# Thread: Does the following vectors generate R^3?

1. ## Does the following vectors generate R^3?

Hi MHF,
I know it's a very basic exercise but as I'm unsure about what I've done, I'd like a check up.
In each case, say if the following vectors generates $\displaystyle \mathbb{R}^3$. If not, characterize implicitly the generated subspace.
1)$\displaystyle (1,0,-1),(1,2,1),(0,-3,2)$.
My attempt : I wrote these 3 vectors as column vectors in a matrix and I calculated the determinant of the matrix. I got that it is different from zero, hence the matrix is invertible and the 3 vectors are linear independent. From it I concluded that they generate $\displaystyle \mathbb{R}^3$.

2)(2,-2,0),(2,-2,1),(-1,4,-3).
My attempt : I did the same as in 1) and got exactly the same conclusion.

3)(1,-2,2),(2,1,3),(0,3,1),(1,2,3).
My attempt : Here it gets interesting. The 4 vectors cannot be all linear independent,but they still might generate $\displaystyle \mathbb{R}^3$. I wrote them as vector columns in a matrix and reduced it. I got that the 3x4 matrix has 3 rows reduced. So the 3 first vectors generates $\displaystyle \mathbb{R}^3$.

4)(1,3,-3),(2,3,-4),(1,-3,1),(3,0,-3).
My attempt : I wrote the vectors as column vectors in a matrix and reduced it. I got that a row of the matrix is null. While 2 rows are reduced, hence the 2 first vectors generate $\displaystyle \mathbb{R}^2$.

Am I right? Did I give the generated subspace implicitly in 4)? By the way for 3) and 4) I wrote an amplied matrix with the column a,b,c. I don't think it was of any use now... or was it useful to describe the subspace generated implicitly?

2. Originally Posted by arbolis
Hi MHF,
I know it's a very basic exercise but as I'm unsure about what I've done, I'd like a check up.
In each case, say if the following vectors generates $\displaystyle \mathbb{R}^3$. If not, characterize implicitly the generated subspace.
1)$\displaystyle (1,0,-1),(1,2,1),(0,-3,2)$.
My attempt : I wrote these 3 vectors as column vectors in a matrix and I calculated the determinant of the matrix. I got that it is different from zero, hence the matrix is invertible and the 3 vectors are linear independent. From it I concluded that they generate $\displaystyle \mathbb{R}^3$.

2)(2,-2,0),(2,-2,1),(-1,4,-3).
My attempt : I did the same as in 1) and got exactly the same conclusion.

3)(1,-2,2),(2,1,3),(0,3,1),(1,2,3).
My attempt : Here it gets interesting. The 4 vectors cannot be all linear independent,but they still might generate $\displaystyle \mathbb{R}^3$. I wrote them as vector columns in a matrix and reduced it. I got that the 3x4 matrix has 3 rows reduced. So the 3 first vectors generates $\displaystyle \mathbb{R}^3$.

4)(1,3,-3),(2,3,-4),(1,-3,1),(3,0,-3).
My attempt : I wrote the vectors as column vectors in a matrix and reduced it. I got that a row of the matrix is null. While 2 rows are reduced, hence the 2 first vectors generate $\displaystyle \mathbb{R}^2$.

Am I right? Did I give the generated subspace implicitly in 4)? By the way for 3) and 4) I wrote an amplied matrix with the column a,b,c. I don't think it was of any use now... or was it useful to describe the subspace generated implicitly?
I am not going to do the computations, I am too lazy, but you can use the rank-nullity theorem. In problem 3 write $\displaystyle A$ to be the $\displaystyle 3\times 4$ matrix consisting of these columns. You want to show that $\displaystyle A\bold{x} = \bold{b}$ has a solution for any $\displaystyle \bold{b}\in \mathbb{R}^3$. Define $\displaystyle T: \mathbb{R}^4 \to \mathbb{R}^3$ by $\displaystyle T(\bold{x}) = A\bold{x}$, this is a linear transformation. We know that $\displaystyle \text{rank}(T) + \text{nullity}(T) = 4$ (the dimension of $\displaystyle \mathbb{R}^4$). The rank of $\displaystyle T$ is the dimension of the image of $\displaystyle T$ i.e. $\displaystyle \text{dim}\left( T[\mathbb{R}^4] \right)$. Thus, $\displaystyle T$ is onto (which is what we want to show) if and only if $\displaystyle \text{rank}(T) = 3$ since the dimension of $\displaystyle \mathbb{R}^3$ is $\displaystyle 3$. It remains that all we need to show is that $\displaystyle \text{nullity}(T) = 1$. Remember $\displaystyle \text{nullity}(T)$ is the dimension of the nullspace of $\displaystyle T$ i.e. $\displaystyle \text{dim}\ker (T)$. Thus, you need to show that the set of solution to $\displaystyle A\bold{x} = \bold{0}$ has one element in its basis. Now procede by Gaussian-Jordan elimination.

3. Originally Posted by ThePerfectHacker
I am not going to do the computations, I am too lazy, but you can use the rank-nullity theorem. In problem 3 write $\displaystyle A$ to be the $\displaystyle 3\times 4$ matrix consisting of these columns. You want to show that $\displaystyle A\bold{x} = \bold{b}$ has a solution for any $\displaystyle \bold{b}\in \mathbb{R}^3$. Define $\displaystyle T: \mathbb{R}^4 \to \mathbb{R}^3$ by $\displaystyle T(\bold{x}) = A\bold{x}$, this is a linear transformation. We know that $\displaystyle \text{rank}(T) + \text{nullity}(T) = 4$ (the dimension of $\displaystyle \mathbb{R}^4$). The rank of $\displaystyle T$ is the dimension of the image of $\displaystyle T$ i.e. $\displaystyle \text{dim}\left( T[\mathbb{R}^4] \right)$. Thus, $\displaystyle T$ is onto (which is what we want to show) if and only if $\displaystyle \text{rank}(T) = 3$ since the dimension of $\displaystyle \mathbb{R}^3$ is $\displaystyle 3$. It remains that all we need to show is that $\displaystyle \text{nullity}(T) = 1$. Remember $\displaystyle \text{nullity}(T)$ is the dimension of the nullspace of $\displaystyle T$ i.e. $\displaystyle \text{dim}\ker (T)$. Thus, you need to show that the set of solution to $\displaystyle A\bold{x} = \bold{0}$ has one element in its basis. Now procede by Gaussian-Jordan elimination.
So this is somewhat what I've done, I mean I reduced the matrix. Precisely $\displaystyle \begin{bmatrix} 1 & 2 & 0 & 1 \\ -2 & 1 & 3 & 2 \\ 2 & 3 & 1 & 3 \end{bmatrix} \Leftrightarrow \begin{bmatrix} 1 & 0 & 0 & \frac{3}{4} \\ 0 & 0 & 1 & \frac{9}{8} \\ 0 & 1 & 0 & \frac{1}{8} \end{bmatrix}$. Finally it means that $\displaystyle \begin{cases} x+\frac{3w}{4} = 0\\ z+\frac{9w}{8} \\ y+\frac{w}{8}=0 \end{cases}$ and we see any vector solution can be written as $\displaystyle w\cdot (-\frac{3}{4}, -\frac{1}{8}, -\frac{9}{8}, 1)$ and we see that the base has a dimension of 1. But I'm not sure I've understood well what you mean by "Thus, you need to show that the set of solution to has one element in its basis.".
The nullspace is the set $\displaystyle \{ \bold{x} \in \mathbb{R}^4 : A\bold{x} = \bold{0}\}$. The nullity of this linear transformation is the dimension of this subspace. You have shown that $\displaystyle \text{nullity}(T)=1$ and this is even to complete the proof as explained above.