Describe the Linear span from the following vectors(3). Find the basis and dimension of that linear subspace.

Question

First and foremost, I'm translating my work from French to English so I apologize if I use incorrect terms.

Question: Describe the Linear span from the following vectors. Find the basis and dimension of that linear subspace.

$U = \{(-3,6,-9),(1,-2,3),(6,-12,18)\}$

1. I found that the determinant from the matrix formed with the $3$ vectors and it gave a determinant of $0$, meaning that the vectors are linearly dependent.

I need to find the vector equation using the following system: \begin{align} -3a + b + 6c &= x\\ 6a -2b -12c &= y\\ -9a +3b +18c &= z \end{align} Solving the system, I came with this answer: $$ U=\{(x;y;z)\in \mathbb{R}^3 | z + 1.5y = 0\} $$ I found it weird that the x disappeared, so I tried solving it another way and came with this answer: $$ U=\{(x;y;z)\in \mathbb{R}^3 | x-y-z = 0\} $$ Both answer apparently work and I am totally confused as to which I have to use to find a basis and the dimension... or does it not matter?

If so, does that mean there exists many answers for the basis? Thank you. I appreciate the help.

Answer 1

The vectors in $U$ are all multiples of each other, so they span just a line.

This means (by the rank-nullity theorem) that the solution set to $xu_1+yu_2+zu_3$ is a two-dimensional subspace of the $(x,y,z)$-space. That's why you can get two different $(x,y,z)$ combinations that both work. But you should be critical of the methods you used to find those two solutions, if both of them failed to tell you that what you got was not the entire solution.

(It also seems that you are a bit confused between what is $(a,b,c)$ and what is $(x,y,z)$, which may or may not explain why something went wrong in your solution.