# fcla / src / section-ILT.xml

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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.

Injective Linear Transformations

As usual, we lead with a definition.

Injective Linear Transformation

Suppose $\ltdefn{T}{U}{V}$ is a linear transformation. Then $T$ is injective if whenever $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$, then $\vect{x}=\vect{y}$.

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.

Examples of Injective Linear Transformations

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

Not injective, Archetype Q

is the linear transformation

Notice that for we have

So we have two vectors from the domain, $\vect{x}\neq\vect{y}$, yet $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$, in violation of . This is another example where you should not concern yourself with how $\vect{x}$ and $\vect{y}$ were selected, as this will be explained shortly. However, do understand why these two vectors provide enough evidence to conclude that $T$ is not injective.

Here's 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}} , 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.

Injective, Archetype R

is the linear transformation

To establish that $R$ is injective we must begin with the assumption that $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$ and somehow arrive at the conclusion that $\vect{x}=\vect{y}$. Here we go, \colvector{0\\0\\0\\0\\0} \lt{T}{\vect{x}}-\lt{T}{\vect{y}}\\ \lt{T}{\colvector{x_1\\x_2\\x_3\\x_4\\x_5}}-\lt{T}{\colvector{y_1\\y_2\\y_3\\y_4\\y_5}}\\ \colvector{-65 x_1 + 128 x_2 + 10 x_3 - 262 x_4 + 40 x_5\\ 36 x_1 - 73 x_2 - x_3 + 151 x_4 - 16 x_5\\ -44 x_1 + 88 x_2 + 5 x_3 - 180 x_4 + 24 x_5\\ 34 x_1 - 68 x_2 - 3 x_3 + 140 x_4 - 18 x_5\\ 12 x_1 - 24 x_2 - x_3 + 49 x_4 - 5 x_5}\\ \colvector{-65 y_1 + 128 y_2 + 10 y_3 - 262 y_4 + 40 y_5\\ 36 y_1 - 73 y_2 - y_3 + 151 y_4 - 16 y_5\\ -44 y_1 + 88 y_2 + 5 y_3 - 180 y_4 + 24 y_5\\ 34 y_1 - 68 y_2 - 3 y_3 + 140 y_4 - 18 y_5\\ 12 y_1 - 24 y_2 - y_3 + 49 y_4 - 5 y_5}\\ \colvector{-65 (x_1-y_1) + 128 (x_2-y_2) + 10 (x_3-y_3) - 262 (x_4-y_4) + 40 (x_5-y_5)\\ 36 (x_1-y_1) - 73 (x_2-y_2) - (x_3-y_3) + 151 (x_4-y_4) - 16 (x_5-y_5)\\ -44 (x_1-y_1) + 88 (x_2-y_2) + 5 (x_3-y_3) - 180 (x_4-y_4) + 24 (x_5-y_5)\\ 34 (x_1-y_1) - 68 (x_2-y_2) - 3 (x_3-y_3) + 140 (x_4-y_4) - 18 (x_5-y_5)\\ 12 (x_1-y_1) - 24 (x_2-y_2) - (x_3-y_3) + 49 (x_4-y_4) - 5 (x_5-y_5)}\\ \begin{bmatrix} \end{bmatrix} \colvector{x_1-y_1\\x_2-y_2\\x_3-y_3\\x_4-y_4\\x_5-y_5}

Now we recognize that we have a homogeneous system of 5 equations in 5 variables (the terms $x_i-y_i$ are the variables), so we row-reduce the coefficient matrix to \begin{bmatrix} \end{bmatrix}

So the only solution is the trivial solution and we conclude that indeed $\vect{x}=\vect{y}$. By , $T$ is injective.

Here's the cartoon for an injective linear transformation. It is meant to suggest that we never have two inputs associated with a single output. Again, the two lonely vectors at the bottom of $V$ have no bearing either way on the injectivity of $T$. Injective Linear Transformation \tikzset{ltvect/.style={shape=circle, minimum size=0.30em, inner sep=0pt, draw, fill=black}} , bend left=20, thick, shorten <=0.1em, shorten >=0.1em}}]]> \draw ( 5em, 8em) circle [x radius=4em, y radius=7em, 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, 10em) {}; \node (u3) [ltvect] at (5em, 7em) {}; \node (v1) [ltvect] at (20em, 13em) {}; \node (v2) [ltvect] at (20em, 10em) {}; \node (v3) [ltvect] at (20em, 7em) {}; \node (v4) [ltvect] at (19em, 3em) {}; \node (v5) [ltvect] at (21em, 3em) {}; \draw[ltedge] (u1) to (v1); \draw[ltedge] (u2) to (v2); \draw[ltedge] (u3) to (v3);

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

Injective, Archetype V

is defined by

To establish that the linear transformation is injective, begin by supposing that two polynomial inputs yield the same output matrix, \lt{T}{a_1+b_1x+c_1x^2+d_1x^3}=\lt{T}{a_2+b_2x+c_2x^2+d_2x^3}

Then \zeromatrix \end{bmatrix}\\ \text{}\\ \begin{bmatrix}

This single matrix equality translates to the homogeneous system of equations in the variables $a_i-b_i$,

This system of equations can be rewritten as the matrix equation \begin{bmatrix} \end{bmatrix} \colvector{(a_1-a_2)\\(b_1-b_2)\\(c_1-c_2)\\(d_1-d_2)}=\colvector{0\\0\\0\\0}

Since the coefficient matrix is nonsingular (check this) the only solution is trivial, i.e. so that so the two inputs must be equal polynomials. By , $T$ is injective.

Kernel of a Linear Transformation

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's the careful definition.

Kernel of a Linear Transformation

Suppose $\ltdefn{T}{U}{V}$ is a linear transformation. Then the kernel of $T$ is the set \krn{T}=\setparts{\vect{u}\in U}{\lt{T}{\vect{u}}=\zerovector}

Kernel of a Linear Transformation $\krn{T}$

Notice that the kernel of $T$ is just the preimage of $\zerovector$, $\preimage{T}{\zerovector}$ (). Here's an example.

Nontrivial kernel, Archetype O

is the linear transformation

To determine the elements of $\complex{3}$ in $\krn{T}$, find those vectors $\vect{u}$ such that $\lt{T}{\vect{u}}=\zerovector$, that is, \colvector{-u_1 + u_2 - 3 u_3\\ -u_1 + 2 u_2 - 4 u_3\\ u_1 + u_2 + u_3\\ 2 u_1 + 3 u_2 + u_3\\ u_1 + 2 u_3 } \colvector{0\\0\\0\\0\\0}

Vector equality () leads us to a homogeneous system of 5 equations in the variables $u_i$,

Row-reducing the coefficient matrix gives \begin{bmatrix} \end{bmatrix}

The kernel of $T$ is the set of solutions to this homogeneous system of equations, which by can be expressed as \krn{T}=\spn{}

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.

Kernel of a Linear Transformation is a Subspace

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation. Then the kernel of $T$, $\krn{T}$, is a subspace of $U$.

We can apply the three-part test of . First $\lt{T}{\zerovector_U}=\zerovector_V$ by , so $\zerovector_U\in\krn{T}$ and we know that the kernel is non-empty.

Suppose we assume that $\vect{x},\,\vect{y}\in\krn{T}$. Is $\vect{x}+\vect{y}\in\krn{T}$? \text{}\\ \text{}

This qualifies $\vect{x}+\vect{y}$ for membership in $\krn{T}$. So we have additive closure.

Suppose we assume that $\alpha\in\complex{\null}$ and $\vect{x}\in\krn{T}$. Is $\alpha\vect{x}\in\krn{T}$? \text{}\\ \text{}

This qualifies $\alpha\vect{x}$ for membership in $\krn{T}$. So we have scalar closure and tells us that $\krn{T}$ is a subspace of $U$.

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

Trivial kernel, Archetype P

is the linear transformation

To determine the elements of $\complex{3}$ in $\krn{T}$, find those vectors $\vect{u}$ such that $\lt{T}{\vect{u}}=\zerovector$, that is, \colvector{ -u_1 + u_2 + u_3\\ -u_1 + 2 u_2 + 2 u_3\\ u_1 + u_2 + 3 u_3\\ 2 u_1 + 3 u_2 + u_3\\ -2 u_1 + u_2 + 3 u_3 } \colvector{0\\0\\0\\0\\0}

Vector equality () leads us to a homogeneous system of 5 equations in the variables $u_i$,

Row-reducing the coefficient matrix gives \begin{bmatrix} \end{bmatrix}

The kernel of $T$ is the set of solutions to this homogeneous system of equations, which is simply the trivial solution $\vect{u}=\zerovector$, so \krn{T}=\set{\zerovector}=\spn{}

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

Kernel and Pre-Image

Suppose $\ltdefn{T}{U}{V}$ is a linear transformation and $\vect{v}\in V$. If the preimage $\preimage{T}{\vect{v}}$ is non-empty, and $\vect{u}\in\preimage{T}{\vect{v}}$ then \preimage{T}{\vect{v}}= \setparts{\vect{u}+\vect{z}}{\vect{z}\in\krn{T}} =\vect{u}+\krn{T}

Let $M=\setparts{\vect{u}+\vect{z}}{\vect{z}\in\krn{T}}$. First, we show that $M\subseteq\preimage{T}{\vect{v}}$. Suppose that $\vect{w}\in M$, so $\vect{w}$ has the form $\vect{w}=\vect{u}+\vect{z}$, where $\vect{z}\in\krn{T}$. Then \text{}\\ \text{} which qualifies $\vect{w}$ for membership in the preimage of $\vect{v}$, $\vect{w}\in\preimage{T}{\vect{v}}$.

For the opposite inclusion, suppose $\vect{x}\in\preimage{T}{\vect{v}}$. Then, \text{}\\

This qualifies $\vect{x}-\vect{u}$ for membership in the kernel of $T$, $\krn{T}$. So there is a vector $\vect{z}\in\krn{T}$ such that $\vect{x}-\vect{u}=\vect{z}$. Rearranging this equation gives $\vect{x}=\vect{u}+\vect{z}$ and so $\vect{x}\in M$. So $\preimage{T}{\vect{v}}\subseteq M$ and we see that $M=\preimage{T}{\vect{v}}$, as desired.

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?

Here is the cartoon which describes the many-to-one behavior of a typical linear transformation. Presume that $\lt{T}{\vect{u}_i}=\vect{v}_i$, for $i=1,2,3$, and as guaranteed by , $\lt{T}{\zerovector_U}=\zerovector_V$. Then four pre-images are depicted, each labeled slightly different. $\preimage{T}{\vect{v}_2}$ is the most general, employing to provide two equal descriptions of the set. The most unusual is $\preimage{T}{\zerovector_V}$ which is equal to the kernel, $\krn{T}$, and hence is a subspace (by ). The subdivisions of the domain, $U$, are meant to suggest the partioning of the domain by the collection of pre-images. It also suggests that each pre-image is of similar size or structure, since each is a shifted copy of the kernel. Notice that we cannot speak of the dimension of a pre-image, since it is almost never a subspace. Also notice that $\vect{x},\,\vect{y}\in V$ are elements of the codomain with empty pre-images. Kernel and Pre-Image \tikzset{ltvect/.style={shape=circle, minimum size=0.30em, inner sep=0pt, draw, fill=black}} , bend left=20, thick, shorten <=0.1em, shorten >=0.1em}}]]> % base generic picture \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); % inputs % fine-tune u_1 to fit \node (u11) [ltvect, label={[label distance=0.1em]80:$\vect{u}_1$}] at (3em, 14em) {}; \node (u12) [ltvect] at (5em, 14em) {}; \node (u13) [ltvect] at (7em, 14em) {}; \node (u21) [ltvect, label=above:$\vect{u}_2$] at (3em, 10em) {}; \node (u22) [ltvect] at (5em, 10em) {}; \node (u23) [ltvect] at (7em, 10em) {}; \node (zeroU1) [ltvect, label=above:$\zerovector_U$] at (3em, 6em) {}; \node (zeroU2) [ltvect] at (5em, 6em) {}; \node (zeroU3) [ltvect] at (7em, 6em) {}; \node (u31) [ltvect, label=above:$\vect{u}_3$] at (3em, 2em) {}; \node (u32) [ltvect] at (5em, 2em) {}; \node (u33) [ltvect] at (7em, 2em) {}; % outputs \node (v1) [ltvect, label=right:$\vect{v}_1$] at (20em, 14em) {}; \node (v2) [ltvect, label=right:$\vect{v}_2$] at (20em, 10em) {}; \node (zeroV) [ltvect, label=right:$\zerovector_V$] at (20em, 6em) {}; \node (o1) [ltvect, label=right:$\vect{x}$] at (18em, 4em) {}; \node (o2) [ltvect, label=right:$\vect{y}$] at (22em, 4em) {}; \node (v3) [ltvect, label=right:$\vect{v}_3$] at (20em, 2em) {}; % associations \draw[ltedge] (u11) to (v1); \draw[ltedge] (u12) to (v1); \draw[ltedge] (u13) to (v1); \draw[ltedge] (u21) to (v2); \draw[ltedge] (u22) to (v2); \draw[ltedge] (u23) to (v2); \draw[ltedge] (zeroU1) to (zeroV); \draw[ltedge] (zeroU2) to (zeroV); \draw[ltedge] (zeroU3) to (zeroV); \draw[ltedge] (u31) to (v3); \draw[ltedge] (u32) to (v3); \draw[ltedge] (u33) to (v3); % preimages \node (pre1) at (4em, 13em) {$\vect{u}_1 + \krn{T}$}; \node (pre2) at (5em, 9em) {$\preimage{T}{\vect{v}_2}=\vect{u}_2 + \krn{T}$}; \node (pre3) at (4.5em, 5em) {$\preimage{T}{\zerovector_V}=\krn{T}$}; \node (pre4) at (5em, 1em) {$\preimage{T}{\vect{v}_3}$}; % banding, x-coordinates are +/- 5*sqrt(3)/2 off midline \node (b11) [minimum size=0em, inner sep=0pt] at (0.669873em, 12em) {}; \node (b12) [minimum size=0em, inner sep=0pt] at (9.330127em, 12em) {}; \node (b21) [minimum size=0em, inner sep=0pt] at ( 0em, 8em) {}; \node (b22) [minimum size=0em, inner sep=0pt] at ( 10em, 8em) {}; \node (b31) [minimum size=0em, inner sep=0pt] at (0.669873em, 4em) {}; \node (b32) [minimum size=0em, inner sep=0pt] at (9.330127em, 4em) {}; \draw [-, bend right=10] (b11) to (b12); \draw [-, bend right=10] (b21) to (b22); \draw [-, bend right=10] (b31) to (b32);

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

Kernel of an Injective Linear Transformation

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation. Then $T$ is injective if and only if the kernel of $T$ is trivial, $\krn{T}=\set{\zerovector}$.

We assume $T$ is injective and we need to establish that two sets are equal (). Since the kernel is a subspace (), $\set{\zerovector}\subseteq\krn{T}$. To establish the opposite inclusion, suppose $\vect{x}\in\krn{T}$. \lt{T}{\vect{x}} \text{}\\ \text{}

We can apply to conclude that $\vect{x}=\zerovector$. Therefore $\krn{T}\subseteq\set{\zerovector}$ and by , $\krn{T}=\set{\zerovector}$.

To establish that $T$ is injective, appeal to and begin with the assumption that $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$. Then \lt{T}{\vect{x}-\vect{y}} \text{}\\

So $\vect{x}-\vect{y}\in\krn{T}$ by and with the hypothesis that the kernel is trivial we conclude that $\vect{x}-\vect{y}=\zerovector$. Then \vect{y} =\vect{y}+\zerovector =\vect{y}+\left(\vect{x}-\vect{y}\right) =\vect{x} thus establishing that $T$ is injective by .

You might begin to think about how would change if the linear transformation is injective, which would make the kernel trivial by .

Not injective, Archetype Q, revisited

We are now in a position to revisit our first example in this section, . In that example, we showed that is not injective by constructing two vectors, which when used to evaluate the linear transformation provided the same output, thus violating . Just where did those two vectors come from?

The key is the vector \vect{z}=\colvector{3\\4\\1\\3\\3} which you can check is an element of $\krn{T}$ for . Choose a vector $\vect{x}$ at random, and then compute $\vect{y}=\vect{x}+\vect{z}$ (verify this computation back in ). Then \text{}\\ \text{}

Whenever the kernel of a linear transformation is non-trivial, we can employ this device and conclude that the linear transformation is not injective. This is another way of viewing . For an injective linear transformation, the kernel is trivial and our only choice for $\vect{z}$ is the zero vector, which will not help us create two different inputs for $T$ that yield identical outputs. For every one of the archetypes that is not injective, there is an example presented of exactly this form.

Not injective, Archetype O

In the kernel of was determined to be \spn{} a subspace of $\complex{3}$ with dimension 1. Since the kernel is not trivial, tells us that $T$ is not injective.

Injective, Archetype P

In it was shown that the linear transformation in has a trivial kernel. So by , $T$ is injective.

Injective Linear Transformations By now, you have probably already figured out how to determine if a linear transformation is injective, and what its kernel is. You may also now begin to understand why Sage calls the null space of a matrix a kernel. Here are two examples, first a reprise of . U = QQ^3 V = QQ^5 x1, x2, x3 = var('x1, x2, x3') outputs = [ -x1 + x2 - 3*x3, -x1 + 2*x2 - 4*x3, x1 + x2 + x3, 2*x1 + 3*x2 + x3, x1 + 2*x3] T_symbolic(x1, x2, x3) = outputs T = linear_transformation(U, V, T_symbolic) T.is_injective() False T.kernel() Vector space of degree 3 and dimension 1 over Rational Field Basis matrix: [ 1 -1/2 -1/2] So we have a concrete demonstration of one half of . Here is the second example, a do-over for , but renamed as S. U = QQ^3 V = QQ^5 x1, x2, x3 = var('x1, x2, x3') outputs = [ -x1 + x2 + x3, -x1 + 2*x2 + 2*x3, x1 + x2 + 3*x3, 2*x1 + 3*x2 + x3, -2*x1 + x2 + 3*x3] S_symbolic(x1, x2, x3) = outputs S = linear_transformation(U, V, S_symbolic) S.is_injective() True S.kernel() Vector space of degree 3 and dimension 0 over Rational Field Basis matrix: [] S.kernel() == U.subspace([]) True And so we have a concrete demonstration of the other half of .

Now that we have , we can return to our discussion from . The .preimage_representative() method of a linear transformation will give us a single element of the pre-image, with no other guarantee about the nature of that element. That is fine, since this is all requires (in addition to the kernel). Remember that not every element of the codomain may have a non-empty pre-image (as indicated in the hypotheses of ). Here is an example using T from above, with a choice of a codomain element that has a non-empty pre-image. TK = T.kernel() v = vector(QQ, [2, 3, 0, 1, -1]) u = T.preimage_representative(v) u (-1, 1, 0) Now the following will create random elements of the preimage of v, which can be verified by the test always returning True. Use the compute cell just below if you are curious what p looks like. p = u + TK.random_element() T(p) == v True p # random (-13/10, 23/20, 3/20) As suggested, some choices of v can lead to empty pre-images, in which case does not even apply. v = vector(QQ, [4, 6, 3, 1, -2]) u = T.preimage_representative(v) Traceback (most recent call last): ... ValueError: element is not in the image The situation is less interesting for an injective linear transformation. Still, pre-images may be empty, but when they are non-empty, they are just singletons (a single element) since the kernel is empty. So a repeat of the above example, with S rather than T, would not be very informative.
Injective Linear Transformations and Linear Independence

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.

Injective Linear Transformations and Linear Independence

Suppose that $\ltdefn{T}{U}{V}$ is an injective linear transformation and is a linearly independent subset of $U$. Then is a linearly independent subset of $V$.

Begin with a relation of linear dependence on $R$ (, ), \text{}\\ \text{}\\ \text{}\\ \text{}\\

Since this is a relation of linear dependence on the linearly independent set $S$, we can conclude that and this establishes that $R$ is a linearly independent set.

Injective Linear Transformations and Bases

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and is a basis of $U$. Then $T$ is injective if and only if is a linearly independent subset of $V$.

Assume $T$ is injective. Since $B$ is a basis, we know $B$ is linearly independent (). Then says that $C$ is a linearly independent subset of $V$.

Assume that $C$ is linearly independent. To establish that $T$ is injective, we will show that the kernel of $T$ is trivial (). Suppose that $\vect{u}\in\krn{T}$. As an element of $U$, we can write $\vect{u}$ as a linear combination of the basis vectors in $B$ (uniquely). So there are are scalars, $\scalarlist{a}{m}$, such that \vect{u}=\lincombo{a}{u}{m}

Then, \zerovector \text{}\\ \text{}\\ \text{}

This is a relation of linear dependence () on the linearly independent set $C$, so the scalars are all zero: $a_1=a_2=a_3=\cdots=a_m=0$. Then \text{}\\ \text{}\\ \text{}

Since $\vect{u}$ was chosen as an arbitrary vector from $\krn{T}$, we have $\krn{T}=\set{\zerovector}$ and tells us that $T$ is injective.

Injective Linear Transformations and Dimension Injective Linear Transformations and Dimension

Suppose that $\ltdefn{T}{U}{V}$ is an injective linear transformation. Then $\dimension{U}\leq\dimension{V}$.

Suppose to the contrary that $m=\dimension{U}>\dimension{V}=t$. Let $B$ be a basis of $U$, which will then contain $m$ vectors. Apply $T$ to each element of $B$ to form a set $C$ that is a subset of $V$. By , $C$ is linearly independent and therefore must contain $m$ distinct vectors. So we have found a set of $m$ linearly independent vectors in $V$, a vector space of dimension $t$, with $m>t$. However, this contradicts , so our assumption is false and $\dimension{U}\leq\dimension{V}$.

Not injective by dimension, Archetype U

The linear transformation in is

Since $\dimension{M_{23}}=6>4=\dimension{\complex{4}}$, $T$ cannot be injective for then $T$ would violate .

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.

Composition of 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.

Composition of Injective Linear Transformations is Injective

Suppose that $\ltdefn{T}{U}{V}$ and $\ltdefn{S}{V}{W}$ are injective linear transformations. Then $\ltdefn{(\compose{S}{T})}{U}{W}$ is an injective linear transformation.

That the composition is a linear transformation was established in , so we need only establish that the composition is injective. Applying , choose $\vect{x}$, $\vect{y}$ from $U$. Then if $\lt{\left(\compose{S}{T}\right)}{\vect{x}}=\lt{\left(\compose{S}{T}\right)}{\vect{y}}$, \text{}\\ \text{ for $S$}\\ \text{ for $T$}

Composition of Injective Linear Transformations 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's 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 transformtion, though we cannot be 100\% certain. Evaluate this cell repeatedly, to see how rarely the result is not injective. E = random_matrix(ZZ, 3, 2, x=-9, y=9) T = linear_transformation(QQ^2, QQ^3, E, side='right') T.is_injective() # random True Our concrete example below was created this way, so here we go. U = QQ^2 V = QQ^3 W = QQ^4 A = matrix(QQ, 3, 2, [[-4, -1], [-3, 3], [ 7, -6]]) B = matrix(QQ, 4, 3, [[ 7, 0, -1], [ 6, 2, 7], [ 3, -1, 2], [-6, -1, 1]]) T = linear_transformation(U, V, A, side='right') T.is_injective() True S = linear_transformation(V, W, B, side='right') S.is_injective() True C = S*T C.is_injective() True
1. Suppose $\ltdefn{T}{\complex{8}}{\complex{5}}$ is a linear transformation. Why can't $T$ be injective?
2. Describe the kernel of an injective linear transformation.
3. should remind you of . Why do we say this?
Each archetype below is a linear transformation. Compute the kernel for each.

, , , , , , , , , , ,
The linear transformation $\ltdefn{T}{\complex{4}}{\complex{3}}$ is not injective. Find two inputs $\vect{x},\,\vect{y}\in\complex{4}$ that yield the same output (that is $\lt{T}{\vect{x}}=\lt{T}{\vect{y}}$). \lt{T}{\colvector{x_1\\x_2\\x_3\\x_4}}= \colvector{ 2x_1+x_2+x_3\\ -x_1+3x_2+x_3-x_4\\ 3x_1+x_2+2x_3-2x_4 } A linear transformation that is not injective will have a non-trivial kernel (), and this is the key to finding the desired inputs. We need one non-trivial element of the kernel, so suppose that $\vect{z}\in\complex{4}$ is an element of the kernel, \colvector{0\\0\\0} =\zerovector =\lt{T}{\vect{z}} =\colvector{ 2z_1+z_2+z_3\\ -z_1+3z_2+z_3-z_4\\ 3z_1+z_2+2z_3-2z_4 } Vector equality leads to the homogeneous system of three equations in four variables, The coefficient matrix of this system row-reduces as \begin{bmatrix} \end{bmatrix} \rref \begin{bmatrix} \end{bmatrix} From this we can find a solution (we only need one), that is an element of $\krn{T}$, \vect{z}=\colvector{-1\\-1\\3\\1} Now, we choose a vector $\vect{x}$ at random and set $\vect{y}=\vect{x}+\vect{z}$, \vect{x} \colvector{2\\3\\4\\-2}+\colvector{-1\\-1\\3\\1} =\colvector{1\\2\\7\\-1} and you can check that \lt{T}{\vect{x}} =\colvector{11\\13\\21} =\lt{T}{\vect{y}} A quicker solution is to take two elements of the kernel (in this case, scalar multiples of $\vect{z}$) which both get sent to $\zerovector$ by $T$. Quicker yet, take $\zerovector$ and $\vect{z}$ as $\vect{x}$ and $\vect{y}$, which also both get sent to $\zerovector$ by $T$. Define the linear transformation \ltdefn{T}{\complex{3}}{\complex{2}},\quad \lt{T}{\colvector{x_1\\x_2\\x_3}}=\colvector{2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3} Find a basis for the kernel of $T$, $\krn{T}$. Is $T$ injective? To find the kernel, we require all $\vect{x}\in\complex{3}$ such that $\lt{T}{\vect{x}}=\zerovector$. This condition is \colvector{2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3}=\colvector{0\\0} This leads to a homogeneous system of two linear equations in three variables, whose coefficient matrix row-reduces to \begin{bmatrix} \end{bmatrix} With two free variables yields the basis for the null space \set{ \colvector{-\frac{5}{2}\\0\\1},\, \colvector{\frac{1}{2}\\1\\0} }

With $\nullity{T}\neq 0$, $\krn{T}\neq\set{\zerovector}$, so says $T$ is not injective.
Let $A = \begin{bmatrix} \end{bmatrix}$ and let $\ltdefn{T}{\complex{5}}{\complex{4}}$ be given by $\lt{T}{\vect{x}}=A\vect{x}$. Is $T$ injective? (Hint: No calculation is required.) By , if a linear transformation $\ltdefn{T}{U}{V}$ is injective, then $\dim(U)\le\dim(V)$. In this case, $\ltdefn{T}{\complex{5}}{\complex{4}}$, and $5=\dimension{\complex{5}} > \dimension{\complex{4}}=4$. Thus, $T$ cannot possibly be injective. Let $\ltdefn{T}{\complex{3}}{\complex{3}}$ be given by $\lt{T}{\colvector{x\\y\\z}} = \colvector{2x + y + z\\ x - y + 2z\\ x + 2y - z}$. Find $\krn{T}$. Is $T$ injective? If $\lt{T}{\colvector{x\\y\\z}} = \zerovector$, then $\colvector{2x + y + z\\x - y + 2z\\x + 2y - z} = \zerovector$. Thus, we have the system . Thus, we are looking for the nullspace of the matrix $A_T = \begin{bmatrix} \end{bmatrix}$. Since $A_T$ row-reduces to $\begin{bmatrix} \end{bmatrix}$, the kernel of $T$ is all vectors where $x = -z$ and $y = z$. Thus, $\krn{T} = \spn{\set{\colvector{ -1\\1\\1}}}$.

Since the kernel is not trivial, tells us that $T$ is not injective.
Let $A = \begin{bmatrix} \end{bmatrix}$ and let $\ltdefn{T}{\complex{4}}{\complex{4}}$ be given by $\lt{T}{\vect{x}}=A\vect{x}$. Find $\krn{T}$. Is $T$ injective? Since $T$ is given by matrix multiplication, $\krn{T} = \nsp{A}$. We have \begin{bmatrix} \end{bmatrix} \begin{bmatrix} \end{bmatrix} The nullspace of $A$ is $\set{\zerovector}$, so the kernel of $T$ is also trivial: $\krn{T} = \set{\zerovector}$. Let $A = \begin{bmatrix} \end{bmatrix}$ and let $\ltdefn{T}{\complex{4}}{\complex{4}}$ be given by $\lt{T}{\vect{x}}=A\vect{x}$. Find $\krn{T}$. Is $T$ injective? Since $T$ is given by matrix multiplication, $\krn{T} = \nsp{A}$. We have \begin{bmatrix} \end{bmatrix} \begin{bmatrix} \end{bmatrix} Thus, a basis for the nullspace of $A$ is $\set{\colvector{-1\\-1\\3\\0}}$, and the kernel is $\krn{T} = \spn{\spn{\colvector{-1\\-1\\3\\0}}}$. Since the kernel is nontrivial, this linear transformation is not injective. Given that the linear transformation $\ltdefn{T}{\complex{3}}{\complex{3}}$, $\lt{T}{\colvector{x\\y\\z}} = \colvector{2x + y\\2y + z\\x + 2z}$ is injective, show directly that $\set{ \lt{T}{\vect{e}_1},\, \lt{T}{\vect{e}_2},\, \lt{T}{\vect{e}_3} }$ is a linearly independent set. We have Let's put these vectors into a matrix and row reduce to test their linear independence. \begin{bmatrix} \end{bmatrix} \begin{bmatrix} \end{bmatrix} so the set of vectors $\set{\lt{T}{\vect{e}_1},\, \lt{T}{\vect{e}_1},\,\lt{T}{\vect{e}_1}}$ is linearly independent. Given that the linear transformation $\ltdefn{T}{\complex{2}}{\complex{3}}$, $\lt{T}{\colvector{x\\y}} = \colvector{x+y\\2x + y\\x + 2y}$ is injective, show directly that $\set{ \lt{T}{\vect{e}_1},\, \lt{T}{\vect{e}_2} }$ is a linearly independent set. We have $\lt{T}{\vect{e}_1} = \colvector{1\\2\\1}$ and $\lt{T}{\vect{e}_2} = \colvector{1\\1\\2}$. Putting these into a matrix as columns and row-reducing, we have \begin{bmatrix} \end{bmatrix} \begin{bmatrix} \end{bmatrix} Thus, the set of vectors $\set{\lt{T}{\vect{e}_1},\,\lt{T}{\vect{e}_2}}$ is linearly independent. Given that the linear transformation $\ltdefn{T}{\complex{3}}{\complex{5}}$, $\lt{T}{\colvector{x\\y\\z}} = \begin{bmatrix} \end{bmatrix} \colvector{x\\y\\z}$ is injective, show directly that $\set{ \lt{T}{\vect{e}_1},\, \lt{T}{\vect{e}_2},\, \lt{T}{\vect{e}_3} }$ is a linearly independent set. We have Let's row reduce the matrix of $T$ to test linear independence. \begin{bmatrix} \end{bmatrix} \begin{bmatrix} \end{bmatrix} so the set of vectors $\set{\lt{T}{\vect{e}_1},\,\lt{T}{\vect{e}_2},\,\lt{T}{\vect{e}_3}}$ is linearly independent. Show that the linear transformation $R$ is not injective by finding two different elements of the domain, $\vect{x}$ and $\vect{y}$, such that $\lt{R}{\vect{x}}=\lt{R}{\vect{y}}$. ($S_{22}$ is the vector space of symmetric $2\times 2$ matrices.) \ltdefn{R}{S_{22}}{P_1}\quad We choose $\vect{x}$ to be any vector we like. A particularly cocky choice would be to choose $\vect{x}=\zerovector$, but we will instead choose Then $\lt{R}{\vect{x}}=9+9x$. Now compute the kernel of $R$, which by we expect to be nontrivial. Setting equal to the zero vector, $\zerovector=0+0x$, and equating coefficients leads to a homogeneous system of equations. Row-reducing the coefficient matrix of this system will allow us to determine the values of $a$, $b$ and $c$ that create elements of the null space of $R$, \begin{bmatrix} \end{bmatrix} \rref \begin{bmatrix} \end{bmatrix} We only need a single element of the null space of this coefficient matrix, so we will not compute a precise description of the whole null space. Instead, choose the free variable $c=2$. Then is the corresponding element of the kernel. We compute the desired $\vect{y}$ as \vect{y}=\vect{x}+\vect{z}= + = Then check that $\lt{R}{\vect{y}}=9+9x$. Suppose $U$ and $V$ are vector spaces. Define the function $\ltdefn{Z}{U}{V}$ by $\lt{T}{\vect{u}}=\zerovector_{V}$ for every $\vect{u}\in U$. Then by , $Z$ is a linear transformation. Formulate a condition on $U$ that is equivalent to $Z$ being an injective linear transformation. In other words, fill in the blank to complete the following statement (and then give a proof): $Z$ is injective if and only if $U$ is . (See , .) Suppose $\ltdefn{T}{U}{V}$ is a linear transformation. For which vectors $\vect{v}\in V$ is $\preimage{T}{\vect{v}}$ a subspace of $U$? Suppose that that $\ltdefn{T}{U}{V}$ and $\ltdefn{S}{V}{W}$ are linear transformations. Prove the following relationship between kernels. \krn{T}\subseteq\krn{\compose{S}{T}} We are asked to prove that $\krn{T}$ is a subset of $\krn{\compose{S}{T}}$. Employing , choose $\vect{x}\in\krn{T}$. Then we know that $\lt{T}{\vect{x}}=\zerovector$. So \text{}\\ \text{} This qualifies $\vect{x}$ for membership in $\krn{\compose{S}{T}}$. Suppose that $A$ is an $m\times n$ matrix. Define the linear transformation $T$ by \ltdefn{T}{\complex{n}}{\complex{m}},\quad \lt{T}{\vect{x}}=A\vect{x} Prove that the kernel of $T$ equals the null space of $A$, $\krn{T}=\nsp{A}$. This is an equality of sets, so we want to establish two subset conditions ().

First, show $\nsp{A}\subseteq\krn{T}$. Choose $\vect{x}\in\nsp{A}$. Check to see if $\vect{x}\in\krn{T}$, \lt{T}{\vect{x}} So by , $\vect{x}\in\krn{T}$ and thus $\nsp{A}\subseteq\krn{T}$.

Now, show $\krn{T}\subseteq\nsp{A}$. Choose $\vect{x}\in\krn{T}$. Check to see if $\vect{x}\in\nsp{A}$, A\vect{x} So by , $\vect{x}\in\nsp{A}$ and thus $\krn{T}\subseteq\nsp{A}$.