91Ó°ÊÓ

Assume that \(U^\perp = \{0\}\). This means that for any vector \(v \in V\), its inner product with any vector \(u \in U\) is equal to 0: \[ \langle u, v\rangle = 0. \] Since \(U\) is a subspace of \(V\), it follows that for any vector \(v \in V\), its inner product with any vector \(u \in U\) is also equal to 0: \[ \langle u, v\rangle = 0. \] Now, consider the vector \(v - u\) (where \(v \in V\) and \(u \in U\)). This vector is orthogonal to all vectors in \(U\), as shown by the following inner product: \[ \langle u', v - u\rangle = \langle u', v\rangle - \langle u', u\rangle = 0, \] for any \(u' \in U\). Since \(V\) is a subspace, the vector \(v - u\) must also be in \(V\). Thus, any vector in \(V\) can be expressed as a sum of vectors in \(U\) and \(U^\perp\) (specifically, as \(v = u + (v - u)\)). Since \(U^\perp = \{0\}\), it follows that any vector in \(V\) can actually be expressed as a vector in \(U\). Therefore, \(U = V\).

Assume that \(U = V\). We need to show that \(U^\perp = \{0\}\). Let \(v \in U^\perp\), which means that \(\langle u, v\rangle = 0\) for all \(u \in U\). Since \(U = V\), it means that \(v\) must also be orthogonal to all vectors in \(V\). We know that the zero vector, \(0\), is always orthogonal to all vectors in any subspace, so it's clear that \(\{0\}\subseteq U^\perp\). Now, if \(v\) is orthogonal to all vectors in \(U\) and is not the zero vector, then it means \(V\) can be decomposed into a direct sum of subspaces \(U\) and \(U^\perp\). But we know that \(U = V\), so such a decomposition is impossible, which implies that the only vector in \(U^\perp\) is the zero vector. Therefore, \(U^\perp = \{0\}\). Both parts have now been proven, so we can conclude that \(U^\perp = \{0\}\) if and only if \(U=V\).