91Ó°ÊÓ

Recall that, by definition, the adjoint of a linear transformation \(T: V \rightarrow W\) is the transformation \(T^*: W \rightarrow V\) satisfying the following property for all \(\textbf{v} \in V\) and \(\textbf{w} \in W\): \[\langle T(\textbf{v}), \textbf{w} \rangle_W = \langle \textbf{v}, T^*(\textbf{w}) \rangle_V\] #a# Step 1: Prove \(T^*~T\) is invertible

If \(T\) is one-to-one, we have that \(\mathrm{Ker}(T)=\{0\}\). Using the adjoint property, let \(\textbf{v} \in \mathrm{Ker}(T^*~T)\), we get \[\langle T^* \mathrm{~T}(\textbf{v}), \textbf{v} \rangle_V = \langle T(\textbf{v}), T(\textbf{v}) \rangle_W = 0\] Therefore, \(T(\textbf{v})=0\). Since \(T\) is one-to-one, \(\textbf{v} = 0\). Thus, \(\mathrm{Ker}(T^* \mathrm{~T}) = \{0\}\).

Since \(\mathrm{Im}(T^* \mathrm{~T})⊆V\). Using the Rank-Nullity Theorem, we have: \[ \mathrm{dim}(V) = \mathrm{dim}\left(\mathrm{Ker}\left(\mathrm{T}^{*} \mathrm{~T}\right)\right) + \mathrm{dim}\left(\mathrm{Im}\left(\mathrm{T}^{*}\mathrm{~T}\right)\right)\] As we have shown that \(\mathrm{Ker}\left(\mathrm{T}^{*} \mathrm{~T}\right)=\{0\}\), we get \(\mathrm{dim}\left(\mathrm{Im}\left(\mathrm{T}^{*}\mathrm{~T}\right)\right)=\mathrm{dim}(V)\). Since \(\mathrm{Ker}\left(\mathrm{T}^{*} \mathrm{~T}\right)=\{0\}\) and \(\mathrm{dim}\left(\mathrm{Im}\left(\mathrm{T}^{*}\mathrm{~T}\right)\right)=\mathrm{dim}(V)\), it follows that \(T^*~T\) is invertible. Step 2: Prove \(T^{\dagger}=(T^* \mathrm{~T})^{-1}~T^*\) By definition of the adjoint, \[\langle T^\dagger T(\textbf{v}), \textbf{v} \rangle_V = \langle (\mathrm{I}_W(\textbf{v})), \textbf{v} \rangle_V = \langle \textbf{v}, \textbf{v} \rangle_V\] Since \(T^*~T\) is invertible, \[\langle (T^* \mathrm{~T})^{-1}~T^*~T(\textbf{v}), \textbf{v} \rangle_V = \langle \textbf{v}, \textbf{v} \rangle_V\] Now, since both the expressions equal \(\langle \textbf{v}, \textbf{v} \rangle_V\), we can conclude that \(T^\dagger = (T^* \mathrm{~T})^{-1}~T^*\). #b# Step 3: Prove \(TT^*\) is invertible

Let \(\textbf{w} \in \mathrm{Ker}(TT^*)\), then using the adjoint property, we have: \[\langle TT^*(\textbf{w}),\textbf{w}\rangle_W = \langle T^*(\textbf{w}),T^*(\textbf{w})\rangle_V = 0.\] Therefore, \(T^*(\textbf{w})=0\). Since \(\mathrm{Ker}(T^*) = \{0\}\), we must have \(\textbf{w}=0\). Thus, \(\mathrm{Ker}(TT^*)=\{0\}\).

Using the Rank-Nullity Theorem again, we have: \[ \mathrm{dim}(W) = \mathrm{dim}\left(\mathrm{Ker}\left(\mathrm{TT}^{*}\right)\right) + \mathrm{dim}\left(\mathrm{Im}\left(\mathrm{TT}^{*}\right)\right)\] Since we have shown that \(\mathrm{Ker}(TT^*) = \{0\}\), we get \(\mathrm{dim}\left(\mathrm{Im}\left(\mathrm{TT}^{*}\right)\right)=\mathrm{dim}(W)\). Since \(\mathrm{Ker}(TT^*)=\{0\}\) and \(\mathrm{dim}\left(\mathrm{Im}\left(\mathrm{TT}^{*}\right)\right)=\mathrm{dim}(W)\), it follows that \(TT^*\) is invertible. Step 4: Prove \(T^{\dagger}=T^*(TT^*)^{-1}\) By definition of the adjoint, \[\langle TT^\dagger(\textbf{w}),\textbf{w}\rangle_W = \langle (\mathrm{I}_V(\textbf{w})), \textbf{w} \rangle_W = \langle \textbf{w}, \textbf{w} \rangle_W\] Since \(TT^*\) is invertible, \[\langle T(T^*(TT^*)^{-1})(\textbf{w}),\textbf{w}\rangle_W = \langle \textbf{w}, \textbf{w} \rangle_W\] Now, since both the expressions equal \(\langle \textbf{w}, \textbf{w} \rangle_W\), we can conclude that \(T^\dagger = T^*(TT^*)^{-1}\).