# fcla / src / section-SLT.xml

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Surjective Linear Transformations

The companion to an injection is a surjection. Surjective linear transformations are closely related to spanning sets and ranges. So as you read this section reflect back on and note the parallels and the contrasts. In the next section, , we will combine the two properties.

Surjective Linear Transformations

As usual, we lead with a definition.

Surjective Linear Transformation

Suppose $\ltdefn{T}{U}{V}$ is a linear transformation. Then $T$ is surjective if for every $\vect{v}\in V$ there exists a $\vect{u}\in U$ so that $\lt{T}{\vect{u}}=\vect{v}$.

Given an arbitrary function, it is possible for there to be an element of the codomain that is not an output of the function (think about the function $y=f(x)=x^2$ and the codomain element $y=-3$). For a surjective function, this never happens. If we choose any element of the codomain ($\vect{v}\in V$) then there must be an input from the domain ($\vect{u}\in U$) which will create the output when used to evaluate the linear transformation ($\lt{T}{\vect{u}}=\vect{v}$). Some authors prefer the term onto where we use surjective, and we will sometimes refer to a surjective linear transformation as a surjection.

Examples of Surjective Linear Transformations

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

Not surjective, Archetype Q

is the linear transformation

We will demonstrate that \vect{v}=\colvector{-1\\2\\3\\-1\\4} is an unobtainable element of the codomain. Suppose to the contrary that $\vect{u}$ is an element of the domain such that $\lt{T}{\vect{u}}=\vect{v}$.

Then =\lt{T}{\colvector{u_1\\u_2\\u_3\\u_4\\u_5}}\\ -2 u_1 + 3 u_2 + 3 u_3 - 6 u_4 + 3 u_5\\ -16 u_1 + 9 u_2 + 12 u_3 - 28 u_4 + 28 u_5\\ -19 u_1 + 7 u_2 + 14 u_3 - 32 u_4 + 37 u_5\\ -21 u_1 + 9 u_2 + 15 u_3 - 35 u_4 + 39 u_5\\ -9 u_1 + 5 u_2 + 7 u_3 - 16 u_4 + 16 u_5}\\ \begin{bmatrix} \end{bmatrix} \colvector{u_1\\u_2\\u_3\\u_4\\u_5}

Now we recognize the appropriate input vector $\vect{u}$ as a solution to a linear system of equations. Form the augmented matrix of the system, and row-reduce to \begin{bmatrix} \end{bmatrix}

With a leading 1 in the last column, tells us the system is inconsistent. From the absence of any solutions we conclude that no such vector $\vect{u}$ exists, and by , $T$ is not surjective.

Again, do not concern yourself with how $\vect{v}$ was selected, as this will be explained shortly. However, do understand why this vector provides enough evidence to conclude that $T$ is not surjective.

Here's a cartoon of a non-surjective linear transformation. Notice that the central feature of this cartoon is that the vector $\vect{v}\in V$ does not have an arrow pointing to it, implying there is no $\vect{u}\in U$ such that $\lt{T}{\vect{u}}=\vect{v}$. Even though this happens again with a second unnamed vector in $V$, it only takes one occurrence to destroy the possibility of surjectivity. Non-Surjective 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}}]]> % 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 \node (u1) [ltvect] at (5em, 13em) {}; \node (u2) [ltvect] at (5em, 11em) {}; \node (u3) [ltvect] at (5em, 8em) {}; \node (u4) [ltvect] at (5em, 6em) {}; % outputs \node (v1) [ltvect] at (20em, 12em) {}; \node (v2) [ltvect] at (20em, 7em) {}; \node (v3) [ltvect] at (19em, 3em) {}; \node (v) [ltvect, label=right:$\vect{v}$] at (21em, 3em) {}; % associations \draw[ltedge] (u1) to (v1); \draw[ltedge] (u2) to (v1); \draw[ltedge] (u3) to (v2); \draw[ltedge] (u4) to (v2);

To show that a linear transformation is not surjective, it is enough to find a single element of the codomain that is never created by any input, as in . However, to show that a linear transformation is surjective we must establish that every element of the codomain occurs as an output of the linear transformation for some appropriate input.

Surjective, Archetype R

is the linear transformation

To establish that $R$ is surjective we must begin with a totally arbitrary element of the codomain, $\vect{v}$ and somehow find an input vector $\vect{u}$ such that $\lt{T}{\vect{u}}=\vect{v}$. We desire, \colvector{-65 u_1 + 128 u_2 + 10 u_3 - 262 u_4 + 40 u_5\\ 36 u_1 - 73 u_2 - u_3 + 151 u_4 - 16 u_5\\ -44 u_1 + 88 u_2 + 5 u_3 - 180 u_4 + 24 u_5\\ 34 u_1 - 68 u_2 - 3 u_3 + 140 u_4 - 18 u_5\\ 12 u_1 - 24 u_2 - u_3 + 49 u_4 - 5 u_5} \colvector{v_1\\v_2\\v_3\\v_4\\v_5}\\ \begin{bmatrix} \end{bmatrix} \colvector{u_1\\u_2\\u_3\\u_4\\u_5} \colvector{v_1\\v_2\\v_3\\v_4\\v_5}

We recognize this equation as a system of equations in the variables $u_i$, but our vector of constants contains symbols. In general, we would have to row-reduce the augmented matrix by hand, due to the symbolic final column. However, in this particular example, the $5\times 5$ coefficient matrix is nonsingular and so has an inverse (, ). \inverse{ \begin{bmatrix} \end{bmatrix} } = \begin{bmatrix} \end{bmatrix} so we find that \colvector{u_1\\u_2\\u_3\\u_4\\u_5} \begin{bmatrix} \end{bmatrix} \colvector{v_1\\v_2\\v_3\\v_4\\v_5}\\ \colvector{-47 v_1 + 92 v_2 + v_3 - 181 v_4 - 14 v_5\\ 27 v_1 - 55 v_2 + \frac{7}{2} v_3 + \frac{221}{2} v_4 + 11 v_5\\ -32 v_1 + 64 v_2 - v_3 - 126 v_4 - 12 v_5\\ 25 v_1 - 50 v_2 + \frac{3}{2} v_3 + \frac{199}{2} v_4 + 9 v_5\\ 9 v_1 - 18 v_2 + \frac{1}{2} v_3 + \frac{71}{2} v_4 + 4 v_5}

This establishes that if we are given any output vector $\vect{v}$, we can use its components in this final expression to formulate a vector $\vect{u}$ such that $\lt{T}{\vect{u}}=\vect{v}$. So by we now know that $T$ is surjective. You might try to verify this condition in its full generality ( evaluate $T$ with this final expression and see if you get $\vect{v}$ as the result), or test it more specifically for some numerical vector $\vect{v}$ (see ).

Here's the cartoon for a surjective linear transformation. It is meant to suggest that for every output in $V$ there is at least one input in $U$ that is sent to the output. (Even though we have depicted several inputs sent to each output.) The key feature of this cartoon is that there are no vectors in $V$ without an arrow pointing to them. Surjective 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}}]]> % base generic picture \draw ( 5em, 8em) circle [x radius=5em, y radius=8em, thick]; \draw (20em, 8em) circle [x radius=4em, y radius=7em, thick]; \node (U) at ( 5em, -1em) {$U$}; \node (V) at (20em, -1em) {$V$}; \draw[->, thick, draw] (U) to node[auto] {$T$} (V); % inputs \node (u1) [ltvect] at (5em, 14em) {}; \node (u2) [ltvect] at (5em, 12em) {}; \node (u3) [ltvect] at (5em, 9em) {}; \node (u4) [ltvect] at (5em, 7em) {}; \node (u5) [ltvect] at (5em, 4em) {}; \node (u6) [ltvect] at (5em, 2em) {}; % outputs \node (v1) [ltvect] at (20em, 12em) {}; \node (v2) [ltvect] at (20em, 8em) {}; \node (v3) [ltvect] at (20em, 4em) {}; % associations \draw[ltedge] (u1) to (v1); \draw[ltedge] (u2) to (v1); \draw[ltedge] (u3) to (v2); \draw[ltedge] (u4) to (v2); \draw[ltedge] (u5) to (v3); \draw[ltedge] (u6) to (v3);

Let's now examine a surjective linear transformation between abstract vector spaces.

Surjective, Archetype V

is defined by

To establish that the linear transformation is surjective, begin by choosing an arbitrary output. In this example, we need to choose an arbitrary $2\times 2$ matrix, say and we would like to find an input polynomial \vect{u}=a+bx+cx^2+dx^3 so that $\lt{T}{\vect{u}}=\vect{v}$. So we have, \end{bmatrix}

Matrix equality leads us to the system of four equations in the four unknowns, $x,y,z,w$, which can be rewritten as a matrix equation, \begin{bmatrix} \end{bmatrix} \colvector{a\\b\\c\\d} = \colvector{x\\y\\z\\w}

The coefficient matrix is nonsingular, hence it has an inverse, \inverse{ \begin{bmatrix} \end{bmatrix} } = \begin{bmatrix} \end{bmatrix} so we have \begin{bmatrix} \end{bmatrix} \colvector{x\\y\\z\\w}\\% \colvector{ x-z-w\\ z+w\\ \frac{1}{2}(x-y-z-w)\\ z }

So the input polynomial $\vect{u}=(x-z-w)+(z+w)x+\frac{1}{2}(x-y-z-w)x^2+zx^3$ will yield the output matrix $\vect{v}$, no matter what form $\vect{v}$ takes. This means by that $T$ is surjective. All the same, let's do a concrete demonstration and evaluate $T$ with $\vect{u}$, \begin{bmatrix} \end{bmatrix}\\ \begin{bmatrix} \end{bmatrix}\\

Range of a Linear Transformation

For a linear transformation $\ltdefn{T}{U}{V}$, the range is a subset of the codomain $V$. Informally, it is the set of all outputs that the transformation creates when fed every possible input from the domain. It will have some natural connections with the column 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.

Range of a Linear Transformation

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

Range of a Linear Transformation $\rng{T}$
Range, Archetype O

is the linear transformation

To determine the elements of $\complex{5}$ in $\rng{T}$, find those vectors $\vect{v}$ such that $\lt{T}{\vect{u}}=\vect{v}$ for some $\vect{u}\in\complex{3}$, -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{-u_1\\-u_1\\u_1\\2 u_1\\ u_1} + \colvector{u_2\\2u_2\\u_2\\3u_2\\ 0} + \colvector{-3u_3\\-4u_3\\u_3\\u_3\\ 2 u_3}\\ u_1\colvector{-1\\-1\\1\\2\\1} + u_2\colvector{1\\2\\1\\3\\ 0} + u_3\colvector{-3\\-4\\1\\1\\2}

This says that every output of $T$ ($\vect{v}$) can be written as a linear combination of the three vectors \colvector{-1\\-1\\1\\2\\1} \colvector{1\\2\\1\\3\\ 0} \colvector{-3\\-4\\1\\1\\2} using the scalars $u_1,\,u_2,\,u_3$. Furthermore, since $\vect{u}$ can be any element of $\complex{3}$, every such linear combination is an output. This means that \rng{T}=\spn{\set{ \colvector{-1\\-1\\1\\2\\1},\, \colvector{1\\2\\1\\3\\ 0},\, \colvector{-3\\-4\\1\\1\\2} }}

The three vectors in this spanning set for $\rng{T}$ form a linearly dependent set (check this!). So we can find a more economical presentation by any of the various methods from and . We will place the vectors into a matrix as rows, row-reduce, toss out zero rows and appeal to , so we can describe the range of $T$ with a basis, \rng{T}=\spn{}

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

Range of a Linear Transformation is a Subspace

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

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

Suppose we assume that $\vect{x},\,\vect{y}\in\rng{T}$. Is $\vect{x}+\vect{y}\in\rng{T}$? If $\vect{x},\,\vect{y}\in\rng{T}$ then we know there are vectors $\vect{w},\,\vect{z}\in U$ such that $\lt{T}{\vect{w}}=\vect{x}$ and $\lt{T}{\vect{z}}=\vect{y}$. Because $U$ is a vector space, additive closure () implies that $\vect{w}+\vect{z}\in U$.

Then \text{}\\

So we have found an input, $\vect{w}+\vect{z}$, which when fed into $T$ creates $\vect{x}+\vect{y}$ as an output. This qualifies $\vect{x}+\vect{y}$ for membership in $\rng{T}$. So we have additive closure.

Suppose we assume that $\alpha\in\complex{\null}$ and $\vect{x}\in\rng{T}$. Is $\alpha\vect{x}\in\rng{T}$? If $\vect{x}\in\rng{T}$, then there is a vector $\vect{w}\in U$ such that $\lt{T}{\vect{w}}=\vect{x}$. Because $U$ is a vector space, scalar closure implies that $\alpha\vect{w}\in U$. Then \text{}\\

So we have found an input ($\alpha\vect{w}$) which when fed into $T$ creates $\alpha\vect{x}$ as an output. This qualifies $\alpha\vect{x}$ for membership in $\rng{T}$. So we have scalar closure and tells us that $\rng{T}$ is a subspace of $V$.

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

Full range, Archetype N

is the linear transformation

To determine the elements of $\complex{3}$ in $\rng{T}$, find those vectors $\vect{v}$ such that $\lt{T}{\vect{u}}=\vect{v}$ for some $\vect{u}\in\complex{5}$, \colvector{ 2 u_1 + u_2 + 3 u_3 - 4 u_4 + 5 u_5\\ u_1 - 2 u_2 + 3 u_3 - 9 u_4 + 3 u_5\\ 3 u_1 + 4 u_3 - 6 u_4 + 5 u_5}\\ \colvector{2u_1\\u_1\\3u_1}+ \colvector{u_2\\-2u_2\\0}+ \colvector{3u_3\\3u_3\\4u_3}+ \colvector{-4u_4\\-9u_4\\-6u_4}+ \colvector{5u_5\\3u_5\\5u_5}\\ u_1\colvector{2\\1\\3}+ u_2\colvector{1\\-2\\0}+ u_3\colvector{3\\3\\4}+ u_4\colvector{-4\\-9\\-6}+ u_5\colvector{5\\3\\5}\\

This says that every output of $T$ ($\vect{v}$) can be written as a linear combination of the five vectors \colvector{5\\3\\5} using the scalars $u_1,\,u_2,\,u_3,\,u_4,\,u_5$. Furthermore, since $\vect{u}$ can be any element of $\complex{5}$, every such linear combination is an output. This means that \rng{T}=\spn{\set{ \colvector{2\\1\\3},\, \colvector{1\\-2\\0},\, \colvector{3\\3\\4},\, \colvector{-4\\-9\\-6},\, \colvector{5\\3\\5} }}

The five vectors in this spanning set for $\rng{T}$ form a linearly dependent set (). So we can find a more economical presentation by any of the various methods from and . We will place the vectors into a matrix as rows, row-reduce, toss out zero rows and appeal to , so we can describe the range of $T$ with a (nice) basis, \rng{T}=\spn{}=\complex{3}

In contrast to injective linear transformations having small (trivial) kernels (), surjective linear transformations have large ranges, as indicated in the next theorem.

Range of a Surjective Linear Transformation

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation. Then $T$ is surjective if and only if the range of $T$ equals the codomain, $\rng{T}=V$.

By , we know that $\rng{T}\subseteq V$. To establish the reverse inclusion, assume $\vect{v}\in V$. Then since $T$ is surjective (), there exists a vector $\vect{u}\in U$ so that $\lt{T}{\vect{u}}=\vect{v}$. However, the existence of $\vect{u}$ gains $\vect{v}$ membership in $\rng{T}$, so $V\subseteq\rng{T}$. Thus, $\rng{T}=V$.

To establish that $T$ is surjective, choose $\vect{v}\in V$. Since we are assuming that $\rng{T}=V$, $\vect{v}\in\rng{T}$. This says there is a vector $\vect{u}\in U$ so that $\lt{T}{\vect{u}}=\vect{v}$, $T$ is surjective.

Not surjective, 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 surjective by constructing a vector in the codomain where no element of the domain could be used to evaluate the linear transformation to create the output, thus violating . Just where did this vector come from?

The short answer is that the vector \vect{v}=\colvector{-1\\2\\3\\-1\\4} was constructed to lie outside of the range of $T$. How was this accomplished? First, the range of $T$ is given by \rng{T}=\spn{}

Suppose an element of the range $\vect{v^*}$ has its first 4 components equal to $-1$, $2$, $3$, $-1$, in that order. Then to be an element of $\rng{T}$, we would have \vect{v^*}=(-1)\colvector{1\\0\\0\\0\\1}+(2)\colvector{0\\1\\0\\0\\-1}+(3) \colvector{0\\0\\1\\0\\-1}+(-1)\colvector{0\\0\\0\\1\\2} =\colvector{-1\\2\\3\\-1\\-8}

So the only vector in the range with these first four components specified, must have $-8$ in the fifth component. To set the fifth component to any other value (say, 4) will result in a vector ($\vect{v}$ in ) outside of the range. Any attempt to find an input for $T$ that will produce $\vect{v}$ as an output will be doomed to failure.

Whenever the range of a linear transformation is not the whole codomain, we can employ this device and conclude that the linear transformation is not surjective. This is another way of viewing . For a surjective linear transformation, the range is all of the codomain and there is no choice for a vector $\vect{v}$ that lies in $V$, yet not in the range. For every one of the archetypes that is not surjective, there is an example presented of exactly this form.

Not surjective, Archetype O

In the range of was determined to be \rng{T}=\spn{} a subspace of dimension 2 in $\complex{5}$. Since $\rng{T}\neq\complex{5}$, says $T$ is not surjective.

Surjective, Archetype N

The range of was computed in to be \rng{T}=\spn{}

Since the basis for this subspace is the set of standard unit vectors for $\complex{3}$ (), we have $\rng{T}=\complex{3}$ and by , $T$ is surjective.

Surjective Linear Transformations The situation in Sage for surjective linear transformations is similar to that for injective linear transformations. One distinction what your text calls the range of a linear transformation is called the image of a transformation, obtained with the .image() method. Sage's term is more commonly used, so you are likey to see it again. As before, two examples, first up is . 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_surjective() False T.image() Vector space of degree 5 and dimension 2 over Rational Field Basis matrix: [ 1 0 -3 -7 -2] [ 0 1 2 5 1] Besides showing the relevant commands in action, this demonstrates one half of . Now a reprise of . U = QQ^5 V = QQ^3 x1, x2, x3, x4, x5 = var('x1, x2, x3, x4, x5') outputs = [2*x1 + x2 + 3*x3 - 4*x4 + 5*x5, x1 - 2*x2 + 3*x3 - 9*x4 + 3*x5, 3*x1 + 4*x3 - 6*x4 + 5*x5] S_symbolic(x1, x2, x3, x4, x5) = outputs S = linear_transformation(U, V, S_symbolic) S.is_surjective() True S.image() Vector space of degree 3 and dimension 3 over Rational Field Basis matrix: [1 0 0] [0 1 0] [0 0 1] S.image() == V True Previously, we have chosen elements of the codomain which have non-empty or empty preimages. We can now explain how to do this predictably. explains that elements of the codomain with non-empty pre-images are precisely elements of the image. Consider the non-surjective linear transformation T from above. TI = T.image() B = TI.basis() B [ (1, 0, -3, -7, -2), (0, 1, 2, 5, 1) ] b0 = B[0] b1 = B[1] Now any linear combination of the basis vectors b0 and b1 must be an element of the image. Moreover, the first two slots of the resulting vector will equal the two scalars we choose to create the linear combination. But most importantly, see that the three remaining slots will be uniquely determined by these two choices. This means there is exactly one vector in the image with these values in the first two slots. So if we construct a new vector with these two values in the first two slots, and make any part of the last three slots even slightly different, we will form a vector that cannot be in the image, and will thus have a preimage that is an empty set. Whew, that is probably worth reading carefully several times, perhaps in conjunction with the example following. in_image = 4*b0 + (-5)*b1 in_image (4, -5, -22, -53, -13) T.preimage_representative(in_image) (-13, -9, 0) outside_image = 4*b0 + (-5)*b1 + vector(QQ, [0, 0, 0, 0, 1]) outside_image (4, -5, -22, -53, -12) T.preimage_representative(outside_image) Traceback (most recent call last): ... ValueError: element is not in the image
Spanning Sets and Surjective Linear Transformations

Just as injective linear transformations are allied with linear independence (, ), surjective linear transformations are allied with spanning sets.

Spanning Set for Range of a Linear Transformation

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and spans $U$. Then spans $\rng{T}$.

We need to establish that $\rng{T}=\spn{R}$, a set equality. First we establish that $\rng{T}\subseteq\spn{R}$. To this end, choose $\vect{v}\in\rng{T}$. Then there exists a vector $\vect{u}\in U$, such that $\lt{T}{\vect{u}}=\vect{v}$ (). Because $S$ spans $U$ there are scalars, $\scalarlist{a}{t}$, such that \vect{u}=\lincombo{a}{u}{t}

Then \vect{v} \text{}\\ \text{}\\ \text{}\\ which establishes that $\vect{v}\in\spn{R}$ (). So $\rng{T}\subseteq\spn{R}$.

To establish the opposite inclusion, choose an element of the span of $R$, say $\vect{v}\in\spn{R}$. Then there are scalars $\scalarlist{b}{t}$ so that \vect{v} \text{}\\ \text{}

This demonstrates that $\vect{v}$ is an output of the linear transformation $T$, so $\vect{v}\in\rng{T}$. Therefore $\spn{R}\subseteq\rng{T}$, so we have the set equality $\rng{T}=\spn{R}$ (). In other words, $R$ spans $\rng{T}$ ().

provides an easy way to begin the construction of a basis for the range of a linear transformation, since the construction of a spanning set requires simply evaluating the linear transformation on a spanning set of the domain. In practice the best choice for a spanning set of the domain would be as small as possible, in other words, a basis. The resulting spanning set for the codomain may not be linearly independent, so to find a basis for the range might require tossing out redundant vectors from the spanning set. Here's an example.

A basis for the range of a linear transformation

Define the linear transformation $\ltdefn{T}{M_{22}}{P_2}$ by =\left(a+2b+8c+d\right)+\left(-3a+2b+5d\right)x+\left(a+b+5c\right)x^2

A convenient spanning set for $M_{22}$ is the basis S=\set{ }

So by , a spanning set for $\rng{T}$ is R }\\

The set $R$ is not linearly independent, so if we desire a basis for $\rng{T}$, we need to eliminate some redundant vectors. Two particular relations of linear dependence on $R$ are

These, individually, allow us to remove $8+5x^2$ and $1+5x$ from $R$ without destroying the property that $R$ spans $\rng{T}$. The two remaining vectors are linearly independent (check this!), so we can write \rng{T}=\spn{\set{1-3x+x^2,\,2+2x+x^2}} and see that $\dimension{\rng{T}}=2$.

Elements of the range are precisely those elements of the codomain with non-empty preimages.

Range and Pre-Image

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation. Then \vect{v}\in\rng{T}\text{ if and only if }\preimage{T}{\vect{v}}\neq\emptyset

If $\vect{v}\in\rng{T}$, then there is a vector $\vect{u}\in U$ such that $\lt{T}{\vect{u}}=\vect{v}$. This qualifies $\vect{u}$ for membership in $\preimage{T}{\vect{v}}$, and thus the preimage of $\vect{v}$ is not empty.

Suppose the preimage of $\vect{v}$ is not empty, so we can choose a vector $\vect{u}\in U$ such that $\lt{T}{\vect{u}}=\vect{v}$. Then $\vect{v}\in\rng{T}$.

Now would be a good time to return to which depicted the pre-images of a non-surjective linear transformation. The vectors $\vect{x},\,\vect{y}\in V$ were elements of the codomain whose pre-images were empty, as we expect for a non-surjective linear transformation from the characterization in .

Surjective Linear Transformations and Bases

Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation and is a basis of $U$. Then $T$ is surjective if and only if is a spanning set for $V$.

Assume $T$ is surjective. Since $B$ is a basis, we know $B$ is a spanning set of $U$ (). Then says that $C$ spans $\rng{T}$. But the hypothesis that $T$ is surjective means $V=\rng{T}$ (), so $C$ spans $V$.

Assume that $C$ spans $V$. To establish that $T$ is surjective, we will show that every element of $V$ is an output of $T$ for some input (). Suppose that $\vect{v}\in V$. As an element of $V$, we can write $\vect{v}$ as a linear combination of the spanning set $C$. So there are are scalars, $\scalarlist{b}{m}$, such that \vect{v}=b_1\lt{T}{\vect{u}_1}+b_2\lt{T}{\vect{u}_2}+b_3\lt{T}{\vect{u}_3}+\cdots+b_m\lt{T}{\vect{u}_m}

Now define the vector $\vect{u}\in U$ by \vect{u}=\lincombo{b}{u}{m}

Then \text{}\\

So, given any choice of a vector $\vect{v}\in V$, we can design an input $\vect{u}\in U$ to produce $\vect{v}$ as an output of $T$. Thus, by , $T$ is surjective.

Surjective Linear Transformations and Dimension Surjective Linear Transformations and Dimension

Suppose that $\ltdefn{T}{U}{V}$ is a surjective linear transformation. Then $\dimension{U}\geq\dimension{V}$.

Suppose to the contrary that 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 a spanning set of $V$ with $m$ or fewer vectors. So we have a set of $m$ or fewer vectors that span $V$, a vector space of dimension $t$, with However, this contradicts , so our assumption is false and $\dimension{U}\geq\dimension{V}$.

Not surjective by dimension, Archetype T

The linear transformation in is

Since $T$ cannot be surjective for then it 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 surjective. and are two more examples of linear transformations that have small domains and big codomains, resulting in an inability to create all possible outputs and thus they are non-surjective linear transformations.

Composition of Surjective 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 surjective linear transformations is again surjective, so we prove that here.

Composition of Surjective Linear Transformations is Surjective

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

That the composition is a linear transformation was established in , so we need only establish that the composition is surjective. Applying , choose $\vect{w}\in W$.

Because $S$ is surjective, there must be a vector $\vect{v}\in V$, such that $\lt{S}{\vect{v}}=\vect{w}$. With the existence of $\vect{v}$ established, that $T$ is surjective guarantees a vector $\vect{u}\in U$ such that $\lt{T}{\vect{u}}=\vect{v}$. Now, \text{}\\

This establishes that any element of the codomain ($\vect{w}$) can be created by evaluating $\compose{S}{T}$ with the right input ($\vect{u}$). Thus, by , $\compose{S}{T}$ is surjective.

Composition of Surjective Linear Transformations As we mentioned in the last section, experimenting with Sage is a worthwhile complement to other methods of learning mathematics. We have purposely avoided providing illustrations of deeper results, such as and , which you should now be equipped to investigate yourself. For completeness, and since composition wil be very important in the next few sections, we will provide an illustration of . Similar to what we did in the previous section, we choose dimensions suggested by , and then use randomly constructed matrices to form a pair of surjective linear transformations. U = QQ^4 V = QQ^3 W = QQ^2 A = matrix(QQ, 3, 4, [[ 3, -2, 8, -9], [-1, 3, -4, -1], [ 3, 2, 8, 3]]) T = linear_transformation(U, V, A, side='right') T.is_surjective() True B = matrix(QQ, 2, 3, [[ 8, -5, 3], [-2, 1, 1]]) S = linear_transformation(V, W, B, side='right') S.is_surjective() True C = S*T C.is_surjective() True
1. Suppose $\ltdefn{T}{\complex{5}}{\complex{8}}$ is a linear transformation. Why can't $T$ be surjective?
2. What is the relationship between a surjective linear transformation and its range?
3. There are many similarities and differences between injective and surjective linear transformations. Compare and contrast these two different types of linear transformations. (This means going well beyond just stating their definitions.)
Each archetype below is a linear transformation. Compute the range for each.

, , , , , , , , , , ,
concludes with an expression for a vector $\vect{u}\in\complex{5}$ that we believe will create the vector $\vect{v}\in\complex{5}$ when used to evaluate $T$. That is, $\lt{T}{\vect{u}}=\vect{v}$. Verify this assertion by actually evaluating $T$ with $\vect{u}$. If you don't have the patience to push around all these symbols, try choosing a numerical instance of $\vect{v}$, compute $\vect{u}$, and then compute $\lt{T}{\vect{u}}$, which should result in $\vect{v}$. The linear transformation $\ltdefn{S}{\complex{4}}{\complex{3}}$ is not surjective. Find an output $\vect{w}\in\complex{3}$ that has an empty pre-image (that is $\preimage{S}{\vect{w}}=\emptyset$.) \lt{S}{\colvector{x_1\\x_2\\x_3\\x_4}}= \colvector{ 2x_1+x_2+3x_3-4x_4\\ x_1+3x_2+4x_3+3x_4\\ -x_1+2x_2+x_3+7x_4 } To find an element of $\complex{3}$ with an empty pre-image, we will compute the range of the linear transformation $\rng{S}$ and then find an element outside of this set.

By we can evaluate $S$ with the elements of a spanning set of the domain and create a spanning set for the range. So \rng{S}=\spn{\set{ \colvector{2\\1\\-1},\, \colvector{1\\3\\2},\, \colvector{3\\4\\1},\, \colvector{-4\\3\\7} }} This spanning set is obviously linearly dependent, so we can reduce it to a basis for $\rng{S}$ using , where the elements of the spanning set are placed as the rows of a matrix. The result is that \rng{S}=\spn{\set{ \colvector{1\\0\\-1},\, \colvector{0\\1\\1} }} Therefore, the unique vector in $\rng{S}$ with a first slot equal to 6 and a second slot equal to 15 will be the linear combination 6\colvector{1\\0\\-1}+15\colvector{0\\1\\1}=\colvector{6\\15\\9} So, any vector with first two components equal to 6 and 15, but with a third component different from 9, such as \vect{w}=\colvector{6\\15\\-63} will not be an element of the range of $S$ and will therefore have an empty pre-image. Another strategy on this problem is to guess. Almost any vector will lie outside the range of $T$, you have to be unlucky to randomly choose an element of the range. This is because the codomain has dimension 3, while the range is much smaller at a dimension of 2. You still need to check that your guess lies outside of the range, which generally will involve solving a system of equations that turns out to be inconsistent.
Determine whether or not the following linear transformation $\ltdefn{T}{\complex{5}}{P_3}$ is surjective: The linear transformation $T$ is surjective if for any $p(x) = \alpha + \beta x + \gamma x^2 + \delta x^3$, there is a vector $\vect{u} = \colvector{a\\b\\c\\d\\e}$ in $\complex{5}$ so that $\lt{T}{\vect{u}} = p(x)$. We need to be able to solve the system This system has an infinite number of solutions, one of which is $a = \alpha$, $b = \beta$, $c = 0$, $d = \gamma$ and $e = \delta - \gamma$, so that \lt{T}{\colvector{\alpha\\ \beta\\0\\ \gamma \\ \delta - \gamma}} Thus, $T$ is surjective, since for every vector $\vect{v} \in P_3$, there exists a vector $\vect{u} \in \complex{5}$ so that $\lt{T}{\vect{u}} = \vect{v}$. Determine whether or not the linear transformation $\ltdefn{T}{P_3}{\complex{5}}$ below is surjective: According to , if a linear transformation $\ltdefn{T}{U}{V}$ is surjective, then $\dimension{U}\ge\dimension{V}$. In this example, $U = P_3$ has dimension 4, and $V = \complex{5}$ has dimension 5, so $T$ cannot be surjective. (There is no way $T$ can expand the domain $P_3$ to fill the codomain $\complex{5}$.) 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 range of $T$, $\rng{T}$. Is $T$ surjective? To find the range of $T$, apply $T$ to the elements of a spanning set for $\complex{3}$ as suggested in . We will use the standard basis vectors (). \rng{T}= \spn{\set{\lt{T}{\vect{e}_1},\,\lt{T}{\vect{e}_2},\,\lt{T}{\vect{e}_3}}}= \spn{\set{\colvector{2\\-4},\,\colvector{-1\\2},\,\colvector{5\\-10}}} Each of these vectors is a scalar multiple of the others, so we can toss two of them in reducing the spanning set to a linearly independent set (or be more careful and apply on a matrix with these three vectors as columns). The result is the basis of the range, \set{\colvector{1\\-2}}

With $\rank{T}\neq 2$, $\rng{T}\neq\complex{2}$, so says $T$ is not surjective.
Let $\ltdefn{T}{\complex{3}}{\complex{3}}$ be given by $\lt{T}{\colvector{a\\b\\c}} = \colvector{a + b + 2c\\ 2c\\ a + b + c}$. Find a basis of $\rng{T}$. Is $T$ surjective? The range of $T$ is \rng{T} Since the vectors $\colvector{1\\0\\1}$ and $\colvector{2\\2\\1}$ are linearly independent (why?), a basis of $\rng{T}$ is $\set{\colvector{1\\0\\1},\colvector{2\\2\\1}}$. Since the dimension of the range is 2 and the dimension of the codomain is 3, $T$ is not surjective. Let $\ltdefn{T}{\complex{3}}{\complex{4}}$ be given by $\lt{T}{\colvector{a\\b\\c}} = \colvector{a + b -c\\ a - b + c\\ -a + b + c\\a + b + c}$. Find a basis of $\rng{T}$. Is $T$ surjective? The range of $T$ is \rng{T} By row reduction (not shown), we can see that the set \set{ \colvector{1\\1\\-1\\1},\, \colvector{1\\-1\\1\\1},\, \colvector{-1\\1\\1\\1} } are linearly independent, so is a basis of $\rng{T}$. Since the dimension of the range is 3 and the dimension of the codomain is 4, $T$ is not surjective. (We should have anticipated that $T$ was not surjective since the dimension of the domain is smaller than the dimension of the codomain.) Let $\ltdefn{T}{\complex{4}}{ M_{2,2}}$ be given by Find a basis of $\rng{T}$. Is $T$ surjective? The range of $T$ is \rng{T} \begin{bmatrix} \end{bmatrix}}{a,b,c,d\in\complexes}\\ }{a,b,c,d\in\complexes}\\ Can you explain the last equality above?

These three matrices are linearly independent, so a basis of $\rng{T}$ is $\set{ (Notice that the range is actually the subspace of symmetric$2 \times 2$matrices in$M_{2,2}$.) Let$\ltdefn{T}{P_2}{P_4}$be given by$\lt{T}{p(x)} = x^2 p(x)$. Find a basis of$\rng{T}$. Is$T$surjective? If we transform the basis of$P_2$, then guarantees we will have a spanning set of$\rng{T}$. A basis of$P_2$is$\set{1, x, x^2}$. If we transform the elements of this set, we get the set$\set{x^2, x^3, x^4}$which is a spanning set for$\rng{T}$. These three vectors are linearly independent, so$\set{x^2, x^3, x^4}$is a basis of$\rng{T}$. Let$\ltdefn{T}{P_4}{P_3}$be given by$\lt{T}{p(x)} = p^\prime(x)$, where$p^\prime(x)$is the derivative. Find a basis of$\rng{T}$. Is$T$surjective? If we transform the basis of$P_4$, then guarantees we will have a spanning set of$\rng{T}$. A basis of$P_4$is$\set{1, x, x^2, x^3, x^4}$. If we transform the elements of this set, we get the set$\set{0, 1, 2x, 3x^2, 4x^3}$which is a spanning set for$\rng{T}$. Reducing this to a linearly independent set, we find that$\{1, 2x, 3x^2, 4x^3\}$is a basis of$\rng{T}$. Since$\rng{T}$and$P_3$both have dimension 4,$T$is surjective. Show that the linear transformation$T$is not surjective by finding an element of the codomain,$\vect{v}$, such that there is no vector$\vect{u}$with$\lt{T}{\vect{u}}=\vect{v}$. \ltdefn{T}{\complex{3}}{\complex{3}},\quad \lt{T}{\colvector{a\\b\\c}}= \colvector{2a+3b-c\\2b-2c\\a-b+2c} We wish to find an output vector$\vect{v}$that has no associated input. This is the same as requiring that there is no solution to the equality \vect{v}=\lt{T}{\colvector{a\\b\\c}}=\colvector{2a+3b-c\\2b-2c\\a-b+2c} =a\colvector{2\\0\\1}+b\colvector{3\\2\\-1}+c\colvector{-1\\-2\\2} In other words, we would like to find an element of$\complex{3}$not in the set Y=\spn{\set{\colvector{2\\0\\1},\,\colvector{3\\2\\-1},\,\colvector{-1\\-2\\2}}} If we make these vectors the rows of a matrix, and row-reduce, provides an alternate description of$Y$, Y=\spn{\set{\colvector{2\\0\\1},\,\colvector{0\\4\\-5}}} If we add these vectors together, and then change the third component of the result, we will create a vector that lies outside of$Y$, say$\vect{v}=\colvector{2\\4\\9}$. 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$V$that is equivalent to$Z$being an surjective linear transformation. In other words, fill in the blank to complete the following statement (and then give a proof):$Z$is surjective if and only if$V$is . (See , .) Suppose that$\ltdefn{T}{U}{V}$and$\ltdefn{S}{V}{W}$are linear transformations. Prove the following relationship between ranges. \rng{\compose{S}{T}}\subseteq\rng{S} This question asks us to establish that one set ($\rng{\compose{S}{T}}$) is a subset of another ($\rng{S}$). Choose an element in the smaller set, say$\vect{w}\in\rng{\compose{S}{T}}$. Then we know that there is a vector$\vect{u}\in U$such that \vect{w}=\lt{\left(\compose{S}{T}\right)}{\vect{u}}=\lt{S}{\lt{T}{\vect{u}}} Now define$\vect{v}=\lt{T}{\vect{u}}$, so that then \lt{S}{\vect{v}}=\lt{S}{\lt{T}{\vect{u}}}=\vect{w} This statement is sufficient to show that$\vect{w}\in\rng{S}$, so$\vect{w}$is an element of the larger set, and$\rng{\compose{S}{T}}\subseteq\rng{S}$. 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 range of$T$equals the column space of$A$,$\rng{T}=\csp{A}$. This is an equality of sets, so we want to establish two subset conditions (). First, show$\csp{A}\subseteq\rng{T}$. Choose$\vect{y}\in\csp{A}$. Then by and there is a vector$\vect{x}\in\complex{n}$such that$A\vect{x}=\vect{y}$. Then \lt{T}{\vect{x}} This statement qualifies$\vect{y}$as a member of$\rng{T}$(), so$\csp{A}\subseteq\rng{T}$. Now, show$\rng{T}\subseteq\csp{A}$. Choose$\vect{y}\in\rng{T}$. Then by , there is a vector$\vect{x}$in$\complex{n}$such that$\lt{T}{\vect{x}}=\vect{y}$. Then A\vect{x} So by and ,$\vect{y}$qualifies for membership in$\csp{A}$and so$\rng{T}\subseteq\csp{A}\$.