91影视

We need to recall the following definitions: - Transposed transformation: \(T^t : U^* \rightarrow V^*\) is the transpose of \(T : V \rightarrow U\), defined as \(T^t(g)=g \circ T\), for any \(g \in U^*\). - Image: The image of a linear transformation is the set of all possible outputs (vectors) based on the given inputs (vectors). \( \operatorname{Im}(T):= \{ T(v) : v \in V \} \). - Kernel: The kernel of a linear transformation is the set of all input vectors that get mapped to the zero vector (element). \( \operatorname{Ker}(T):= \{ v \in V : T(v) = 0 \} \). - Annihilator: The annihilator of a subspace W is the set of all linear functionals in V* that send every element of W to zero. \((W)^0 = \{ l \in V^* : l(w) = 0, \forall w \in W\}\).

Let \(l \in \operatorname{Im}(T^t)\), which means that there exists a functional \(g \in U^*\) such that \(T^t(g) = l\). By definition of the transpose, \(T^t(g)=g \circ T\). Now, consider any \(v\in \operatorname{Ker}(T)\). Since T is linear, we have: \[ T(v)=0 \] Now we'll evaluate \(l\) to an element of kernel of \(T\): \[l(v) = g(T(v)) = g(0) = 0\]. This shows that if \(l \in \operatorname{Im}(T^t)\), then it annihilates any element in the kernel of T. Thus, \[ \operatorname{Im}(T^t) \subseteq (\operatorname{Ker}(T))^0. \]

Let's suppose \(l\) is a functional that annihilates the kernel of T: \[l(v) = 0, \forall v \in \operatorname{Ker}(T)\] Since V has a finite dimension, we can extend the basis \(\{v_1, \dots, v_n\}\) of \(\operatorname{Ker}(T)\) to a basis for V. Let \(\{v_1, \dots, v_n, v_{n+1}, \dots, v_m\}\) be the basis of V and \(\{u_1, \dots, u_m\}\) be a basis of U. Now we will define a functional \(g \in U^*\), such that \(g(T(v_i)) = l(v_i)\), for all \(1 \le i \le m\). Using the fact that \(T(v_i) = 0\) for \(1 \le i \le n\), we define g: \[g(T(v_i)) = \left\{ \begin{aligned} &0, && \text{for}\ 1 \le i \le n \\ &l(v_i), && \text{for}\ n < i \le m \\ \end{aligned} \right.\] Since g is a functional in U*, we have: \[ l(v_i) = g \circ T(v_i) = (T^t(g))(v_i), \forall i = 1, \dots, m \] Since these relations hold for the basis vectors, they hold for all vectors in V. Therefore, \( l = T^t(g) \), and so if a functional \(l\) annihilates the kernel of T, it lies in the image of T岬�. This means: \[ (\operatorname{Ker}(T))^0 \subseteq \operatorname{Im}(T^t). \]

After demonstrating the two previous inclusions, we can conclude that: \[ \operatorname{Im}(T^t) = (\operatorname{Ker}(T))^0. \] This completes the proof that the image of the transposed transformation T岬� is equal to the annihilator of the kernel of T.