# Showing that a given set of vectors forms a basis

## Homework Statement

show that

$$\left(\begin{array}{cc}2 & -1\\-1 & 1\end{array}\right)$$

forms a basis for R^2

## The Attempt at a Solution

ok...my instructor said he wants me to show that they are linearly independant and to show that they span to form a basis...not just by a theorm

im trying to show that the c_{1} and c_{2} are zero to say that they are but i get

2c1-c2=0 and c2=c1 which does not work

my instructor does not want me to do the determinant way

i think i can show that it spans R^2 but i just need help showing that it is linear independant

## Answers and Replies

anyone?

Dick
Homework Helper
I think you mean to say that you want to show the two column vectors [2,-1] and [-1,1] are linearly independent. You probably don't want to write that as a 2x2 matrix. But are there any solutions to 2c1-c2=0 and c2=c1 that aren't c1=0 and c2=0? BTW don't bump your problem after only 20 minutes.

sorry bout the bump...

no...c1 or c2 must be zero to satisfy those equations, which would make them linearly independant?

pls tell me if next part is right...

if i say that 2c_1 - c2 = x1 and -c1+c2 = x2.....since this tells me how to get c's....this set spans R^2, thus forming a basis?

Dick
Homework Helper
You're forgiven for the bump, just give people time to collect their thoughts. And yes, they are linearly independent and yes, 2c_1 - c2 = x1 and -c1+c2 = x2 can be solved for any x1 and x2. So they span. I think you get it. Happy?

almost =) perhaps you can take a gander at my other thread about the runga kutta method?

thank you

Dick
Homework Helper
almost =) perhaps you can take a gander at my other thread about the runga kutta method?

thank you

I'm with everyone else on that one. If you are going to do a numerical integration, I think you need a numerical initial value.

thats the thing...i went to two different professors today and asked for help, one of which was the author of the book. he said its solveable and that you don't need an initial condition to solve, and another professor said that you do, because you can have so many different solutions on that x value that you need the y value to start

Dick