# Thread: Linear Basis

1. ## Linear Basis

I am trying to explain why S is not a basis for R3.

S={(1,1,2),(0,2,1)}
S is a basis if it spans and is linearly independent.

I have determined that S is linearly independent from c1(1,1,2)+c2(0,2,1)=0;c1=c2=0
The next step is to determine if it spans. I can't use the determinant since it is not a square matrix. Could someone help me how to determine if a non-square matrix spans or not?

2. Dimension of $\displaystyle R^3$ is 3, so there are only 3-elements basis that are basis for $\displaystyle R^3$

3. How does that show that S is not a basis of R3 since S shows 3 elements?

4. I mean 3-vectors basis in stead of 3-elements

5. A basis for a vector space always has three properties:
1) The vectors in the basis are independent.
2) The vectors span the space.
3) The number of vectors in the basis is equal to the dimension of the space.

And any two of those is sufficient to prove the third. That's why it is sufficient to observe that there are only two vectors in that set while $\displaystyle R^3$ has dimension three.

However, you may not have had that yet. It is best to show this is not a basis by showing it does not span $\displaystyle R^3$. Any vector in the span of {(1,1,2), (0,2,1)} can be written as a(1,1,2)+ b(0,2,1)= (a, a+ 2b, 2a+ b). Can you find a member of $\displaystyle R^3$ that is NOT of that form? Try something of the form, (0, 0, x). What must a and b equal?