Intersection of Two Subspaces is a Subspace but no need of Union

Intersection and Union of Subspaces

Intersection of Two Subspaces is a Subspace

Proof: Let \( S \) and \( T \) be two subspaces of a vector space \( V \) over a field \( F \). We shall prove that \( S \cap T \) is also a subspace of \( V \).

Let \( \alpha, \beta \in F \) and \( x, y \in S \cap T \). Then \( x, y \in S \) and \( x, y \in T \). Since \( S \) and \( T \) are subspaces of \( V \), we have: \[ \alpha x + \beta y \in S \] \[ \alpha x + \beta y \in T \] Therefore, \[ \alpha x + \beta y \in S \cap T \]

Thus, \( \alpha x + \beta y \in S \cap T \) for all \( \alpha, \beta \in F \) and \( x, y \in S \cap T \). Hence, \( S \cap T \) is a subspace of \( V \).

Union of Two Subspaces Need Not Be a Subspace

Now, by giving an example we shall prove that the union of two subspaces need not be a subspace.

Example: Let \( V = \mathbb{R}^3 = \{(x, y, z) : x, y, z \in \mathbb{R}\} \) be a vector space over the field \( \mathbb{R} \) of all real numbers.

Let \( S = \{(x, y, 0) : x, y \in \mathbb{R}\} \) and \( T = \{(x, 0, z) : x, z \in \mathbb{R}\} \). Then \( S \) and \( T \) are subspaces of \( V \).

But \( S \cup T = \{(x, y, 0), (x, 0, z) : x, y, z \in \mathbb{R}\} \) is not a subspace of \( V \) because for \( \alpha, \beta \in \mathbb{R} \) and \( a = (x_1, y_1, 0) \), \( b = (x_2, 0, z_2) \in S \cup T \), we have: \[ \alpha a + \beta b = \alpha (x_1, y_1, 0) + \beta (x_2, 0, z_2) = (\alpha x_1, \alpha y_1, 0) + (\beta x_2, 0, \beta z_2) = (\alpha x_1 + \beta x_2, \alpha y_1, \beta z_2) \]

This result is not necessarily in \( S \cup T \).

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