This experiment is designed to help students understand the concept of a subspace of a vector space, a foundational idea in linear algebra, explaining how certain subsets of vectors, such as lines and planes through the origin in R³, can themselves form a vector space under the same operations of addition and scalar multiplication as that of the vector space. The experiment emphasizes visualization by exploring subspaces of R, R² and R³. This helps build intuition about the structure and properties of subspaces in higher dimensions.
Notation: Vector space over the field F≡R or C is denoted by V, where R is the set of real numbers and C is the set of complex numbers.

1. Definition: A non-empty subset W of V is called a subspace if it is a vector space with respect to the addition and scalar multiplication defined on V.

2. Characterization: A set W is a subspace of V if and only if
i. W⊆V
ii. 0∈W
iii. x+y∈W; where x∈W, y∈W and
iv. α.x∈W; where x∈W, α∈F.

3. Example - I:
i. V is a subspace of itself.
ii. {0} is a subspace of V.
iii. {0} is the only subspace of the vector space {0} over F.

4. Example - II:
(i) W₁={(x, 0, 0, …, 0)∈Rⁿ: x∈R} is a subspace of the vector space Rⁿ over R.
Reason: Clearly W₁⊆ Rⁿ and (0, 0, 0…, 0)∈W₁. Further, (x, 0, 0, …, 0)+ (y, 0, 0, …, 0)=(x+y, 0, 0, …, 0)∈W₁; where (x, 0, 0, …, 0) ∈W₁, (y, 0, 0, …, 0)∈W₁ and α.(x, 0, 0, …, 0)=(α.x, 0, 0, …, 0)∈W₁; where (x, 0, 0, …, 0)∈W₁, α∈F. Using the characterization given in Section2, we have the required result.
(ii) W₂={(x, x, 0)∈R³ : x∈R} is a subspace of the vector space R³ over R.
Reason: It is the same as discussed in Example (i) above.
(iii) W₃={(x, x, x) ∈ R³ : x∈R} is a subspace of the vector space R³ over R.
Reason: It is the same as discussed in Example (i) above.
(iv) Let M^{2 x 2} be the collection of 2x2 matrices with real entries. Then M^{2 x 2} forms a vector space over R with respect to matrix addition and matrix scalar multiplication and W₄ = {$\begin{bmatrix} a & b \\ c & 0 \end{bmatrix}$ ∈ M^{2 x 2}: a, b, c ∈R} is a subspace of the vector space M^{2 x 2}.
Reason: It is the same as discussed in Example (i) above.
(v) W₅={(x+2, x, x)∈R³ : x∈R} is not a subspace of the vector space R³ over R in view of the result in Section 2, as it does not contain (0, 0, 0) which is the zero of the vector space R³ over R . Thus it may be noted that a subset of a vector space need not to be its subspace.

5. Definition: The subspaces V and {0} of V are called improper subspaces of V and subspaces other than V and {0} are called proper.

6. Subspaces of R: Subspaces of the vector space R over R are precisely {0} and R. That is,
(i) W≡{0} is a subspace of the vector space R over R,
(ii) W≡R itself is a subspace of the vector space R over R and
(iii) If W is a subspace of the vector space R over R, then W={0} or R.
Proof:
(i) Clearly W⊆R and 0∈W. Let x, y∈W and α∈R. Then x=y=0. Hence x+y=0∈W and α.x=0∈W. Hence W≡{0} is a subspace.
(ii) Clearly W⊆R and 0∈W. Let x, y∈R and α∈R. Hence x+y∈R and α.x∈R. Hence W≡R is a subspace.

(iii) Let W be a subspace and W≠{0}. We prove that W=R. Clearly W⊆R. Let x∈R. Then for y∈W, y≠0, we have x=(x/y).y. Since (x/y).y∈W, x∈W. This shows that R⊆W. Hence R=W.

7. Subspaces of R²: Subspaces of the vector space R² over R are precisely {(0, 0)}, R and lines passing through the origin. That is,
(i) W≡{(0,0)} is a subspace of the vector space R² over R,
(ii) W≡R² is a subspace of the vector space R² over R,
(iii) If W is a line passing through the origin, then W is a subspace of the vector space R² over R and
(iv) If W is a subspace of the vector space R² over R, then W={(0, 0)} or R² or a line passing through the origin.
Proof:
(i) It is similar to that of 6(i).
(ii) It is similar to that of 6(ii).
(iii) Clearly W≡{(x, y)∈R²: ax+by=0}, for some a, b∈R, such that either a≠0 or b≠0. Notice that W is the line y=mx or x=0, where m∈R. Clearly W⊆R² and (0, 0)∈W. Let (r, s), (t, u)∈W. Hence ar+bs=0, at+bu=0. Then (r, s) + (t, u)=(r+t, s+u)∈W, since a(r+t)+b(s+u)=[(ar+bs)+(at+bu)])=0. Also α.(r, s)=(α.r, α.s)∈W, since a(α.r)+b(α.s)=0. Hence W is a subspace.

(iv) Let W be a subspace of R² other than {(0, 0)} and R². We prove that W is a line passing through the origin. Let (c, d)∈W such that (c, d) ≠ 0. We claim that W is the line L passing through (c, d) and origin. Then L={(x, y)∈R² : dx-cy=0}. Clearly L⊆W and we prove that W⊆L. Let (m, n)∈W such that (m, n)≠(0, 0). To the contrary, let (m, n)∉L. For (x, y)∈R²; one can obtain α, β∈R such that (x, y)=α.(m, n)+β.(c, d). Then (x, y)∈W. Hence W≡R², a contradiction. This completes the proof.

8.Properties of subspaces:
Let W₁ and W₂ be the subspaces of V. Then,
i. Intersection of W₁ and W₂ that is, W₁ ∩ W₂ is also a subspace of V.

ii. Union of W₁ and W₂ that is, W₁ ⋃ W₂ may or may not be a subspace of V.
iii. Union of W₁ and W₂ that is, W₁ ⋃ W₂ is a subspace of V if and only if W₁ ⊆ W₂ or W₂ ⊆ W₁.
iv. W₁ + W₂≡{x+y: x∈W₁, y∈W₂} is a subspace of V.
v. W₁ + W₂ = span (W₁ ⋃ W₂).
vi. W₁ + W₂ is the smallest subspace containing W₁ and W₂.