Coordinate Systems

tl;dr

Let $B = {b_{1}, b_{2}, b_{3}, b_{4}, \dots b_{n},}$ be a basis for $R^{n}$ .

To find the coordinates of a vector $v$ with respect to $B$ , let $P = [b_{1}, b_{2}, b_{3}, b_{4}, \dots b_{n},]$ then,

[v]_{B} = P^{- 1} \times v

This takes us from standard coordinates to B-coordinates.

B_{2} \leftarrow B_{1} P = B_{2}^{- 1} C^{- 1} C \times B_{1} B

Standard coordinates in $R^{2}$ ?

Standard basis in $R^{2}$ , $S = {e_{1}, e_{2}} = {[10], [01]}$ .

Coordinates are labels that we assign vectors (based on a given basis).

~~When we say $v = [23] = 2 e_{1} + 3 e_{2}$ , we can represent that graphically…~~

But what if we picked other basis that aren’t the standard?

Coordinates using bases that are not the standard basis

$B = {[1 - 1], [10]}$ .

We know this is a basis because they are linearly independent, the Determinant isn’t zero $1 - 1 10 = (0) - (- 1) = 1$ , thus they are linearly independent.

Let’s find the coordinates of $v = [23]$ in terms of the basis of B.

[23] = a [1 - 1] + b [10] [v]_{B} = coordinates of v relative to the basis B The coordinates of v relative to B

Because the basis is linearly independent, the determinant isn’t zero. We can get $[a b]$ via the inverse of the basis times $v$ .

[a b] [a b] = [1 - 1 10]^{- 1} [23] = \frac{1}{Determinant 1 \cdot 0 - ( - 1 ) ( 1 )} [01 - 1 1] [23] = [- 3 5]

vishwiki

Explorer

Coordinate Systems

tl;dr

Coordinates using bases that are not the standard basis

Graph View