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.
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
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