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:

This answer is restricted. Please login to view the answer of this question.

Login Now

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,

* 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 R3

If you found any type of error on the answer then please mention on the comment or report an answer or submit your new answer.
Leave your Answer:

Click here to submit your answer.

Discussion
0 Comments
  Loading . . .