Question

Let \(V \stackrel{T}{\rightarrow} U \stackrel{S}{\rightarrow} W\) be linear transformations. a. If \(S T\) is one-to-one, show that \(T\) is one-to-one and that \(\operatorname{dim} V \leq \operatorname{dim} U\) b. If \(S T\) is onto, show that \(S\) is onto and that \(\operatorname{dim} W \leq \operatorname{dim} U\)

Short answer

a. If \(ST\) is one-to-one, \(T\) is one-to-one and \(\operatorname{dim} V \leq \operatorname{dim} U\). b. If \(ST\) is onto, \(S\) is onto and \(\operatorname{dim} W \leq \operatorname{dim} U\).

Step-by-step solution

Understand the Given Composition

In the problem, we have two linear transformations, with \(T: V \to U\) and \(S: U \to W\), which implies that \(ST\) is a composition of these two transformations, taking \(V\) into \(W\). The problem requires us to analyze the implications when \(ST\) is either one-to-one or onto.

Analyze the One-to-One Case for ST

To show that \(T\) is one-to-one if \(ST\) is one-to-one, assume \(T(v_1) = T(v_2)\) for some \(v_1, v_2 \in V\). Then, \(S(T(v_1)) = S(T(v_2))\). Since \(ST\) is one-to-one, it follows that \(v_1 = v_2\). Thus, \(T\) is one-to-one.

Relate Dimensions with One-to-One Transformation

If \(T\) is one-to-one, then the null space of \(T\) contains only the zero vector, so \(\text{dim}\, V \leq \text{dim}\, U\) by the injective property, meaning no vector in \(V\) is mapped to the same vector in \(U\).

Analyze the Onto Case for ST

To show that \(S\) is onto if \(ST\) is onto, take any vector \(w \in W\). Since \(ST\) is onto, there exists \(v \in V\) such that \(ST(v) = w\). Thus, \(S(T(v)) = w\), and hence \(S\) is onto because \(T(v) \in U\).

Relate Dimensions with Onto Transformation

If \(S\) is onto, every vector in \(W\) is an image of some vector in \(U\). The rank-nullity theorem implies \(\text{dim}\, W \leq \text{dim}\, U\), since the range of \(U\) must span \(W\), meaning the dimension of \(W\) cannot exceed that of \(U\).

Key concepts

One-to-one function

A one-to-one function, also called an injective function, ensures that each element in the domain maps to a unique element in the codomain. This means no two distinct elements in the domain map to the same element in the codomain. If we consider a linear transformation such as our linear map \(T: V \to U\), for \(T\) to be one-to-one, it must satisfy the condition that if \(T(v_1) = T(v_2)\), then \(v_1 = v_2\).
This statement essentially means that in the context of vectors, no two different vectors in \(V\) can end up at the same point in \(U\).
For an injective linear transformation, the null space, which is the set of all vectors mapped to the zero vector, contains only the zero vector itself. This is crucial because it assures us that the kernel (null space) of \(T\) is trivial, confirming that the transformation is injective. Determining that a transformation is one-to-one can significantly help in further analysis of systems and functions in linear algebra.
In our example, if the composition \(ST\) is one-to-one, it directly indicates that \(T\) is one-to-one, because, in order for the final mapping to \(W\) through \(S\) to be unique, the initial mapping performed by \(T\) must also be unique. Thus, by knowing \(ST\) is injective, \(T\) is confirmed as one-to-one, and hence \(\text{dim} V \leq \text{dim} U\).

Onto function

An onto function, or surjective function, covers the entire codomain. In other words, every element in the codomain is mapped from at least one element in the domain. For a linear transformation like \(S: U \to W\), \(S\) is considered onto if for every vector \(w\) in \(W\), there exists at least one vector \(u\) in \(U\) such that \(S(u) = w\).
This property is crucial as it confirms that the mapping fully spans the entire target space \(W\). In a composition such as \(ST\), if this composition is onto, it implies that the transformation \(S\) itself is onto. This is because, for \(ST\) to successfully cover all of \(W\), \(S\) must ensure every element in \(W\) has a preimage in \(U\).
In practical terms, having an onto function means that there are no "unmapped" elements in the codomain, which is particularly important in solving equations because having an onto mapping guarantees that every possible output can be reached from some input in the domain.
When \(ST\) is surjective, it ensures that \(S\) is onto, and the dimension of the target space \(W\) is at most the dimension of \(U\). Thus, \(\text{dim} W \leq \text{dim} U\), ensuring the complete coverage of the vectors in \(W\).

Dimension theorem

The Dimension Theorem is a fundamental concept in linear algebra that relates the dimensions of different subspaces of a vector space. If \(T: V \to U\) is a linear transformation, the Dimension Theorem tells us how the dimensions of the null space and range (image) relate to the dimension of \(V\).
Specifically, if \(T\) is a linear map, then:

• The dimension of the domain \(V\) is equal to the sum of the dimensions of the null space (kernel) and the range (image).


• Mathematically, this can be expressed as \(\text{dim} V = \text{dim Null}(T) + \text{dim Range}(T)\).

In context, the null space is the set of all vectors that the transformation maps to the zero vector, while the range consists of all vectors that can be obtained as outputs of the transformation.
Why is this significant? Because it allows us to understand how transformations "squash," "stretch," or rearrange elements of vector spaces. Knowing how to compute these dimensions helps in understanding the structure of a space and its subspaces.
In the example where \(T\) is one-to-one, according to the Dimension Theorem, having a trivial null space (i.e., only the zero vector) implies that \(\text{dim Range}(T) = \text{dim} V\), which helps in proving \(\text{dim} V \leq \text{dim} U\).

Rank-nullity theorem

The Rank-Nullity Theorem is an essential tool that provides a bridge between the algebraic and geometric perspectives of linear transformations. This theorem links the dimension of a vector space \(V\) to the sum of two critical components: the rank and the nullity of a transformation.
If you have a linear transformation \(T: V \rightarrow U\), the Rank-Nullity Theorem states:

• \(\text{dim} V = \text{rank}(T) + \text{nullity}(T)\)

Here, the rank of \(T\) refers to the dimension of the image of \(T\), while the nullity refers to the dimension of the kernel of \(T\).
This theorem is powerful because it provides a quantitative measure of how the transformation changes the vector space. Specifically, it asserts that the total dimension of the domain equals the sum of the dimensions of the range and the kernel. This way, the rank and nullity together account for every dimension in the original space.
In our particular problem, using the rank to determine \(\text{dim} W \leq \text{dim} U\) comes into play when considering onto functions. If \(St\) is onto, it ensures that the rank of the transformation is equal to the dimension of \(W\). The theorem tells us about the overlap of these transformations within the constraints of dimension, confirming essential properties of the transformations at hand.