Coordinates Let V be a vector space with dim(V ) = n, and B = {v1, v2, . . . , vn} 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) = [v]B It was shown in Sec. 3.4 that this map is well defined, that is,

Question

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.

T (v) =[latex] \left[V\right] _{B} [/latex]

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.

Accepted Answer

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 [latex]c_{1} , . . . , c_{n}[/latex] and [latex]d_{1} , . . . , d_{n}[/latex] such that

u [latex] = c_{1} v_{1} +· · ·+c_{n} v_{n}[/latex] and v [latex] = d_{1} v_{1} +· · ·+d_{n} v_{n} [/latex]

Applying T to the vector ku + v gives

T (ku + v) = T[latex] ((kc_{1} + d_{1} )v_{1} +· · ·+(kc_{n} + d_{n} )v_{n} )[/latex]

=[latex] \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} [/latex]

= kT (u) + T (v)

Therefore, we have shown that the mapping T is a linear transformation.

Question 4.1.7: Coordinates Let V be a vector space with dim(V ) = n, and B ...

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