Exercise-12

LADR

Exercise 1

Prove that the union of two subspaces of \(V\) is a subspace of \(V\) if and only if one of the subspaces is contained in the other.

Solution

Let \(V_1\) and \(V_2\) be two subspaces of \(V\). We will prove the two implications in step-1 and step-2 respectively.

Step-1

Assume that \(V_1 \cup V_2\) is a subspace.

Step-1.1

If \(V_1 \subset V_2\) or \(V_1 \supset V_2\) then we are done. If not, without loss of generality, let \(x\) be an element of \(V_1\) that is not present in \(V_2\). Let \(y \in V_2\).

Step-1.2

Since \(x, y \in V_1 \cup V_2\) and \(V_1 \cup V_2\) is a subspace, \(x + y \in V_1 \cup V_2\).

Step-1.3

Now, \(x + y \in V_1\) or \(x + y \in V_2\).

Step-1.4

If \(x + y \in V_2\), then \((x + y) - y = x \in V_2\), which is not possible by the assumption that we have started off with.

Step-1.5

Therefore, \(x + y \in V_1\). From this, it follows that \((x + y) - x = y \in V_1\).

Step-1.6

We have shown that \(y \in V_2 \implies y \in V_1\). Therefore, \(V_2 \subset V_1\).

Step-2

Now for the other direction. If \(V_1 \subset V_2\), then \(V_1 \cup V_2 = V_2\), which is a subspace. A similar reasoning holds if \(V_2 \subset V_1\).

Footnotes

  1. Exercise-12, Exercises 1C, Page-25, Linear Algebra Done Right, Fourth Edition, Sheldon Axler↩︎