Question

Given linear transformations \(V \stackrel{T}{\rightarrow} W \stackrel{S}{\rightarrow} U\) a. If \(S\) and \(T\) are both one-to-one, show that \(S T\) is one-to-one. b. If \(S\) and \(T\) are both onto, show that \(S T\) is onto.

Short answer

(a) If \(S\) and \(T\) are one-to-one, then \(ST\) is one-to-one. (b) If \(S\) and \(T\) are onto, then \(ST\) is onto.

Step-by-step solution

Understanding One-to-One Transformation

A linear transformation \(T : V \rightarrow W\) is one-to-one if for every pair of distinct vectors \(\mathbf{v_1}, \mathbf{v_2} \in V\), it holds that \(T(\mathbf{v_1}) eq T(\mathbf{v_2})\). Equivalently, \(T\) is one-to-one if the kernel of \(T\), \(\text{Ker}(T)\), is the zero vector, meaning \(\text{Ker}(T) = \{\mathbf{0}\}\).

Showing One-to-One for Composite Transformation

Given that \(T : V \rightarrow W\) is one-to-one, \(\text{Ker}(T) = \{\mathbf{0}\}\). Since \(S : W \rightarrow U\) is also one-to-one, \(\text{Ker}(S) = \{\mathbf{0}\}\). For \(ST : V \rightarrow U\) to be one-to-one, we need to show that \(\text{Ker}(ST) = \{\mathbf{0}\}\). Assume \(ST(\mathbf{v}) = \mathbf{0}\), which implies \(S(T(\mathbf{v})) = \mathbf{0}\). Because \(S\) is one-to-one, \(T(\mathbf{v}) = \mathbf{0}\). From \(\text{Ker}(T) = \{\mathbf{0}\}\), it follows that \(\mathbf{v} = \mathbf{0}\). Thus, \(\text{Ker}(ST) = \{\mathbf{0}\}\), showing \(ST\) is one-to-one.

Understanding Onto Transformation

A linear transformation \(T : V \rightarrow W\) is onto if for every vector \(\mathbf{w} \in W\), there exists a vector \(\mathbf{v} \in V\) such that \(T(\mathbf{v}) = \mathbf{w}\). This means that the range of \(T\), \(\text{Im}(T)\), is the whole space \(W\).

Showing Onto for Composite Transformation

Since \(T\) is onto, \(\text{Im}(T) = W\), meaning for every vector \(\mathbf{w} \in W\), there exists \(\mathbf{v} \in V\) such that \(T(\mathbf{v}) = \mathbf{w}\). Furthermore, since \(S\) is onto, \(\text{Im}(S) = U\), so for every \(\mathbf{u} \in U\), there exists \(\mathbf{w} \in W\) such that \(S(\mathbf{w}) = \mathbf{u}\). Therefore, for any \(\mathbf{u} \in U\), we can find a \(\mathbf{v} \in V\) such that \(ST(\mathbf{v}) = S(T(\mathbf{v})) = \mathbf{u}\). This shows that \(ST\) is onto.

Key concepts

One-to-One Transformation

A linear transformation is labeled as one-to-one (also injective) when distinct vectors in the input map to distinct vectors in the output. This means no two different input vectors create the same output vector. To determine if a transformation \(T: V \rightarrow W\) is one-to-one, we examine the kernel of \(T\), denoted \(\text{Ker}(T)\). The kernel is the set of all vectors in \(V\) that map to the zero vector in \(W\). In other words, \(\text{Ker}(T) = \{ \mathbf{v} \in V \mid T(\mathbf{v}) = \mathbf{0} \} \). If this kernel is just the zero vector, i.e., \(\text{Ker}(T) = \{\mathbf{0}\}\), then \(T\) is one-to-one.
Here’s what to keep in mind:

• Different inputs should yield different outputs.
• The kernel being only the zero vector signifies the transformation is injective.
• Any multi-step transformation (like \(ST\)) preserves this property if each step is one-to-one.

To understand it better, think of a one-to-one transformation as a function where if you imagine inputting two different items, you’ll always output different results. There's a uniqueness preserved in the relating input and output.

Onto Transformation

An onto (or surjective) transformation is a scenario where a linear transformation maps every possible element in the output space. For a transformation \(T : V \rightarrow W\), it is onto if every vector \(\mathbf{w} \in W\) can be written as \(T(\mathbf{v})\) for at least one \(\mathbf{v} \in V\). This means the image, or the range, of \(T\) is all of \(W\).
Here are key points on onto transformations:

• An onto transformation covers the entire output space.
• Every output is accounted for with an input.
• In composite transformations (like \(ST\)), the entire output space is reached so long as each part is onto.

Imagine onto transformations as filling up every spot in the range \(W\). Like fitting keys into locks, every lock has a key in the set \(V\) that can open it. This ensures completeness and coverage.

Kernel and Image in Linear Algebra

The concepts of kernel and image are central to understanding how transformations behave in linear algebra. The kernel (\(\text{Ker}(T)\)) of a linear transformation \(T: V \rightarrow W\) includes all vectors in \(V\) that turn into the zero vector in \(W\) upon transformation. Meanwhile, the image (\(\text{Im}(T)\)) consists of all possible outputs, or the span of the transformation.
Here’s a compact breakdown:

• The kernel relates to the one-to-one nature of a transformation.
• If the kernel is only the zero vector, the transformation is one-to-one.
• The image relates to how much of the output space the transformation covers.
• If the image is the whole output space, the transformation is onto.

By observing these two components, you obtain a detailed insight into the functionality of a transformation. Consider them as diagnostics tools - the kernel ensures no overlapping inputs in your transformation, while the image verifies every possible outcome is achievable, completing the picture of transformation behavior.