This page titled 5.5: One-to-One and Onto Transformations is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. 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MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, 9.8: The Kernel and Image of a Linear Map, [ "article:topic", "kernel", "license:ccby", "showtoc:no", "authorname:kkuttler", "licenseversion:40", "source@https://lyryx.com/first-course-linear-algebra" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FLinear_Algebra%2FA_First_Course_in_Linear_Algebra_(Kuttler)%2F09%253A_Vector_Spaces%2F9.08%253A_The_Kernel_and_Image_of_a_Linear_Map, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Kernel and Image of a Linear Transformation, 9.9: The Matrix of a Linear Transformation, Definition \(\PageIndex{1}\): Kernel and Image, Proposition \(\PageIndex{1}\): Kernel and Image as Subspaces, Example \(\PageIndex{1}\): Kernel and Image of a Transformation, Example \(\PageIndex{2}\): Kernel and Image of a Linear Transformation, Theorem \(\PageIndex{1}\): Dimension of Kernel + Image, Definition \(\PageIndex{2}\): Rank of Linear Transformation, Theorem \(\PageIndex{2}\): Subspace of Same Dimension, Corollary \(\PageIndex{1}\): One to One and Onto Characterization, Example \(\PageIndex{3}\): One to One Transformation, source@https://lyryx.com/first-course-linear-algebra. linear algebra noun : a branch of mathematics that is concerned with mathematical structures closed under the operations of addition and scalar multiplication and that includes the theory of systems of linear equations, matrices, determinants, vector spaces, and linear transformations Example Sentences Then if \(\vec{v}\in V,\) there exist scalars \(c_{i}\) such that \[T(\vec{v})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})\nonumber \] Hence \(T\left( \vec{v}-\sum_{i=1}^{r}c_{i}\vec{v}_{i}\right) =0.\) It follows that \(\vec{v}-\sum_{i=1}^{r}c_{i}\vec{v}_{i}\) is in \(\ker \left( T\right)\). Hence by Definition \(\PageIndex{1}\), \(T\) is one to one. How will we recognize that a system is inconsistent? In looking at the second row, we see that if \(k=6\), then that row contains only zeros and \(x_2\) is a free variable; we have infinite solutions. \[\left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\nonumber \]. The first two rows give us the equations \[\begin{align}\begin{aligned} x_1+x_3&=0\\ x_2 &= 0.\\ \end{aligned}\end{align} \nonumber \] So far, so good. Every linear system of equations has exactly one solution, infinite solutions, or no solution. 1. I'm having trouble with some true/false questions in my linear algebra class and was hoping someone could help me out. Finally, consider the linear system \[\begin{align}\begin{aligned} x+y&=1\\x+y&=2.\end{aligned}\end{align} \nonumber \] We should immediately spot a problem with this system; if the sum of \(x\) and \(y\) is 1, how can it also be 2? First here is a definition of what is meant by the image and kernel of a linear transformation. GSL is a standalone C library, not as fast as any based on BLAS. B. Introduction to linear independence (video) | Khan Academy Suppose then that \[\sum_{i=1}^{r}c_{i}\vec{v}_{i}+\sum_{j=1}^{s}a_{j}\vec{u}_{j}=0\nonumber \] Apply \(T\) to both sides to obtain \[\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})+\sum_{j=1}^{s}a_{j}T(\vec{u} _{j})=\sum_{i=1}^{r}c_{i}T(\vec{v}_{i})= \vec{0}\nonumber \] Since \(\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{r})\right\}\) is linearly independent, it follows that each \(c_{i}=0.\) Hence \(\sum_{j=1}^{s}a_{j}\vec{u }_{j}=0\) and so, since the \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s}\right\}\) are linearly independent, it follows that each \(a_{j}=0\) also. In this case, we have an infinite solution set, just as if we only had the one equation \(x+y=1\). The coordinates \(x, y\) (or \(x_1\),\(x_2\)) uniquely determine a point in the plan. Linear Algebra Introduction | Linear Functions, Applications and Examples Therefore, recognize that \[\left [ \begin{array}{r} 2 \\ 3 \end{array} \right ] = \left [ \begin{array}{rr} 2 & 3 \end{array} \right ]^T\nonumber \]. Consider \(n=3\). It is also a good practice to acknowledge the fact that our free variables are, in fact, free. 3 Answers. A special case was done earlier in the context of matrices. Let us learn how to . For the specific case of \(\mathbb{R}^3\), there are three special vectors which we often use. In fact, \(\mathbb{F}_m[z]\) is a finite-dimensional subspace of \(\mathbb{F}[z]\) since, \[ \mathbb{F}_m[z] = \Span(1,z,z^2,\ldots,z^m). In the next section, well look at situations which create linear systems that need solving (i.e., word problems). Now we have seen three more examples with different solution types. Let \(T:V\rightarrow W\) be a linear transformation where \(V,W\) are vector spaces. The first variable will be the basic (or dependent) variable; all others will be free variables. Putting the augmented matrix in reduced row-echelon form: \[\left [\begin{array}{rrr|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right ] \rightarrow \cdots \rightarrow \left [\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right ].\nonumber \]. Linear Algebra - GeeksforGeeks You may have previously encountered the \(3\)-dimensional coordinate system, given by \[\mathbb{R}^{3}= \left\{ \left( x_{1}, x_{2}, x_{3}\right) :x_{j}\in \mathbb{R}\text{ for }j=1,2,3 \right\}\nonumber \]. To find two particular solutions, we pick values for our free variables. A linear system will be inconsistent only when it implies that 0 equals 1. To find the solution, put the corresponding matrix into reduced row echelon form. Find the solution to the linear system \[\begin{array}{ccccccc} x_1&+&x_2&+&x_3&=&1\\ x_1&+&2x_2&+&x_3&=&2\\ 2x_1&+&3x_2&+&2x_3&=&0\\ \end{array}. Again, there is no right way of doing this (in fact, there are \(\ldots\) infinite ways of doing this) so we give only an example here. By definition, \[\ker(S)=\{ax^2+bx+c\in \mathbb{P}_2 ~|~ a+b=0, a+c=0, b-c=0, b+c=0\}.\nonumber \]. Hence \(S \circ T\) is one to one. The above examples demonstrate a method to determine if a linear transformation \(T\) is one to one or onto. We will first find the kernel of \(T\). Let \(T:\mathbb{P}_1\to\mathbb{R}\) be the linear transformation defined by \[T(p(x))=p(1)\mbox{ for all } p(x)\in \mathbb{P}_1.\nonumber \] Find the kernel and image of \(T\). We write our solution as: \[\begin{align}\begin{aligned} x_1 &= 3-2x_4 \\ x_2 &=5-4x_4 \\ x_3 & \text{ is free} \\ x_4 & \text{ is free}. It consists of all numbers which can be obtained by evaluating all polynomials in \(\mathbb{P}_1\) at \(1\). What exactly is a free variable? This is a fact that we will not prove here, but it deserves to be stated. By picking two values for \(x_3\), we get two particular solutions. We further visualize similar situations with, say, 20 equations with two variables. Recall that to find the matrix \(A\) of \(T\), we apply \(T\) to each of the standard basis vectors \(\vec{e}_i\) of \(\mathbb{R}^4\). ( 6 votes) Show more. Returning to the original system, this says that if, \[\left [ \begin{array}{cc} 1 & 1 \\ 1 & 2\\ \end{array} \right ] \left [ \begin{array}{c} x\\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \], then \[\left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \]. If is a linear subspace of then (). Precisely, \[\begin{array}{c} \vec{u}=\vec{v} \; \mbox{if and only if}\\ u_{j}=v_{j} \; \mbox{for all}\; j=1,\cdots ,n \end{array}\nonumber \] Thus \(\left [ \begin{array}{rrr} 1 & 2 & 4 \end{array} \right ]^T \in \mathbb{R}^{3}\) and \(\left [ \begin{array}{rrr} 2 & 1 & 4 \end{array} \right ]^T \in \mathbb{R}^{3}\) but \(\left [ \begin{array}{rrr} 1 & 2 & 4 \end{array} \right ]^T \neq \left [ \begin{array}{rrr} 2 & 1 & 4 \end{array} \right ]^T\) because, even though the same numbers are involved, the order of the numbers is different. Thus by Lemma 9.7.1 \(T\) is one to one. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. We can think as above that the first two coordinates determine a point in a plane. Notice that there is only one leading 1 in that matrix, and that leading 1 corresponded to the \(x_1\) variable. A linear system is inconsistent if it does not have a solution. In other words, \(A\vec{x}=0\) implies that \(\vec{x}=0\). \end{aligned}\end{align} \nonumber \]. If a consistent linear system of equations has a free variable, it has infinite solutions. \end{aligned}\end{align} \nonumber \], (In the second particular solution we picked unusual values for \(x_3\) and \(x_4\) just to highlight the fact that we can.). Look back to the reduced matrix in Example \(\PageIndex{1}\). To prove that \(S \circ T\) is one to one, we need to show that if \(S(T (\vec{v})) = \vec{0}\) it follows that \(\vec{v} = \vec{0}\). \end{aligned}\end{align} \nonumber \], Find the solution to a linear system whose augmented matrix in reduced row echelon form is, \[\left[\begin{array}{ccccc}{1}&{0}&{0}&{2}&{3}\\{0}&{1}&{0}&{4}&{5}\end{array}\right] \nonumber \], Converting the two rows into equations we have \[\begin{align}\begin{aligned} x_1 + 2x_4 &= 3 \\ x_2 + 4x_4&=5.\\ \end{aligned}\end{align} \nonumber \], We see that \(x_1\) and \(x_2\) are our dependent variables, for they correspond to the leading 1s. Recall that if \(S\) and \(T\) are linear transformations, we can discuss their composite denoted \(S \circ T\). Observe that \[T \left [ \begin{array}{r} 1 \\ 0 \\ 0 \\ -1 \end{array} \right ] = \left [ \begin{array}{c} 1 + -1 \\ 0 + 0 \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \] There exists a nonzero vector \(\vec{x}\) in \(\mathbb{R}^4\) such that \(T(\vec{x}) = \vec{0}\). \(T\) is onto if and only if the rank of \(A\) is \(m\). You can think of the components of a vector as directions for obtaining the vector. If there are no free variables, then there is exactly one solution; if there are any free variables, there are infinite solutions. Let \(\vec{z}\in \mathbb{R}^m\). Linear Equations - Definition, Formula, Graph, Examples - Cuemath Consider the reduced row echelon form of an augmented matrix of a linear system of equations. Look also at the reduced matrix in Example \(\PageIndex{2}\). It is like you took an actual arrow, and moved it from one location to another keeping it pointing the same direction. That gives you linear independence. For convenience in this chapter we may write vectors as the transpose of row vectors, or \(1 \times n\) matrices. Draw a vector with its tail at the point \(\left( 0,0,0\right)\) and its tip at the point \(\left( a,b,c\right)\). Second, we will show that if \(T(\vec{x})=\vec{0}\) implies that \(\vec{x}=\vec{0}\), then it follows that \(T\) is one to one. We have infinite choices for the value of \(x_2\), so therefore we have infinite solutions. Next suppose \(T(\vec{v}_{1}),T(\vec{v}_{2})\) are two vectors in \(\mathrm{im}\left( T\right) .\) Then if \(a,b\) are scalars, \[aT(\vec{v}_{2})+bT(\vec{v}_{2})=T\left( a\vec{v}_{1}+b\vec{v}_{2}\right)\nonumber \] and this last vector is in \(\mathrm{im}\left( T\right)\) by definition. As we saw before, there is no restriction on what \(x_3\) must be; it is free to take on the value of any real number. You see that the ordered triples correspond to points in space just as the ordered pairs correspond to points in a plane and single real numbers correspond to points on a line. PDF LINEAR ALGEBRA. Part 0 Definitions. F R C Fn F A F linear, if for all A Now suppose we are given two points, \(P,Q\) whose coordinates are \(\left( p_{1},\cdots ,p_{n}\right)\) and \(\left( q_{1},\cdots ,q_{n}\right)\) respectively. The numbers \(x_{j}\) are called the components of \(\vec{x}\). Determinant, invertible matrices, and rank - Help with true/false Each vector, \(\overrightarrow{0P}\) and \(\overrightarrow{AB}\) has the same length (or magnitude) and direction. Any point within this coordinate plane is identified by where it is located along the \(x\) axis, and also where it is located along the \(y\) axis. Therefore, the reader is encouraged to employ some form of technology to find the reduced row echelon form. The standard form for linear equations in two variables is Ax+By=C. (So if a given linear system has exactly one solution, it will always have exactly one solution even if the constants are changed.) The concept will be fleshed out more in later chapters, but in short, the coefficients determine whether a matrix will have exactly one solution or not. The following examines what happens if both \(S\) and \(T\) are onto. If \(k\neq 6\), there is exactly one solution; if \(k=6\), there are infinite solutions. 3.Now multiply the resulting matrix in 2 with the vector x we want to transform. for a finite set of \(k\) polynomials \(p_1(z),\ldots,p_k(z)\). We answer this question by forming the augmented matrix and starting the process of putting it into reduced row echelon form. A vector ~v2Rnis an n-tuple of real numbers. These matrices are linearly independent which means this set forms a basis for \(\mathrm{im}(S)\). Now, consider the case of \(\mathbb{R}^n\) for \(n=1.\) Then from the definition we can identify \(\mathbb{R}\) with points in \(\mathbb{R}^{1}\) as follows: \[\mathbb{R} = \mathbb{R}^{1}= \left\{ \left( x_{1}\right) :x_{1}\in \mathbb{R} \right\}\nonumber \] Hence, \(\mathbb{R}\) is defined as the set of all real numbers and geometrically, we can describe this as all the points on a line. We can picture all of these solutions by thinking of the graph of the equation \(y=x\) on the traditional \(x,y\) coordinate plane. Try plugging these values back into the original equations to verify that these indeed are solutions. In this video I work through the following linear algebra problem: For which value of c do the following 2x2 matrices commute?A = [ -4c 2; -4 0 ], B = [ 1. Theorem 5.1.1: Matrix Transformations are Linear Transformations. They are given by \[\vec{i} = \left [ \begin{array}{rrr} 1 & 0 & 0 \end{array} \right ]^T\nonumber \] \[\vec{j} = \left [ \begin{array}{rrr} 0 & 1 & 0 \end{array} \right ]^T\nonumber \] \[\vec{k} = \left [ \begin{array}{rrr} 0 & 0 & 1 \end{array} \right ]^T\nonumber \] We can write any vector \(\vec{u} = \left [ \begin{array}{rrr} u_1 & u_2 & u_3 \end{array} \right ]^T\) as a linear combination of these vectors, written as \(\vec{u} = u_1 \vec{i} + u_2 \vec{j} + u_3 \vec{k}\). Legal. Similarly, since \(T\) is one to one, it follows that \(\vec{v} = \vec{0}\). Linear Algebra - Span of a Vector Space - Datacadamia