Why does this happen in linear transformation?

Question

I'm learning linear algebra here from online MOOC's and have just can't understand this. When we want to say rotate a vector by some angle we just multiply it by a transformation matrix to get the resultant vector. Now why do we multiply a vector as viewed from the transformed basis by the same transformation matrix to get the vector as viewed from the normal basis vectors ?

Answer 1

This doesn't just hold for rotations; any invertible matrix will work.

Let $M$ be an invertible $n \times n$ matrix, and $S = (e_1, \ldots, e_n)$ be the standard basis. Then $B = (Me_1, \ldots, Me_n)$ is also a basis, consisting of the columns of $M$.

Consider the map that takes a coordinate vector $[v]_B$ and produces $[v]_S$, i.e. the vector written as usual in standard coordinates. This map is defined as follows: $$\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} \mapsto a_1 Me_1 + \ldots + a_n M e_n = M(a_1 e_1 + \ldots + a_n e_n) = M\begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix}.$$ So, the map that "undoes" this map, i.e. the one that takes a vector $[v]_S$ to $[v]_B$ must be characterised by multiplication by $M^{-1}$. Therefore, $$M[v]_B = MM^{-1}[v]_S = [v]_S,$$ as required.

Answer 2

When in doubt, go back to definitions. The coordinates of a vector $\mathbf v$ relative to some basis $\mathcal B = \{\mathbf b_1,\dots,\mathbf b_n\}$ are the coefficients $c_i$ in the unique linear combination $c_1\mathbf b_1+\cdots+c_n\mathbf b_n$ that equals $\mathbf v$. So, when you perform a change of basis from $\mathcal B$ to $\mathcal B'$, you’re basically solving this equation with $\mathbf v$ and the $\mathbf b_i$ expressed as coordinates relative to the original basis $\mathcal B$, i.e., $$c_1[\mathbf b_1]_{\mathcal B}+\cdots+c_n[\mathbf b_n]_{\mathcal B} = [\mathbf v]_{\mathcal B}$$ or in matrix form $$B[\mathbf v]_{\mathcal B'}=[\mathbf v]_{\mathcal B}.$$ The $\mathbf b_i$ are linearly independent, so $B$ is invertible and the solution to this equation is $B^{-1}[\mathbf v]_{\mathcal B}$.

This is really no different in principle from what you might have learned in basic algebra courses about transforming graphs of functions. Suppose you shift the origin of a Cartesian coordinate system two units right. The coordinates of this new origin in the original coordinate system are obviously $(2,0)$—you’ve added $2$ to the $x$-coordinate. However, coordinates in the shifted system are related to the original coordinates via the equation $x'=x-2$. That is, you subtract $2$ from a point’s $x$-coordinate to get its $x'$-coordinate—the inverse of what you did to obtain the new origin.