$ with $B \in \mathbb{R}^{n \times n}$. The matrix $A$ is defined with respect to the old basis, meaning that the operation $A x'$ will not(!) yield the desired result. To obtain your desired result, you transform your vector into its original form, i.e. $(B^{-1}x')$ and THEN do the matrix multiplication $A (B^{-1} x')$. However, the resulting vector will again be with respect to the OLD basis, therefore you transform it such that it is described with respect to the NEW basis. The whole operation thus looks like
$$ B \cdot A \cdot (B^{-1} x') = (BAB^{-1})x' $$
Consequently, your matrix with respect to the NEW basis is of the form $A' = (BAB^{-1})$.
$${}$$
A remark to the second question:
Under a change of basis $A' = BAB^{-1}$, the determinant of the matrix $A'$ transforms as follows:
$$ \det A' \enspace = \enspace \det \big( B A B^{-1} \big) \enspace = \enspace \det B \cdot \det A \cdot \det B^{-1} \enspace = \enspace \det B \cdot \det A \cdot \frac{1}{\det B} \enspace = \enspace \det A $$
The determinante is therefore invariant under a change of basis.
