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.
Best Answer
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.