Question-58
Let \(\displaystyle V\) be a vector space and let \(\displaystyle S_{1}\) and \(\displaystyle S_{2}\) be subspaces of \(\displaystyle V\) such that \(\displaystyle S_{1} \cap S_{2} =\{0\}\). We have a vector \(\displaystyle v\in V\) and vectors \(\displaystyle u_{1} ,v_{1} \in S_{1}\) and \(\displaystyle u_{2} ,v_{2} \in S_{2}\) such that the following holds:
\[\begin{equation*} \begin{aligned} v & =v_{1} +v_{2} =u_{1} +u_{2} \end{aligned} \end{equation*}\]
Which of the following statements are true?
Let us consider the difference \(\displaystyle ( v_{1} +v_{2}) -( u_{1} +u_{2})\):
\[ \begin{equation*} \begin{aligned} 0 & =( v_{1} +v_{2}) -( u_{1} +u_{2})\\ & =( v_{1} -u_{1}) +( v_{2} -u_{2})\\ v_{1} -u_{1} & =u_{2} -v_{2} \end{aligned} \end{equation*} \]
Since \(\displaystyle v_{1} ,u_{1} \in S_{1}\) and \(\displaystyle S_{1}\) is a subspace, \(\displaystyle v_{1} -u_{1} \in S_{1}\). A similar argument shows that \(\displaystyle u_{2} -v_{2} \in S_{2}\). Since these two vectors are equal and since they appear in both sets, \(\displaystyle v_{1} -u_{1} =u_{2} -v_{2} \in S_{1} \cap S_{2}\). But the only element of \(\displaystyle S_{1} \cap S_{2}\) is the zero vector. It follows that \(\displaystyle v_{1} =u_{1}\) and \(\displaystyle v_{2} =u_{2}\).