# Define subspace of a vector space. Let $$H = \left \{ \begin{pmatrix}s\\ t\\ 0\end{pmatrix}:s,t \in R \right \}$$. Show that H is a subspace of:

Let V be a vector space over a field k, then a non empty subset w of vector space v is saied to be subspace of v if it statisfies the following properties

1. ∀ u, v ∈ W,   u + v ∈ w  [Closure under addition]
2. ∀ u ∈ W,   c ∈ w, cu ∈ w  [Closure under scalar multiplication]
3. The zero vector is in w i.e. 0 ∈ w

Given,

H = $$\left\{ \begin{bmatrix}s\\ t\\ 0\end{bmatrix} \right.$$     s, t ∈ R

To show H is a subspace of R3. First we check, if H is a subsset of R3.

Obviously, H ⊆ R3

Now,

Let H1 = $$\begin{bmatrix}s1\\ t1\\ 0\end{bmatrix}$$, H2 = $$\begin{bmatrix}s2\\ t2\\ 0\end{bmatrix}$$  ∈ H

∴ H1 + H2 = $$\begin{bmatrix}s1 + s2\\ t1 + t2\\ 0\end{bmatrix}$$ also ∈ H

So, it is closed under addition

* Closure under scalar multiplication

Here, H = $$\begin{bmatrix}s\\ t\\ 0\end{bmatrix}$$

Let, c ∈ H

∴ cH = c $$\begin{bmatrix}s\\ t\\ 0\end{bmatrix}$$ = $$\begin{bmatrix}cs\\ ct\\ 0\end{bmatrix}$$ ∈ H

so, it closed under scalar multiplication

* Obviously

0 = $$\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$$ ∈ H

Hence, H is a subset of R3