How can I find the bases and generated subspace for a given set of vectors?

[kacete]

Homework Statement

From the course of Linear Algebra and Analytic Geometry

I need to find the dimension and two different bases of subspace R³ generated by vectors (1,2,3), (4,5,6), (7,8,9).

Homework Equations

None.

The Attempt at a Solution

I tried

(a,b,c)=α₁(1,2,3)+α₂(4,5,6)+α₃(7,8,9)

which became

a = α₁ + 4α₂ + 7α₃
b = 2α₁ + 5α₂ + 8α₃
c = 3α₁ + 6α₂ + 9α₃

which (by Gaussian elimination) became an undetermined system with free variable α₃.

4. The solution given by teacher
dim=2, example of bases={(1,0,-1),(0,1,2)} or {(2,1,0),(-1,0,1)}

I don't want the solution, I just want to understand the mechanics on how to find the bases and the generated subspace. If someone could explain it to me, thank you.

 

[Mark44]

The system of equations you show would be used to find the span of your three vectors. Instead of equations that start with a= , b=, and c=, put 0 in for all three of those variables. You should end up with a row of zeroes and two nonzero rows.

What did you end up with when you row-reduced your matrix?

 

[kacete]

So, I should have used the homogeneous system A . x = 0 ? Being A a matrix. Hmm...
I ended up with (after Gaussian elimination):

[ 1  4  7 | a      ]
[ 0 -3 -6 | b-2a   ]
[ 0  0  0 | c-a-2b ]

 

[Mark44]

If I can backpedal a bit, your work is fine. For the system represented by your augmented matrix to be consistent, it must be that c - a - 2b = 0.

or

a = -2b + c
b =    b
c =          c

I added the 2nd and 3rd equations above so that I can get some vectors out of the equation c - a - 2b = 0. The equations I added are obviously true for all values of b and c, respectively.

Any vector [a b c]^T is a linear combination of [-2 1 0]^T and [1 0 1]^T. These come from setting b = 1, c = 0 and then b = 0, c = 1.

Different pairs of choices for b and c will give you different pairs of vectors for your basis.

Hope that helps.