Let $V = \{ (x_1, x_2, x_3,x_4)\in R^4: x_1-2x_2+x_4 = 0, 2x_1-3x_2+x_3 = 0 \}$
I am trying to find a basis for V. Subtracting the constraints from each other yields $x_4= x_2/2+x_3/2$ this means that we have 3 degrees of freedom so the number of elements in our basis should be 3.
$(1,0,0,0), (0,1,0,1/2), (0,0,1,1/2)$ are three linear independent vectors with span of V.
Then is this a basis for V? Am I allowed to subtract the constraint equations like so?
Thanks.
Best Answer
Take
$$u=(1,0,-2,-1)$$ $$v=(0, 1, 3, 2)$$
and you are done.
Every vector in V has a representation with these two vectors, as you can check with ease. And from the first two components of u and v, you see, u and v are linear independet.
You have two equations in four unknowns, so rank is two. You can't find more then two linear independent vectors.