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Ok, so I'm a bit curios as to how you can prove a set does not span a vector space.

For example, let ${S}$ be the vector set \begin{bmatrix} 1\\ 0\\ 0\\ 0\\ \end{bmatrix}

\begin{bmatrix} 0\\ 1\\ 0\\ 0\\ \end{bmatrix}

\begin{bmatrix} 0\\ 0\\ 1\\ 0\\ \end{bmatrix}

\begin{bmatrix} 0\\ 0\\ 0\\ 1\\ \end{bmatrix}

So how would you prove that it is not in ${R^3}$.


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These can actually be in $R^3$. – Juris Apr 18 '14 at 8:39
Really? How so? I assumed by adding making the last one be [0,0,1,1] I was preventing this. Uggg... Can this set span ${R^3}$ because there are only 3 unknowns? – boidkan Apr 18 '14 at 8:41
Any three literally independent vectors should be able to span $R^3$ – Nikhil Mahajan Apr 18 '14 at 8:45
I don't get it. These are vectors with $4$ components, hence elements of $\mathbb R^4$. Elements of $\mathbb R^3$ are vectors with $3$ components. Of course no element of $\mathbb R^4$ is an element of $\mathbb R^3$. So what do you mean by "it is not in $R^3$"? – Christoph Apr 18 '14 at 8:48
@boidkan: do you know the words "isomorphism" and "homomorphism"? You seem to be asking whether the space spanned by those 4-vectors is a subset of $\mathbb{R}^3$, which of course it is not since none of them is an element of $\mathbb{R}^3$. What Christoph proves in his answer is that no subspace of $\mathbb{R}^3$ is 4-dimensional, which you can relate to your 4-vectors by saying that the space they span ($\mathbb{R}^4$)is not isomorphic to any subspace of $\mathbb{R}^3$. Or say that there's no injective homomorphism from $\mathbb{R}^4$ to $\mathbb{R}^3$. – Steve Jessop Apr 18 '14 at 12:18
up vote 3 down vote accepted

In your last comment you said:

Basically what I was trying to ask is how to prove that a set that is 4 dimensional not be in $R^3$.

A $4$-dimensional subspace of $\mathbb R^3$ would have a basis consisting of $4$ elements, by the definition of dimension. For the sake of argument, assume we have a $4$-dimensional subspace $U\subseteq \mathbb R^3$ and $U$ has a basis $\{v_1,v_2,v_3,v_4\}$. So we have $4$ linearly independent vectors in $\mathbb R^3$ now. Since $\{v_1,v_2,v_3,v_4\}$ is linearly independent, the subset $\{v_1,v_2,v_3\}$ is linearly independent as well. Now any set of $3$ linearly independent vectors in $\mathbb R^3$ actually spans $\mathbb R^3$, i.e. $$ \mathbb R^3 = \operatorname{span}(v_1, v_2, v_3) = \left\{\, a_1 v_1 + a_2 v_2 + a_3 v_3 \in \mathbb R^3 \mid a_1,a_2,a_3\in\mathbb R\,\right\}. $$ Since $v_4\in\mathbb R^3$, it follows that $$ v_4 = a_1 v_1 + a_2 v_2 + a_3 v_3 $$ for some $a_1,a_2,a_3\in\mathbb R$. But then $\{v_1,v_2,v_3,v_4\}$ is not linearly independent, so we reached a contradiction.

It follows that $\mathbb R^3$ can not have a subspace of dimension $4$.

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In your case it's easy, because a base for $\mathbb{R}^4$ cannot have less than $4$ elements (previous version of the question had $|S| = 3$). Nevertheless, suppose that we have additional vector $s_3 = [1,1,0,0]^T$, that is

$$S = \Big\{[1,0,0,0]^T, [0,1,0,0]^T, [0,0,1,1]^T, [1,1,0,0]^T\Big\}$$

Let $v = [0,0,1,0]^T$. If $S$ spans $\mathbb{R}^4$ then $$v = a_0s_0 + a_1s_1 + a_2s_2 + a_3s_3$$

for some $a_0,\ldots,a_3 \in \mathbb{R}$. However, considering the coordinates one by one, we get the following set of equalities

$$ \begin{cases} a_0+a_3 = 0, \\ a_1+a_3 = 0, \\ a_2 = 1, \\ a_2 = 0. \end{cases} $$

While there is no problem with first two, the second two imply a contradiction, i.e. $1 = a_2 = 0$, and so $v$ cannot be expressed as a linear combination of $S$, hence $S$ does not span $\mathbb{R}^4$.

I hope this helps $\ddot\smile$

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I edited my question but this answers the general concept I was asking so I will accept this as an answer. Thanks! – boidkan Apr 18 '14 at 8:50

Plug the vectors in as rows in a matrix, then row-reduce to find a basis for the row space. Remember the row space of a matrix is the subspace spanned by the initial row vectors. If you end up with one or more rows of zeros after row-reduction, then that indicates that your initial row vectors were not linearly independent.

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You simply have to check how many linearly independent vectors you have. If you have 3, you span $\mathbb{R}^3$, if you have four - they span $\mathbb{R}^4$. If you have $n$ vectors you can be sure they can't span more than $\mathbb{R}^n$.

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As I said in the comments, a three dimensional subspace of $\mathbb R^4$ is different from $\mathbb R^3$. Please take that into account. – Christoph Apr 18 '14 at 9:09

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