I’m a high school student studying Linear Algebra for Neural Networks and am on Linear Transformations and Basis Change.
When using Khan Academy, it said in regards from changing basis and linear transformations within the same dimension (Rn->Rn) that to obtain a matrix D that transforms some vector x we use this ‘formula’
$$ D = C^{-1}AC $$ Where C is the change of basis transformation matrix, and A is the original transformation matrix.
But, in the book ‘Mathematics for Machine Learning’, it speaks about transforming between dimensions $\mathbb{R^n}$ to some $\mathbb{R^m}$ and using 4 bases ”B, B’, C & C’”. Then it states to find the new transformation matrix A’ which I’m guessing is similar to what Khan Academy references as D, we use: $$ A’ = T^{-1}AS $$ Where, quoted from the book, ‘Here, S of $\Bbb{R^{nxn}}$ is the trans. Matrix of IDv that maps coordinates with respect to B’ onto coordinates with respect to B, and T of $\Bbb{R^{mxm}}$ is the trans. Matrix of IDw that maps coordinates with respect to C’ onto coordinates with respect to C”. Why are we dealing with 4 bases now?
Apologies if I’m explaining this poorly. I’d appreciate if anyone could help me on this. If you need an image of what I’m talking about in the book I can, I just didn’t want to clutter the original post too much.
Thanks, Justin
If you have two basis changes, say $S:\mathbb R^n\dashrightarrow\mathbb R^n$ and $T:\mathbb R^m\dashrightarrow\mathbb R^m$, where we use $\dashrightarrow$ to indicate basis change to get no confused with a linear transformation arrow, and you have a linear transformation $A:\mathbb R^n\to\mathbb R^m$, then the assignment is given by symbols $v\mapsto Av$. So, if we involve the basis changes then we have $$Av=(AS)S^{-1}v$$ which explain how the old components of $Av$ are related (via the $AS$ matrix) with the new components of $v$. But $Av=TT^{-1}Av=ASS^{-1}v$ brings the basis change $T$ into the play, then $$T^{-1}Av=(T^{-1}AS)S^{-1}v.$$ Here we can see how the new components of $Av$ under the basis change $T$ are fasten with the new components of $v$ under the change $S$ through the matrix $T^{-1}AS$.