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Chapter 11: Problem 35

Suppose \(T: V \rightarrow U\) is linear and \(V\) has finite dimension. Prove that \(\operatorname{Im} T^{t}=(\operatorname{Ker} T)^{0}\).

Short Answer

Expert verified
Given a linear transformation T from V to U, we need to prove that the image of the transposed transformation T岬� is equal to the annihilator of the kernel of T. We can follow these steps: 1. Recall the definitions of the transposed transformation, image, kernel, and annihilator. 2. Show that if a functional is in the image of T岬�, then it annihilates the kernel of T. 3. Show that if a functional annihilates the kernel of T, then it is in the image of T岬�. Following these steps, we concluded 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.

Step by step solution

01

Recall Definitions

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. \((W)^0 = \{ l \in V^* : l(w) = 0, \forall w \in W\}\).
02

Proving that if a functional is in the image of T岬� then it annihilates the kernel of T

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, \[ \operatorname{Im}(T^t) \subseteq (\operatorname{Ker}(T))^0. \]
03

Proving that if a functional annihilates the kernel of T, then it is in the image of T岬�

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: \[ (\operatorname{Ker}(T))^0 \subseteq \operatorname{Im}(T^t). \]
04

Concluding that Im(T岬�) = (Ker(T))鈦�

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Transpose of a Linear Map
The concept of a transpose in linear maps adds a fascinating layer to linear transformations and their properties. When dealing with a linear transformation \( T: V \rightarrow U \), where \( V \) and \( U \) are vector spaces, the transpose of \( T \) is denoted as \( T^t \). This is a transformation that maps from the dual space \( U^* \) to the dual space \( V^* \). The dual space of a vector space \( X \) consists of all linear functionals that map \( X \) into the field over which the vector space is defined.

The transpose is defined by the relation \( T^t(g) = g \circ T \) for any \( g \in U^* \). This means, given a functional \( g \), the action of \( T^t \) on \( g \) is equivalent to applying \( g \) after \( T \), effectively capturing how functionals transform under \( T \). The utility here lies in translating geometric transformations into analytic form, which proves invaluable in theoretical analysis, such as understanding the image and kernel as seen in the given problem.
Finite Dimensional Vector Spaces
A vector space is termed finite-dimensional if it has a finite basis, meaning there exists a finite set of vectors that can span the entire space. In other words, every element within the space can be expressed as a finite linear combination of these basis vectors. The number of such vectors in any basis is referred to as the dimension of the space.

Understanding finite-dimensional vector spaces is crucial because they simplify many mathematical procedures. They allow for the application of techniques like matrix representation of linear transformations, which is the foundation for most linear algebra computations. Methods involving basis transformations and coordinate changes heavily rely on the concept of finite dimensions, ensuring more manageable computation and visualization.
Annihilator
The annihilator plays a significant role in the intersection of linear algebra and functional analysis. Given a subspace \( W \) of a vector space \( V \), the annihilator of \( W \), denoted \( (W)^0 \), is the set of all linear functionals in \( V^* \) that map every vector in \( W \) to zero. Formally, \( (W)^0 = \{ l \in V^* \: l(w) = 0, \text{ for all } w \in W \}.\)

The annihilator provides insight into the orthogonality properties of the underlying spaces. In the context of our exercise, understanding how the image of \( T^t \) is related to the kernel of \( T \) through the annihilator highlights the duality and symmetry intrinsic to linear transformations, reinforcing the structure and properties of vector spaces.
Kernel and Image in Linear Algebra
In linear algebra, the kernel and the image are fundamental concepts associated with linear transformations. The kernel (null space) of a transformation \( T: V \rightarrow U \) is the set of vectors in \( V \) that are mapped to the zero vector in \( U \). Denoted as \( \operatorname{Ker}(T) \), it provides critical information about the injectivity of the transformation.

On the other hand, the image (range) of \( T \) is the set of all vectors \( T(v) \) resulting from application of \( T \) on vectors from \( V \). Expressed as \( \operatorname{Im}(T) \), this set tells us about the surjectivity and the span achieved by the transformation.

Understanding these concepts is essential for exploring the dimensionality of transformations and solving systems of linear equations, and they significantly overlap with broader topics like rank-nullity theorem and the fundamental theorem of linear maps.

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Most popular questions from this chapter

Let \(V\) be a vector space over \(\mathbf{R}\). Let \(\phi_{1}, \phi_{2} \in V^{*}\) and suppose \(\sigma: V \rightarrow \mathbf{R}\), defined by \(\sigma(v)=\phi_{1}(v) \phi_{2}(v)\), also belongs to \(V^{*}\). Show that either \(\phi_{1}=\mathbf{0}\) or \(\phi_{2}=\mathbf{0}\).

Let \(W\) be a subspace of \(V\). For any linear functional \(\phi\) on \(W\), show that there is a linear functional \(\sigma\) on \(V\) such that \(\sigma(w)=\phi(w)\) for any \(w \in W\); that is, \(\phi\) is the restriction of \(\sigma\) to \(W\).

Suppose \(T_{1}: U \rightarrow V\) and \(T_{2}: V \rightarrow W\) are linear. Prove that \(\left(T_{2} \circ T_{1}\right)^{t}=T_{1}^{t} \circ T_{2}^{t}\).

Suppose \(\phi, \sigma \in V^{*}\) and that \(\phi(v)=0\) implies \(\sigma(v)=0\) for all \(v \in V .\) Show that \(\sigma=k \phi\) for some scalar \(k\).

Prove Theorem 11.1: Suppose \(\left\\{v_{1}, \ldots, v_{n}\right\\}\) is a basis of \(V\) over \(K .\) Let \(\phi_{1}, \ldots, \phi_{n} \in V^{*}\) be defined by \(\phi_{i}\left(v_{j}\right)=0\) for \(i \neq j\), but \(\phi_{i}\left(v_{j}\right)=1\) for \(i=j .\) Then \(\left\\{\phi_{1}, \ldots, \phi_{n}\right\\}\) is a basis of \(V^{*}\) We first show that \(\left\\{\phi_{1}, \ldots, \phi_{n}\right\\}\) spans \(V^{*} .\) Let \(\phi\) be an arbitrary element of \(V^{*}\), and suppose $$ \phi\left(v_{1}\right)=k_{1}, \quad \phi\left(v_{2}\right)=k_{2}, \quad \ldots, \quad \phi\left(v_{n}\right)=k_{n} $$ Set \(\sigma=k_{1} \phi_{1}+\cdots+k_{n} \phi_{n} .\) Then $$ \begin{aligned} \sigma\left(v_{1}\right) &=\left(k_{1} \phi_{1}+\cdots+k_{n} \phi_{n}\right)\left(v_{1}\right)=k_{1} \phi_{1}\left(v_{1}\right)+k_{2} \phi_{2}\left(v_{1}\right)+\cdots+k_{n} \phi_{n}\left(v_{1}\right) \\ &=k_{1} \cdot 1+k_{2} \cdot 0+\cdots+k_{n} \cdot 0=k_{1} \end{aligned} $$ Similarly, for \(i=2, \ldots, n\), $$ \sigma\left(v_{i}\right)=\left(k_{1} \phi_{1}+\cdots+k_{n} \phi_{n}\right)\left(v_{i}\right)=k_{1} \phi_{1}\left(v_{i}\right)+\cdots+k_{i} \phi_{i}\left(v_{i}\right)+\cdots+k_{n} \phi_{n}\left(v_{i}\right)=k_{i} $$ Thus, \(\phi\left(v_{i}\right)=\sigma\left(v_{i}\right)\) for \(i=1, \ldots, n .\) Because \(\phi\) and \(\sigma\) agree on the basis vectors, \(\phi=\sigma=k_{1} \phi_{1}+\cdots+k_{n} \phi_{n} .\) Accordingly, \(\left\\{\phi_{1}, \ldots, \phi_{n}\right\\}\) spans \(V^{*}\) It remains to be shown that \(\left\\{\phi_{1}, \ldots, \phi_{n}\right\\}\) is linearly independent. Suppose $$ a_{1} \phi_{1}+a_{2} \phi_{2}+\cdots+a_{n} \phi_{n}=\mathbf{0} $$ Applying both sides to \(v_{1}\), we obtain $$ \begin{aligned} 0 &=\mathbf{0}\left(v_{1}\right)=\left(a_{1} \phi_{1}+\cdots+a_{n} \phi_{n}\right)\left(v_{1}\right)=a_{1} \phi_{1}\left(v_{1}\right)+a_{2} \phi_{2}\left(v_{1}\right)+\cdots+a_{n} \phi_{n}\left(v_{1}\right) \\ &=a_{1} \cdot 1+a_{2} \cdot 0+\cdots+a_{n} \cdot 0=a_{1} \end{aligned} $$ Similarly, for \(i=2, \ldots, n\) $$ 0=\mathbf{0}\left(v_{i}\right)=\left(a_{1} \phi_{1}+\cdots+a_{n} \phi_{n}\right)\left(v_{i}\right)=a_{1} \phi_{1}\left(v_{i}\right)+\cdots+a_{i} \phi_{i}\left(v_{i}\right)+\cdots+a_{n} \phi_{n}\left(v_{i}\right)=a_{i} $$ That is, \(a_{1}=0, \ldots, a_{n}=0 .\) Hence, \(\left\\{\phi_{1}, \ldots, \phi_{n}\right\\}\) is linearly independent, and so it is a basis of \(V^{*}\).
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