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Login NowLet 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

- ∀ u, v ∈ W, u + v ∈ w [Closure under addition]
- ∀ u ∈ W, c ∈ w, cu ∈ w [Closure under scalar multiplication]
- 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 R^{3}. First we check, if H is a subsset of R^{3}.

Obviously, H ⊆ R^{3}

Now,

*** Closure Under Addition**

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 R^{3}

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