# Is any linear combination of arbitrary elements in a vector space also arbitrary?

Assuming $K$ is some vector space, is it valid to say the following:

If $a, b, c \in K$ are arbitrary and $\gamma$ and $\phi$ are scalars, then $a+b$, $a+c$, $a+b+c$, $\gamma a$, $\gamma b$, $\gamma a+\phi b$, etc. are also arbitrary.

Put another way, is any linear combination of arbitrary vectors in a vector space also arbitrary? I am mainly interested in the case that $K$ is a vector space, but does it necessarily have to be a vector space?

I am probably over-thinking my introductory linear algebra homework problem, but the following is the context I'd like to know this for:

Show that

$$\begin{pmatrix}1\\1\\0\\0\end{pmatrix}, \begin{pmatrix}1\\0\\1\\0\end{pmatrix}, \begin{pmatrix}1\\0\\0\\1\end{pmatrix}, \begin{pmatrix}0\\1\\1\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\\1\end{pmatrix}, \begin{pmatrix}0\\0\\1\\1\end{pmatrix}$$

span $\mathbb{C}^4$; i.e., every vector on $\mathbb{C}^4$ can be written as a linear combination of these vectors. Which collections of those six vectors form a basis for $\mathbb{C}^4$?

You might say, well just show those column vectors span $\mathbb{C}^4$ by simply row reducing a matrix of those column vectors to show there is a trivial solution and thus these are independent vectors and so on. I technically haven't learned that in this class yet, so I would like to be more "rigorous" and prove it by showing that there exists scalars along with those vectors that form a linear combination for all vectors in $\mathbb{C}^4$.

My proof would go something like this:

Let $x_1, x_2, x_3, x_4 \in{\mathbb{C}}$ be arbitrary. Thus, $x_1+x_2+x_3$, $x_1+x_4$, and $x_2+x_4$ are also arbitrary elements of $\mathbb{C}$.

This means that $x=\begin{pmatrix}x_1+x_2+x_3\\x_1+x_4\\x_2+x_4\\x_3\end{pmatrix} \in{\mathbb{C}^4}$ is also arbitrary.

So a linear combination of $x$ with the given vectors is simply

$x_1\begin{pmatrix}1\\1\\0\\0\end{pmatrix}+ x_2\begin{pmatrix}1\\0\\1\\0\end{pmatrix}+ x_3\begin{pmatrix}1\\0\\0\\1\end{pmatrix}+ x_4\begin{pmatrix}0\\1\\1\\0\end{pmatrix}+ 0\begin{pmatrix}0\\1\\0\\1\end{pmatrix}+ 0\begin{pmatrix}0\\0\\1\\1\end{pmatrix}$

Therefore, the given vectors span $\mathbb{C}^4$ for all $x_i$. $\blacksquare$

For the second question, I'd argue that any 4 of those given vectors can be shown to span $\mathbb{C}^4$ in a similar manner. So there are ${6}\choose{4}$ ways 4 of those vectors form a basis for $\mathbb{C}^4$.

• Could you define what you mean by arbitrary? – YoTengoUnLCD Jul 15 '15 at 2:43
• I might be abusing that word in my example, but I'm taking it to mean that if $z\in{Z}$ is arbitrary , then whatever result I get for $z$ extends to all elements of the set $Z$. Would you say "$z\in{Z}$ is arbitrary" is the functional equivalent of "$\forall z \in{Z}$"? – gbrlrz017 Jul 15 '15 at 2:48
• I see what you mean, I would avoid using that word altogether, it causes confusion. Also, there aren't 6 choose 4 ways to pick those vectors to form a basis, pick vectors 2,3,4 and 5. Do those form a basis? However, there are infinite ways to form a basis from those vectors, let $x_1=6y_1 \in \Bbb C$, and you have a different basis than your original one. – YoTengoUnLCD Jul 15 '15 at 2:58
• You can't just generalize from a single example willy-nilly...maybe in science...but in math you need to show how being true for $z$ logically implies it holds for all Z – user237392 Jul 15 '15 at 2:59
• @ YoTengoUnLCD @ Bey, would you say my proof at the end is essentially correct? If it is, I think a better way of making the final conclusion is if I prove that there is a unique linear combination involving the vectors 1, 2, 3, and 4, so these four vectors form a basis. After this I'd say that there are ${6}\choose{4}$ possible $\bf{unordered}$ bases involving any 4 out of those 6 vectors. – gbrlrz017 Jul 15 '15 at 3:21

Your concept of "arbitrary" makes sense as long as you're talking about one value at a time; then it can be made rigorous, e.g., by saying "if $x_1$ traverses all of $\mathbb C$, so does $x_2+x_1$".

However, you then start talking about several things at a time, and at that point you're begging the question, since what you're supposed to show is precisely that these things independently traverse all $\mathbb C$ (that they're "independently arbitrary"), and that's not at all obvious. For instance, there are six different sums $x_i+x_j$ for $1\le i\le j\le 4$, and they're all individually "arbitrary" in that they traverse all of $\mathbb C$ as the $x_i$ do, but that doesn't mean you can span a six-dimensional vector space using them as coefficients; it will still be just four-dimensional because there are really only four coefficients.