Show that the following set of vectors is a subspace of $\mathbb{R^4}$

Show that the following set of vectors is a subspace of $\mathbb{R^4}$:

The set of all linear combinations of the vectors $(1,0,1,0)$ and $(0,1,0,1)$.

I understand you need to show closure under addition, scalar multiplication and that it's non-empty, but I'm not sure how to set it all out and how to write the set described above in set notation. I'd appreciate it if someone could show me, thanks.

• The set of all linear combinations of the vectors $(1,0,1,0)$ and $(0,1,0,1)$ can be written as $\text{span}\left( (1,0,1,0),(0,1,0,1)\right) = \left\{(a,b,a,b)~:~a\in\Bbb R,~b\in\Bbb R\right\}$ Commented Aug 12, 2016 at 3:53
• In general, the "set of all linear combinations of vectors $u_1,u_2,\dots,u_n$" can be rewritten as: $\text{span}(u_1,u_2,\dots,u_n)=\left\{c_1u_1+c_2u_2+\dots+c_nu_n~:~\forall i~(c_i\in\Bbb K)\right\}$ where $\Bbb K$ is whatever scalar field you are working in. In your specific case, $\text{span}((1,0,1,0),(0,1,0,1))=\{c_1(1,0,1,0)+c_2(0,1,0,1)~:~c_1\in\Bbb R,c_2\in\Bbb R\}$ $=\{(c_1,0,c_1,0)+(0,c_2,0,c_2)~:~c_1\in\Bbb R,c_2\in \Bbb R\} = \{(c_1,c_2,c_1,c_2)~:~c_1\in\Bbb R,c_2\in\Bbb R\}$ Commented Aug 12, 2016 at 4:02

The answer by M10687 provides a very quick and excellent way to determine that this is a subspace. As I interpreted that you wanted to see explicitly the verification of the axioms, I wrote up how one would go about verifying the axioms directly...

To check if a set is a subspace of a vector space, you need to check that it is closed under addition, closed under scalar multiplication, and contains $0$ (and non-emptiness, technically, but if we show the set contains $0$, it cannot be empty).

To write this specific set in set notation, we can write: $\{a(1,0,1,0) + b(0,1,0,1) : a, b \in \mathbb{R}\}$, or, like in JMoravitz's comment: $\{(a,b,a,b) : a, b \in \mathbb{R}\}$.

Let's start by showing that the zero vector is in this set:

If we take $a = b = 0$, then we form the linear combination: $0(1,0,1,0) + 0(0,1,0,1) = (0,0,0,0)$ Since $0$ can be written as a linear combination of $(1,0,1,0), (0,1,0,1)$, zero is contained in this set.

We next show vector addition: Consider two arbitrary linear combinations of these vectors, $a(0,1,0,1) + b(1,0,1,0)$ and $c(0,1,0,1) + d(1,0,1,0)$. We add them together:

$$a(0,1,0,1) + b(1,0,1,0) + c(0,1,0,1) + d(1,0,1,0) = (a+c)(0,1,0,1) + (b+d)(1,0,1,0)$$

which is indeed a linear combination of the desired vectors (remember, $a+c$ and $b+d$ are scalars)

Finally, scalar multiplication: Consider an arbitrary vector in this set, $a(0,1,0,1) + b(1,0,1,0)$ and an arbitrary scalar $c$, then:

$$c(a(0,1,0,1) + b(1,0,1,0)) = ca(0,1,0,1) + cb(1,0,1,0)$$

which we can see is in the set.

After non-emptiness is shown (this should be pretty clear), we have that the set is a subspace.

In general, the span of any set of vectors in $V$ is a subspace of $V$. See here. The comment by @JMoravitz then completes the proof.