Question 4.1.7: Coordinates Let V be a vector space with dim(V ) = n, and B ...
Coordinates Let V be a vector space with dim(V ) = n, and B = \left\{v_{1} , v_{2} , . . . , v_{n}\right\} an ordered basis for V. Let T: V → R^{n} be the map that sends a vector v in V to its coordinate vector in R^{n} relative to B. That is,
T (v) = \left[V\right] _{B}
It was shown in Sec. 3.4 that this map is well defined, that is, the coordinate vector of v relative to B is unique. Show that the map T is also a linear transformation.
Let u and v be vectors in V and let k be a scalar. Since B is a basis, there are unique sets of scalars c_{1} , . . . , c_{n} and d_{1} , . . . , d_{n} such that
u = c_{1} v_{1} +· · ·+c_{n} v_{n} and v = d_{1} v_{1} +· · ·+d_{n} v_{n}
Applying T to the vector ku + v gives
T (ku + v) = T ((kc_{1} + d_{1} )v_{1} +· · ·+(kc_{n} + d_{n} )v_{n} )
= \begin{bmatrix} kc_{1}+ d_{1} \\ kc_{2}+ d_{2} \\ \vdots \\ kc_{n}+ d_{n} \end{bmatrix} = k \begin{bmatrix} c_{1} \\ c_{2} \\ \vdots \\ c_{n} \end{bmatrix}+ \begin{bmatrix} d_{1} \\ d_{2} \\ \vdots \\ d_{n} \end{bmatrix}
= kT (u) + T (v)
Therefore, we have shown that the mapping T is a linear transformation.