We generally write our solution with the dependent variables on the left and independent variables and constants on the right. This follows from the definition of matrix multiplication. 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. Rank is thus a measure of the "nondegenerateness" of the system of linear equations and linear transformation . The corresponding augmented matrix and its reduced row echelon form are given below. \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. Using this notation, we may use \(\vec{p}\) to denote the position vector of point \(P\). Find a basis for \(\mathrm{ker} (T)\) and \(\mathrm{im}(T)\). Here we dont differentiate between having one solution and infinite solutions, but rather just whether or not a solution exists. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Since \(S\) is one to one, it follows that \(T (\vec{v}) = \vec{0}\). You may have previously encountered the \(3\)-dimensional coordinate system, given by \[\mathbb{R}^{3}= \left\{ \left( x_{1}, x_{2}, x_{3}\right) :x_{j}\in \mathbb{R}\text{ for }j=1,2,3 \right\}\nonumber \]. If there are no free variables, then there is exactly one solution; if there are any free variables, there are infinite solutions. Therefore by the above theorem \(T\) is onto but not one to one. Definition 9.8.1: Kernel and Image The only vector space with dimension is {}, the vector space consisting only of its zero element.. Properties. Let nbe a positive integer and let R denote the set of real numbers, then Rnis the set of all n-tuples of real numbers. Let \(T: \mathbb{M}_{22} \mapsto \mathbb{R}^2\) be defined by \[T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ]\nonumber \] Then \(T\) is a linear transformation. By looking at the matrix given by \(\eqref{ontomatrix}\), you can see that there is a unique solution given by \(x=2a-b\) and \(y=b-a\). 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. You can prove that \(T\) is in fact linear. B. A First Course in Linear Algebra (Kuttler), { "9.01:_Algebraic_Considerations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.02:_Spanning_Sets" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.03:_Linear_Independence" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.04:_Subspaces_and_Basis" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.05:_Sums_and_Intersections" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.06:_Linear_Transformations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "9.07:_Isomorphisms" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", 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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. Putting the augmented matrix in reduced row-echelon form: \[\left [\begin{array}{rrr|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right ] \rightarrow \cdots \rightarrow \left [\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right ].\nonumber \]. Accessibility StatementFor more information contact us atinfo@libretexts.org. Hence by Definition \(\PageIndex{1}\), \(T\) is one to one. We have been studying the solutions to linear systems mostly in an academic setting; we have been solving systems for the sake of solving systems. A system of linear equations is consistent if it has a solution (perhaps more than one). 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. Suppose first that \(T\) is one to one and consider \(T(\vec{0})\). 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 \[\left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \]. When a consistent system has only one solution, each equation that comes from the reduced row echelon form of the corresponding augmented matrix will contain exactly one variable. The complex numbers are both a real and complex vector space; we have = and = So the dimension depends on the base field. Is \(T\) onto? The second important characterization is called onto. This section is devoted to studying two important characterizations of linear transformations, called one to one and onto. As an extension of the previous example, consider the similar augmented matrix where the constant 9 is replaced with a 10. We can picture all of these solutions by thinking of the graph of the equation \(y=x\) on the traditional \(x,y\) coordinate plane. As in the previous example, if \(k\neq6\), we can make the second row, second column entry a leading one and hence we have one solution. 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 \(\vec{v} = \vec{0}\). We start with a very simple example. Most modern geometrical concepts are based on linear algebra. More succinctly, if we have a leading 1 in the last column of an augmented matrix, then the linear system has no solution. Here we consider the case where the linear map is not necessarily an isomorphism. We need to prove two things here. Therefore, well do a little more practice. It is common to write \(T\mathbb{R}^{n}\), \(T\left( \mathbb{R}^{n}\right)\), or \(\mathrm{Im}\left( T\right)\) to denote these vectors. Given vectors \(v_1,v_2,\ldots,v_m\in V\), a vector \(v\in V\) is a linear combination of \((v_1,\ldots,v_m)\) if there exist scalars \(a_1,\ldots,a_m\in\mathbb{F}\) such that, \[ v = a_1 v_1 + a_2 v_2 + \cdots + a_m v_m.\], The linear span (or simply span) of \((v_1,\ldots,v_m)\) is defined as, \[ \Span(v_1,\ldots,v_m) := \{ a_1 v_1 + \cdots + a_m v_m \mid a_1,\ldots,a_m \in \mathbb{F} \}.\], Let \(V\) be a vector space and \(v_1,v_2,\ldots,v_m\in V\). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. By picking two values for \(x_3\), we get two particular solutions. This page titled 1.4: Existence and Uniqueness of Solutions is shared under a CC BY-NC 3.0 license and was authored, remixed, and/or curated by Gregory Hartman et al. Similarly, since \(T\) is one to one, it follows that \(\vec{v} = \vec{0}\). The rank of \(A\) is \(2\). \[\left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\nonumber \]. You may recall this example from earlier in Example 9.7.1. \\ \end{aligned}\end{align} \nonumber \]. First, we will consider what Rn looks like in more detail. Then the image of \(T\) denoted as \(\mathrm{im}\left( T\right)\) is defined to be the set \[\left\{ T(\vec{v}):\vec{v}\in V\right\}\nonumber \] In words, it consists of all vectors in \(W\) which equal \(T(\vec{v})\) for some \(\vec{v}\in V\). Group all constants on the right side of the inequality. \[\mathrm{ker}(T) = \left\{ \left [ \begin{array}{cc} s & s \\ t & -t \end{array} \right ] \right\} = \mathrm{span} \left\{ \left [ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right ] \right\}\nonumber \] It is clear that this set is linearly independent and therefore forms a basis for \(\mathrm{ker}(T)\). Example: Let V = Span { [0, 0, 1], [2, 0, 1], [4, 1, 2]}. For this reason we may write both \(P=\left( p_{1},\cdots ,p_{n}\right) \in \mathbb{R}^{n}\) and \(\overrightarrow{0P} = \left [ p_{1} \cdots p_{n} \right ]^T \in \mathbb{R}^{n}\). Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. 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. It is asking whether there is a solution to the equation \[\left [ \begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array} \right ] \left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\nonumber \] This is the same thing as asking for a solution to the following system of equations. CLAPACK is the library which uder the hood uses very high-performance BLAS library, as do other libraries, like ATLAS. Note that while the definition uses \(x_1\) and \(x_2\) to label the coordinates and you may be used to \(x\) and \(y\), these notations are equivalent. Therefore \(x_1\) and \(x_3\) are dependent variables; all other variables (in this case, \(x_2\) and \(x_4\)) are free variables. This is not always the case; we will find in this section that some systems do not have a solution, and others have more than one. Now, imagine taking a vector in \(\mathbb{R}^n\) and moving it around, always keeping it pointing in the same direction as shown in the following picture. Every linear system of equations has exactly one solution, infinite solutions, or no solution. If the consistent system has infinite solutions, then there will be at least one equation coming from the reduced row echelon form that contains more than one variable. Find the position vector of a point in \(\mathbb{R}^n\). However, actually executing the process by hand for every problem is not usually beneficial. 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The two vectors would be linearly independent. The easiest way to find a particular solution is to pick values for the free variables which then determines the values of the dependent variables. Here we consider the case where the linear map is not necessarily an isomorphism. The reduced row echelon form of the corresponding augmented matrix is, \[\left[\begin{array}{ccc}{1}&{1}&{0}\\{0}&{0}&{1}\end{array}\right] \nonumber \]. (lxn) matrix and (nx1) vector multiplication. We can also determine the position vector from \(P\) to \(Q\) (also called the vector from \(P\) to \(Q\)) defined as follows. We conclude this section with a brief discussion regarding notation. Let \(S:\mathbb{P}_2\to\mathbb{M}_{22}\) be a linear transformation defined by \[S(ax^2+bx+c) = \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \mbox{ for all } ax^2+bx+c\in \mathbb{P}_2.\nonumber \] Prove that \(S\) is one to one but not onto. Consider the reduced row echelon form of the augmented matrix of a system of linear equations.\(^{1}\) If there is a leading 1 in the last column, the system has no solution. \nonumber \] There are obviously infinite solutions to this system; as long as \(x=y\), we have a solution. \[\begin{aligned} \mathrm{im}(T) & = \{ p(1) ~|~ p(x)\in \mathbb{P}_1 \} \\ & = \{ a+b ~|~ ax+b\in \mathbb{P}_1 \} \\ & = \{ a+b ~|~ a,b\in\mathbb{R} \}\\ & = \mathbb{R}\end{aligned}\] Therefore a basis for \(\mathrm{im}(T)\) is \[\left\{ 1 \right\}\nonumber \] Notice that this is a subspace of \(\mathbb{R}\), and in fact is the space \(\mathbb{R}\) itself. If \(x+y=0\), then it stands to reason, by multiplying both sides of this equation by 2, that \(2x+2y = 0\). Accessibility StatementFor more information contact us atinfo@libretexts.org. This page titled 5.1: Linear Span is shared under a not declared license and was authored, remixed, and/or curated by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling. By Proposition \(\PageIndex{1}\), \(A\) is one to one, and so \(T\) is also one to one. Then \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies \(\vec{x}=\vec{0}\). 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). \[\left[\begin{array}{cccc}{1}&{1}&{1}&{1}\\{1}&{2}&{1}&{2}\\{2}&{3}&{2}&{0}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{1}&{0}\\{0}&{1}&{0}&{0}\\{0}&{0}&{0}&{1}\end{array}\right] \nonumber \]. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Suppose \(A = \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ]\) is such a matrix. Two F-vector spaces are called isomorphic if there exists an invertible linear map between them. Suppose that \(S(T (\vec{v})) = \vec{0}\). This is as far as we need to go. To show that \(T\) is onto, let \(\left [ \begin{array}{c} x \\ y \end{array} \right ]\) be an arbitrary vector in \(\mathbb{R}^2\). as a standard basis, and therefore = More generally, =, and even more generally, = for any field. Let \(P=\left( p_{1},\cdots ,p_{n}\right)\) be the coordinates of a point in \(\mathbb{R}^{n}.\) Then the vector \(\overrightarrow{0P}\) with its tail at \(0=\left( 0,\cdots ,0\right)\) and its tip at \(P\) is called the position vector of the point \(P\). These are of course equivalent and we may move between both notations. (lxm) and (mxn) matrices give us (lxn) matrix. If is a linear subspace of then (). It is also widely applied in fields like physics, chemistry, economics, psychology, and engineering. Which one of the following statements is TRUE about every. Let \(m=\max(\deg p_1(z),\ldots,\deg p_k(z))\). If \(\mathrm{ rank}\left( T\right) =m,\) then by Theorem \(\PageIndex{2}\), since \(\mathrm{im} \left( T\right)\) is a subspace of \(W,\) it follows that \(\mathrm{im}\left( T\right) =W\). 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. Learn linear algebra for freevectors, matrices, transformations, and more. In the previous section, we learned how to find the reduced row echelon form of a matrix using Gaussian elimination by hand. The standard form for linear equations in two variables is Ax+By=C. Draw a vector with its tail at the point \(\left( 0,0,0\right)\) and its tip at the point \(\left( a,b,c\right)\). Performing the same elementary row operation gives, \[\left[\begin{array}{ccc}{1}&{2}&{3}\\{3}&{k}&{10}\end{array}\right]\qquad\overrightarrow{-3R_{1}+R_{2}\to R_{2}}\qquad\left[\begin{array}{ccc}{1}&{2}&{3}\\{0}&{k-6}&{1}\end{array}\right] \nonumber \]. \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.). \[\left[\begin{array}{cccc}{0}&{1}&{-1}&{3}\\{1}&{0}&{2}&{2}\\{0}&{-3}&{3}&{-9}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{2}&{2}\\{0}&{1}&{-1}&{3}\\{0}&{0}&{0}&{0}\end{array}\right] \nonumber \], Now convert this reduced matrix back into equations.
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