Matrices of linear transformation, help with what this basis means.

Question

I don't understand this, doesn't $\mathbb{R}^n$, have the standard vectors as basis? How can $x$ then have the basis $B$ if it's in $\mathbb{R}$. The book just says that $B$ is some arbitrary base.

Answer 1

Any $n$-dimensional vector space is isomorphic to $\mathbb{R}^n$. An easy and very useful isomorphism is the coordinate isomorphism with respect to some basis $B$:

Consider a vector $v= \sum_{i=1}^n v_i b_i$ where $\{b_i\}=B$ and $v_i\in\mathbb{R}$. The coordinate isomorphism with respect to $B$ is then the map $v\mapsto (v_1,...,v_n)^t$. So you just map a vector to the column vector containing its coefficients with respect to $B$ and I assume that this column vector is what they mean by $[x]_B$.