# Span of vectors with more entries in each vector than the amount of vectors

I can easily visualize why $$\begin{bmatrix} 1\\ 0 \end{bmatrix}$$ and $$\begin{bmatrix} 0\\ 1 \end{bmatrix}$$ spans $$R^2$$. I have learned that we need at least two linearly independent vectors to span $$R^2$$, $$3$$ to span $$R^3$$ and so on... This is easy to visualize when the entries in the vectors are the same as the amount of total vectors.

But what happens if we for instance have the two vectors $$\begin{bmatrix} 1\\ 0 \\ 1 \\ 0 \end{bmatrix}$$ and $$\begin{bmatrix} 0\\1\\0\\1 \end{bmatrix}$$.

Clearly, the two vectors are linearly independent, but do they still only span $$R^2$$? If we add the two vectors we produce $$\begin{bmatrix} 1\\ 1\\1\\1 \end{bmatrix}$$ which is a vector in $$R^4$$, right?

So my question boils down to what is the span of $$\begin{bmatrix} 1\\ 0 \\ 1 \\ 0 \end{bmatrix}$$ and $$\begin{bmatrix} 0\\1\\0\\1 \end{bmatrix}$$, and in general what is the span of a set of vectors with more entries in each vector than vectors in total?

The span of the vectors you mention is a subspace of the vector space to which they belong. So $$(1,0,1,0)$$ and $$(0,1,0,1)$$ span a two dimensional subspace (since as you note the vectors are linearly independent) of $$\mathbb{R}^4$$. It is not $$\mathbb{R}^2$$ exactly as this subspace contains $$4$$-tuples, but it is `equivalen' (in some sense) to $$\mathbb{R}^2$$ if this is of interest to you.

So in general a span of $$m linearly independent vectors in $$\mathbb{R}^n$$ is a $$m$$-dimensional subspace of $$\mathbb{R}^n$$.

Your vectors live in $$\mathbb{R}^4$$, not in $$\mathbb{R}^2$$, and therefore it is out of the question that they span $$\mathbb{R}^2$$. Since they live in $$\mathbb{R}^4$$, they span a subspace of $$\mathbb{R}^4$$, whish happens to be$$\left\{\begin{bmatrix}x&y&x&y\end{bmatrix}^T\,\middle|\,x,y\in\mathbb{R}\right\}.$$

• I really don't understand downvotes for such kind of OP and answers! – user Sep 29 '18 at 13:32
• @gimusi Me neither. – José Carlos Santos Sep 29 '18 at 13:34
• @gimusi: I can understand downvotes for the question. It is a question about a comparatively advanced concept (spans) when the OP appears to be fundamentally confused about a more basic concept (linear spaces). Such questions are difficult to answer - and even if a direct answer is attempted, it doesn't necessarily remove the original confusion. – Meni Rosenfeld Sep 29 '18 at 20:35
• @MeniRosenfeld I can't know in advance all the lacking in knowledge by the asker. I usually give the strictly necessary information and then if the OP requires some further explanation I add that or I discuss that by comments. See for example OP there is a line of answer and 3 hours of discussion. – user Sep 29 '18 at 20:39
• @gimusi: I didn't say this kind of question is impossible to address. You did fine work there, and answers here are also good. But the fact that the linked discussion took so long kind of strengthens my point. It took a lot of effort to carry the point across, because the OP tried to bite more than they could chew. Anyway: I also didn't say such questions are necessarily inappropriate, and I didn't downvote myself. It's inevitable that some misunderstandings will only manifest when tackling harder material. But I do understand why someone might want to downvote them. – Meni Rosenfeld Sep 29 '18 at 21:54

You are right the two vectors are linearly independent and therefore they span a $$2$$ dimensional subspace in $$\mathbb{R^4}$$ that is in parametric form

$$\begin{bmatrix} x\\ y \\ z \\ w \end{bmatrix}=s\cdot \begin{bmatrix} 1\\ 0 \\ 1 \\ 0 \end{bmatrix}+t\cdot \begin{bmatrix} 0\\ 1 \\ 0 \\ 1 \end{bmatrix}=\begin{bmatrix} s\\ t \\ s \\ t \end{bmatrix}$$

with $$s,t \in \mathbb{R}$$, or in cartesian form

• $$x=z$$
• $$y=w$$