Does the following vectors generate R^3?

• Feb 21st 2009, 08:52 AM
arbolis
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?
• Feb 21st 2009, 10:46 AM
ThePerfectHacker
Quote:

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.
• Feb 21st 2009, 11:21 AM
arbolis
Quote:

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 http://www.mathhelpforum.com/math-he...45072a0e-1.gif 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.