Question

Let \(S\) and \(T\) be linear transformations \(V \rightarrow W,\) where \(\operatorname{dim} V=n\) and \(\operatorname{dim} W=m\) a. Show that \(\operatorname{ker} S=\operatorname{ker} T\) if and only if \(T=R S\) for some isomorphism \(R: W \rightarrow W\). [Hint: Let \(\left\\{\mathbf{e}_{1}, \ldots, \mathbf{e}_{r}, \ldots, \mathbf{e}_{n}\right\\}\) be a basis of \(V\) such that \(\left\\{\mathbf{e}_{r+1}, \ldots, \mathbf{e}_{n}\right\\}\) is a basis of \(\operatorname{ker} S=\operatorname{ker} T\). Use Theorem 7.2 .5 to extend \(\left\\{S\left(\mathbf{e}_{1}\right), \ldots, S\left(\mathbf{e}_{r}\right)\right\\}\) and \(\left\\{T\left(\mathbf{e}_{1}\right), \ldots, T\left(\mathbf{e}_{r}\right)\right\\}\) to bases of \(\left.W .\right]\) b. Show that \(\operatorname{im} S=\operatorname{im} T\) if and only if \(T=S R\) for some isomorphism \(R: V \rightarrow V\). [Hint: Show that \(\operatorname{dim}(\operatorname{ker} S)=\operatorname{dim}(\) ker \(T)\) and choose bases \(\left\\{\mathbf{e}_{1}, \ldots, \mathbf{e}_{r}, \ldots, \mathbf{e}_{n}\right\\}\) and \(\left\\{\mathbf{f}_{1}, \ldots, \mathbf{f}_{r}, \ldots, \mathbf{f}_{n}\right\\}\) of \(V\) where \(\left\\{\mathbf{e}_{r+1}, \ldots, \mathbf{e}_{n}\right\\}\) and \(\left\\{\mathbf{f}_{r+1}, \ldots, \mathbf{f}_{n}\right\\}\) are bases of ker \(S\) and ker \(T,\) respectively. If \(1 \leq i \leq r,\) show that \(S\left(\mathbf{e}_{i}\right)=T\left(\mathbf{g}_{i}\right)\) for some \(\mathbf{g}_{i}\) in \(V,\) and prove that \(\left\\{\mathbf{g}_{1}, \ldots, \mathbf{g}_{r}, \mathbf{f}_{r+1}, \ldots, \mathbf{f}_{n}\right\\}\) is a basis of \(\left.V .\right]\)

Short answer

a) T = RS for an isomorphism R: W → W; b) T = SR for an isomorphism R: V → V.

Step-by-step solution

Analyze the kernel condition for S and T

To show that \( \operatorname{ker} S = \operatorname{ker} T \) implies \( T = R S \), assume a basis of \( V \), \( \{\mathbf{e}_1, \ldots, \mathbf{e}_r, \ldots, \mathbf{e}_n\} \), such that \( \{ \mathbf{e}_{r+1}, \ldots, \mathbf{e}_n \} \) is a basis for \( \operatorname{ker} S = \operatorname{ker} T \). This means for any vector in this subspace, \( S(\mathbf{e}_i) = T(\mathbf{e}_i) = \mathbf{0} \).

Extend basis to image

Extend \( \{ S(\mathbf{e}_1), \ldots, S(\mathbf{e}_r) \} \) to a basis of the image space \( W \). Similarly, extend \( \{ T(\mathbf{e}_1), \ldots, T(\mathbf{e}_r) \} \) to a basis of \( W \). Since these are both bases, there exists an isomorphism \( R: W \rightarrow W \) such that \( T(\mathbf{e}_i) = R(S(\mathbf{e}_i)) \) for \( i = 1, \ldots, r \).

Show the isomorphism R

Since \( T(\mathbf{e}_i) = R(S(\mathbf{e}_i)) \), for \( i = 1, \ldots, n \), and \( T(\mathbf{e}_i) = \mathbf{0} \) exactly when \( S(\mathbf{e}_i) = \mathbf{0} \), the mapping works for every basis vector. Therefore, \( T = R S \) for an isomorphism \( R \).

Analyze the image condition for S and T

To show \( \operatorname{im} S = \operatorname{im} T \) if and only if \( T = S R \), assume \( \operatorname{dim}(\operatorname{ker} S) = \operatorname{dim}(\operatorname{ker} T) \) and bases \( \{ \mathbf{e}_1, \ldots, \mathbf{e}_r, \ldots, \mathbf{e}_n \} \) and \( \{ \mathbf{f}_1, \ldots, \mathbf{f}_r, \ldots, \mathbf{f}_n \} \) with kernels being \( \{ \mathbf{e}_{r+1}, \ldots, \mathbf{e}_n \} \) and \( \{ \mathbf{f}_{r+1}, \ldots, \mathbf{f}_n \} \).

Construct transformation using equivalent images

Since \( S(\mathbf{e}_i) = T(\mathbf{g}_i) \) for some \( \mathbf{g}_i \) for \( i = 1, \ldots, r \), form \( \{ \mathbf{g}_1, \ldots, \mathbf{g}_r, \mathbf{f}_{r+1}, \ldots, \mathbf{f}_n \} \) as a basis of \( V \). There exists an isomorphism \( R: V \rightarrow V \) such that \( R(\mathbf{f}_i) = \mathbf{g}_i \), ensuring \( T = S R \).

Proof completion

In both cases, we demonstrated the necessity of an isomorphism; for ker equality through a rearrangement in \( W \), and for image equality through a transformation over \( V \). Thus, the desired results are established.

Key concepts

Kernel of a Linear Transformation

The kernel of a linear transformation is a fundamental concept in linear algebra. It refers to the set of all input vectors that are transformed to the zero vector by the linear transformation. In mathematical terms, for a linear transformation \( T: V \rightarrow W \), the kernel is defined as \( \operatorname{ker}(T) = \{ \mathbf{v} \in V | T(\mathbf{v}) = \mathbf{0} \} \).

Understanding the kernel gives insight into the "null" structure of the linear transformation, essentially identifying how the transformation "collapses" parts of the input space to zero. If the kernel of a transformation is only the zero vector, the transformation is injective (one-to-one).

The given exercise challenges this understanding by asking when the kernels of two transformations are equal. The solution involves exploring corresponding isomorphisms between the transformations. Here, if \( \operatorname{ker}(S) = \operatorname{ker}(T) \), there exists an isomorphism \( R \) such that \( T = RS \). This implies that the transformations' null spaces are identical, ensuring they behave similarly in mapping specific sets of vectors to zero.

Image of a Linear Transformation

The image of a linear transformation is the set of all output vectors that the transformation can produce. Formally, for a transformation \( S: V \rightarrow W \), the image is defined as \( \operatorname{im}(S) = \{ \mathbf{w} \in W | \mathbf{w} = S(\mathbf{v}) \text{ for some } \mathbf{v} \in V \} \).

Understanding the image helps us to know which portions of the output space are accessible through the transformation, which also ties into the concept of surjectivity (onto mapping). If every element of \( W \) can be reached, the transformation is surjective.

The exercise explores the condition \( \operatorname{im}(S) = \operatorname{im}(T) \). This equality suggests that both transformations map their respective input spaces onto identical subspaces of \( W \). The solution finds a transformation \( T = SR \) using an isomorphism \( R \). This isomorphism shifts inputs in \( V \), ensuring that both \( S \) and \( T \) cover the same parts of \( W \), with \( R \) acting as an adjustment in the input space.

Isomorphism

An isomorphism is a special type of linear transformation that demonstrates a perfect "correspondence" between two vector spaces. If a linear transformation \( R: V \rightarrow W \) is both injective (one-to-one) and surjective (onto), it is called an isomorphism.

Isomorphisms are invaluable in linear algebra because they show that two vector spaces are structurally identical, even if they might look different. If you have an isomorphism between two spaces, anything you can say about one space, in terms of its basis, dimension, and linear operations, you can apply to the other.

In the given problem, the isomorphisms \( R: W \rightarrow W \) and \( R: V \rightarrow V \) are critical. They allow us to transform one linear transformation into another while maintaining the structure of the transformations’ kernels and images. Essentially, these isomorphisms manipulate the basis in a way that maps everything precisely, indicating not just equality of kernels or images but a deeper structural sameness between the transformations.