Prove that the span: $\bigl\{[x,y,z]\in\Bbb R^3:y=3x\bigr\}$.

Question

Given the following vectors, $[1, 3, 3]$, $[0, 0, 1]$ & $[1, 3, 1]$ as referenced to this question in the other forum "Find the span of a set of vectors", how do i prove that the span of the three vectors equivalent to $\bigl\{[x,y,z]\in\Bbb R^3:y=3x\bigr\}$?

I'm having a conceptual block here, but the best I can think of is to reduce the vectors into their row-echelon form. However, its row-echelon form: $$\begin{bmatrix}1&0&1\\ 0&1&-2\\ 0&0&0 \end{bmatrix}$$ consists one zero row, and thus the linear system is not always consistent which means the span $≠$ $R^3$. Am i right? Nevertheless, how do i prove that the span: $\bigl\{[x,y,z]\in\Bbb R^3:y=3x\bigr\}$?

Thanks.

Answer 1

Since the row echelon form has a zero row, we know that one of the vectors can be written as a linear combination of the others (that is, one vector lies in the span of the other two).

By inspection, we see that $(1,3,3)^T = (1,3,1)^T + 2 (0,0,1)^T$ so we can reduce the problem to finding the span of the other two vectors, $S = \operatorname{sp} \{ (1,3,1)^T , (0,0,1)^T \}$.

We can write $S= \{ x (1,3,1)^T + w (0,0,1)^T | x,w \in \mathbb{R} \} $. Since the $w$ is arbitrary, we can write \begin{eqnarray} S &=& \{ x (1,3,1)^T - x(0,0,1)^T + (w-x) (0,0,1)^T | x,w \in \mathbb{R} \} \\ &=& \{ x (1,3,0)^T + z(0,0,1)^T | x,z \in \mathbb{R} \} \\ &=& \{ (x,3x,z)^T| x,z \in \mathbb{R} \} \end{eqnarray}

Answer 2

You might be confusing a spanning set with a base. Being a base is stronger condition, base is linearly independent spanning set, while in general spanning sets need not be linearly independent. Row-echelon form will give you a base which doesn't need to coincide with given spanning set (and usually it won't).

By definition, $$\operatorname{span}_kS = \{\sum_{i=1}^n\alpha_iv_i\mid n\in\mathbb N, \alpha_i \in k, v_i\in S, i = 1,\ldots, n\}$$

and we need to prove that $$\operatorname{span}_{\mathbb R}\{(1,3,3),(0,0,1),(1,3,1)\} = \{(x,y,z)\in\mathbb R^3\mid y = 3x\}.$$

To prove one inclusion, we need to prove that $$\alpha(1,3,3)+\beta(0,0,1)+\gamma(1,3,1) \in \{(x,y,z)\in\mathbb R^3\mid y = 3x\},\ \forall \alpha,\beta,\gamma \in\mathbb R,$$ but, since the set on RHS is a subspace, it is enough to prove that $$(1,3,3),(0,0,1),(1,3,1) \in \{(x,y,z)\in\mathbb R^3\mid y = 3x\}$$ and all linear combinations will be automatically contained in the subspace. These vectors are obviously all vectors in $\mathbb R^3$ and satisfy equation $y = 3x$, so we have:

$$\operatorname{span}_{\mathbb R}\{(1,3,3),(0,0,1),(1,3,1)\} \subseteq \{(x,y,z)\in\mathbb R^3\mid y = 3x\}.$$

To prove the other inclusion, we need to prove that any vector $(x,y,z)\in\mathbb R^3$ such that $y = 3x$ can be written as linear combination of $(1,3,3),(0,0,1),(1,3,1)$, i.e. we need to find scalars $\alpha,\beta,\gamma\in\mathbb R$ such that

$$(x,3x,z) = \alpha(1,3,3)+\beta(0,0,1)+\gamma(1,3,1)$$

or, equivalently, solve linear system

$$\left(\begin{array}{c c c | c} 1 & 0 & 1 & x\\ 3 & 0 & 3 & 3x\\ 3 & 1 & 1 & z \end{array}\right).$$

Now, all solutions are given by $(x,z-3x,0)+t(-1,2,1),\ t\in\mathbb R,$ but we only need one, for example by letting $t = 0$, we can see that $$(x,3x,z) = x(1,3,3)+(z-3x)(0,0,1)+0(1,3,1).$$ This proves that $$\{(x,y,z)\in\mathbb R^3\mid y = 3x\}\subseteq \operatorname{span}_{\mathbb R}\{(1,3,3),(0,0,1),(1,3,1)\}.$$