Injective Linear Transformations

Some linear transformations possess one, or both, of two key properties, which go by the names injective and surjective. We will see that they are closely related to ideas like linear independence and spanning, and subspaces like the null space and the column space. In this section we will define an injective linear transformation and analyze the resulting consequences. The next section will do the same for the surjective property. In the final section of this chapter we will see what happens when we have the two properties simultaneously.

1. Suppose $\ltdefn{T}{\complex{8}}{\complex{5}}$ is a linear transformation. Why is $T$ not injective?
2. Describe the kernel of an injective linear transformation.

As usual, we lead with a definition.

Given an arbitrary function, it is possible for two different inputs to yield the same output (think about the function $f(x)=x^2$ and the inputs $x=3$ and $x=-3$). For an injective function, this never happens. If we have equal outputs ($\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$) then we must have achieved those equal outputs by employing equal inputs ($\vect{x}=\vect{y}$). Some authors prefer the term one-to-one where we use injective, and we will sometimes refer to an injective linear transformation as an injection.

It is perhaps most instructive to examine a linear transformation that is not injective first.

Here is a cartoon of a non-injective linear transformation. Notice that the central feature of this cartoon is that $\lt{T}{\vect{u}}=\vect{v}=\lt{T}{\vect{w}}$. Even though this happens again with some unnamed vectors, it only takes one occurrence to destroy the possibility of injectivity. Note also that the two vectors displayed in the bottom of $V$ have no bearing, either way, on the injectivity of $T$. Non-Injective Linear Transformation \tikzset{ltvect/.style={shape=circle, minimum size=0.30em, inner sep=0pt, draw, fill=black}} \tikzset{ltedge/.style={->, bend left=20, thick, shorten <=0.1em, shorten >=0.1em}} \draw ( 5em, 8em) circle [x radius=5em, y radius=8em, thick]; \draw (20em, 8em) circle [x radius=5em, y radius=8em, thick]; \node (U) at ( 5em, -1em) {$U$}; \node (V) at (20em, -1em) {$V$}; \draw[->, thick, draw] (U) to node[auto] {$T$} (V); \node (u1) [ltvect] at (5em, 13em) {}; \node (u2) [ltvect] at (5em, 11em) {}; \node (u) [ltvect, label=left:$\vect{u}$] at (5em, 8em) {}; \node (w) [ltvect, label=left:$\vect{w}$] at (5em, 6em) {}; \node (v1) [ltvect] at (20em, 12em) {}; \node (v) [ltvect, label=right:$\vect{v}$] at (20em, 7em) {}; \node (v2) [ltvect] at (19em, 3em) {}; \node (v3) [ltvect] at (21em, 3em) {}; \draw[ltedge] (u1) to (v1); \draw[ltedge] (u2) to (v1); \draw[ltedge] (u) to (v); \draw[ltedge] (w) to (v);

To show that a linear transformation is not injective, it is enough to find a single pair of inputs that get sent to the identical output, as in . However, to show that a linear transformation is injective we must establish that this coincidence of outputs never occurs. Here is an example that shows how to establish this.

Let us now examine an injective linear transformation between abstract vector spaces.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

For a linear transformation $\ltdefn{T}{U}{V}$, the kernel is a subset of the domain $U$. Informally, it is the set of all inputs that the transformation sends to the zero vector of the codomain. It will have some natural connections with the null space of a matrix, so we will keep the same notation, and if you think about your objects, then there should be little confusion. Here is the careful definition.

Notice that the kernel of $T$ is just the preimage of $\zerovector$, $\preimage{T}{\zerovector}$ (). Here is an example.

We know that the span of a set of vectors is always a subspace (), so the kernel computed in is also a subspace. This is no accident, the kernel of a linear transformation is always a subspace.

Let us compute another kernel, now that we know in advance that it will be a subspace.

Our next theorem says that if a preimage is a nonempty set then we can construct it by picking any one element and adding on elements of the kernel.

This theorem, and its proof, should remind you very much of . Additionally, you might go back and review . Can you tell now which is the only preimage to be a subspace?

The next theorem is one we will cite frequently, as it characterizes injections by the size of the kernel.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

There is a connection between injective linear transformations and linearly independent sets that we will make precise in the next two theorems. However, more informally, we can get a feel for this connection when we think about how each property is defined. A set of vectors is linearly independent if the only relation of linear dependence is the trivial one. A linear transformation is injective if the only way two input vectors can produce the same output is in the trivial way, when both input vectors are equal.

Condensed — the full section is in Beezer, A First Course in Linear Algebra.

Notice that the previous example made no use of the actual formula defining the function. Merely a comparison of the dimensions of the domain and codomain are enough to conclude that the linear transformation is not injective. and are two more examples of linear transformations that have big domains and small codomains, resulting in collisions of outputs and thus are non-injective linear transformations.

In we saw how to combine linear transformations to build new linear transformations, specifically, how to build the composition of two linear transformations (). It will be useful later to know that the composition of injective linear transformations is again injective, so we prove that here.

One way to use Sage is to construct examples of theorems and verify the conclusions. Sometimes you will get this wrong: you might build an example that does not satisfy the hypotheses, or your example may not satisfy the conclusions. This may be because you are not using Sage properly, or because you do not understand a definition or a theorem, or in very limited cases you may have uncovered a bug in Sage (which is always the preferred explanation!). But in the process of trying to understand a discrepancy or unexpected result, you will learn much more, both about linear algebra and about Sage. And Sage is incredibly patient it will stay up with you all night to help you through a rough patch.

Let us illustrate the above in the context of . The hypotheses indicate we need two injective linear transformations. Where will get two such linear transformations? Well, the contrapositive of tells us that if the dimension of the domain exceeds the dimension of the codomain, we will never be injective. So we should at a minimum avoid this scenario. We can build two linear transformations from matrices created randomly, and just hope that they lead to injective linear transformations. Here is an example of how we create examples like this. The random matrix has single-digit entries, and almost always will lead to an injective linear transformation, though we cannot be absolutely certain. Evaluate this cell repeatedly, to see how rarely the result is not injective.

Our concrete example below was created this way, so here we go.