Question-62
Let \(\displaystyle S_{1}\) and \(\displaystyle S_{2}\) be two subspaces of \(\displaystyle \mathbb{R}^{n}\). Select all true statements.
Option-1: \(\displaystyle S_{1} \subseteq S_{2}\)
Let \(\displaystyle B_{1}\) be a basis for \(\displaystyle S_{1}\). We can extend \(\displaystyle B_{1}\) to form a basis \(\displaystyle B_{2}\) for \(\displaystyle S_{2}\). \(\displaystyle |B_{1} |\leqslant |B_{2} |\), from which it follows that \(\displaystyle \text{dim}( S_{1}) \leqslant \text{dim}( S_{2})\).
Option-2: \(\displaystyle \text{dim}( S_{1}) \leqslant \text{dim}( S_{2})\)
As a counter example, choose \(\displaystyle n=2\), \(\displaystyle S_{1} =\text{span}\{( 1,0)\}\) and \(\displaystyle S_{2} =\text{span}\{( 0,1)\}\). This choice satisfies the premise. Clearly, \(\displaystyle S_{1} \not \subseteq S_{2}\).
Option-3
The previous argument holds here as well
Option-4
Let \(\displaystyle B\) be a basis of \(\displaystyle S_{1}\). We have \(\displaystyle B\subset S_{1} \subseteq S_{2}\). Since \(B\) is a basis of \(S_1\), it is linearly independent. This means that \(\displaystyle B\) is a linearly independent subset of \(S_2\). Since the dimensions of \(S_1\) and \(S_2\) are equal, we also know that \(|B| = \text{dim}(S_2)\). Every linearly independent subset of a vector space that has the same number of elements as its dimension is a basis. Therefore, \(\displaystyle B\) is also a basis of \(\displaystyle S_{2}\). It follows that \(\displaystyle S_{1} =S_{2}\) since \(\text{span}(B) = S_1 = S_2\).