Finding x in R^2 with Basis B & Coordinate Vector [x]_B

[UrbanXrisis]

Consider the basis B of R^2 consisting of vectors

\left(\begin{array}{c}-6 & -2 \end{array}\right) and \left(\begin{array}{c}1 & 3 \end{array}\right)

find x in R^2 whose coordinate vector relative to the basis B is

[x]_B = \left(\begin{array}{c}6 & 5 \end{array}\right)

I need to find x

i'm not sure what a coordinate vector is but here's how I would solve this:

-x6+y=6
-x2+3y=5

solving for x and y, i get x = \left(\begin{array}{c}-.8125 & 1.125 \end{array}\right)

this is not correct and I'm not sure how to find x

 

[AKG]

In general, if {v,w} is an ordered basis for your space, then any vector x in your space can be expressed as a linear combination of v and w, i.e. there exists numbers a and b such that x = av + bw. Then, with respect to the ordered basis {v,w}, the coordinate vector of x would be (a b)^T.

Note:

\left[\left(\begin{array}{c}-6 & -2 \end{array}\right)\right]_B = \left(\begin{array}{c}1 & 0 \end{array}\right)

\left[\left(\begin{array}{c}1 & 3 \end{array}\right)\right]_B = \left(\begin{array}{c}0 & 1 \end{array}\right)

 

[UrbanXrisis]

what was done to show that \left[\left(\begin{array}{c}-6 & -2 \end{array}\right)\right]_B = \left(\begin{array}{c}1 & 0 \end{array}\right)?

 

[AKG]

Look at:

In general, if {v,w} is an ordered basis for your space, then any vector x in your space can be expressed as a linear combination of v and w, i.e. there exists numbers a and b such that x = av + bw. Then, with respect to the ordered basis {v,w}, the coordinate vector of x would be (a b)T.

I found a to be 1, and b to be 0, when taking x to be v, i.e. the first vector in your ordered basis.

 

[AKG]

Just reread this section of your textbook until you become more comfortable with it. Try the problems as well, but if you're still getting stuck with the basic ones, go back and reread.