In this case, we only have one equation, \[x_1+x_2=1 \nonumber \] or, equivalently, \[\begin{align}\begin{aligned} x_1 &=1-x_2\\ x_2&\text{ is free}. Hence \(S \circ T\) is one to one. Example: Let V = Span { [0, 0, 1], [2, 0, 1], [4, 1, 2]}. The above examples demonstrate a method to determine if a linear transformation \(T\) is one to one or onto. Suppose the dimension of \(V\) is \(n\). Since \(0\neq 4\), we have a contradiction and hence our system has no solution. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. We can write the image of \(T\) as \[\mathrm{im}(T) = \left\{ \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] \right\}\nonumber \] Notice that this can be written as \[\mathrm{span} \left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} -1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\nonumber \], However this is clearly not linearly independent. If a consistent linear system of equations has a free variable, it has infinite solutions. Note that this proposition says that if \(A=\left [ \begin{array}{ccc} A_{1} & \cdots & A_{n} \end{array} \right ]\) then \(A\) is one to one if and only if whenever \[0 = \sum_{k=1}^{n}c_{k}A_{k}\nonumber \] it follows that each scalar \(c_{k}=0\). 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. Thus by Lemma 9.7.1 \(T\) is one to one. We can picture all of these solutions by thinking of the graph of the equation \(y=x\) on the traditional \(x,y\) coordinate plane. Here we consider the case where the linear map is not necessarily an isomorphism. Find the solution to the linear system \[\begin{array}{ccccccc} & &x_2&-&x_3&=&3\\ x_1& & &+&2x_3&=&2\\ &&-3x_2&+&3x_3&=&-9\\ \end{array}. Since the unique solution is \(a=b=c=0\), \(\ker(S)=\{\vec{0}\}\), and thus \(S\) is one-to-one by Corollary \(\PageIndex{1}\). \[\left[\begin{array}{cccc}{1}&{1}&{1}&{5}\\{1}&{-1}&{1}&{3}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{1}&{4}\\{0}&{1}&{0}&{1}\end{array}\right] \nonumber \], Converting these two rows into equations, we have \[\begin{align}\begin{aligned} x_1+x_3&=4\\x_2&=1\\ \end{aligned}\end{align} \nonumber \] giving us the solution \[\begin{align}\begin{aligned} x_1&= 4-x_3\\x_2&=1\\x_3 &\text{ is free}.\\ \end{aligned}\end{align} \nonumber \]. When we learn about s and s, we will see that under certain circumstances this situation arises. Then. \end{aligned}\end{align} \nonumber \]. Definition 5.1.3: finite-dimensional and Infinite-dimensional vector spaces. We need to know how to do this; understanding the process has benefits. A First Course in Linear Algebra (Kuttler), { "4.01:_Vectors_in_R" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.02:_Vector_Algebra" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.03:_Geometric_Meaning_of_Vector_Addition" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.04:_Length_of_a_Vector" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.05:_Geometric_Meaning_of_Scalar_Multiplication" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.06:_Parametric_Lines" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.07:_The_Dot_Product" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.08:_Planes_in_R" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.09:_The_Cross_Product" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.10:_Spanning_Linear_Independence_and_Basis_in_R" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.11:_Orthogonality" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.12:_Applications" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "4.E:_Exercises" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Systems_of_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Matrices" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Determinants" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_R" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Linear_Transformations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "06:_Complex_Numbers" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "07:_Spectral_Theory" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "08:_Some_Curvilinear_Coordinate_Systems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "09:_Vector_Spaces" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "10:_Some_Prerequisite_Topics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, [ "article:topic", "position vector", "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)%2F04%253A_R%2F4.01%253A_Vectors_in_R, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Definition \(\PageIndex{1}\) THe Position Vector, Definition \(\PageIndex{2}\) Vectors in \(\mathbb{R}^n\), source@https://lyryx.com/first-course-linear-algebra. There is no right way of doing this; we are free to choose whatever we wish. It turns out that the matrix \(A\) of \(T\) can provide this information. Here we will determine that \(S\) is one to one, but not onto, using the method provided in Corollary \(\PageIndex{1}\). The third component determines the height above or below the plane, depending on whether this number is positive or negative, and all together this determines a point in space. 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. Now we want to find a way to describe all matrices \(A\) such that \(T(A) = \vec{0}\), that is the matrices in \(\mathrm{ker}(T)\). Above we showed that \(T\) was onto but not one to one. Consider the system \[\begin{align}\begin{aligned} x+y&=2\\ x-y&=0. Let \(T: \mathbb{R}^k \mapsto \mathbb{R}^n\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^m\) be linear transformations. Furthermore, since \(T\) is onto, there exists a vector \(\vec{x}\in \mathbb{R}^k\) such that \(T(\vec{x})=\vec{y}\). Legal. Recall that for an \(m\times n\) matrix \(% A,\) it was the case that the dimension of the kernel of \(A\) added to the rank of \(A\) equals \(n\). Let \(T:\mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. Putting the augmented matrix in reduced row-echelon form: \[\left [\begin{array}{rrr|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right ] \rightarrow \cdots \rightarrow \left [\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right ].\nonumber \]. Theorem 5.1.1: Matrix Transformations are Linear Transformations. However, actually executing the process by hand for every problem is not usually beneficial. So the span of the plane would be span (V1,V2). Look back to the reduced matrix in Example \(\PageIndex{1}\). We can verify that this system has no solution in two ways. Consider as an example the following diagram. The rank of \(A\) is \(2\). 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}. From Proposition \(\PageIndex{1}\), \(\mathrm{im}\left( T\right)\) is a subspace of \(W.\) By Theorem 9.4.8, there exists a basis for \(\mathrm{im}\left( T\right) ,\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{r})\right\} .\) Similarly, there is a basis for \(\ker \left( T\right) ,\left\{ \vec{u} _{1},\cdots ,\vec{u}_{s}\right\}\). Then \[T \left [ \begin{array}{cc} a & b \\ c & d \end{array} \right ] = \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \] The values of \(a, b, c, d\) that make this true are given by solutions to the system \[\begin{aligned} a - b &= 0 \\ c + d &= 0 \end{aligned}\] The solution is \(a = s, b = s, c = t, d = -t\) where \(s, t\) are scalars. \[\begin{align}\begin{aligned} x_1 &= 4\\ x_2 &=1 \\ x_3 &= 0 . You may recall this example from earlier in Example 9.7.1. Accessibility StatementFor more information contact us atinfo@libretexts.org. Here, the two vectors are dependent because (3,6) is a multiple of the (1,2) (or vice versa): . That gives you linear independence. Thus every point \(P\) in \(\mathbb{R}^{n}\) determines its position vector \(\overrightarrow{0P}\). \[\left [ \begin{array}{rr|r} 1 & 1 & a \\ 1 & 2 & b \end{array} \right ] \rightarrow \left [ \begin{array}{rr|r} 1 & 0 & 2a-b \\ 0 & 1 & b-a \end{array} \right ] \label{ontomatrix}\] You can see from this point that the system has a solution. Performing the same elementary row operation gives, \[\left[\begin{array}{ccc}{1}&{2}&{3}\\{3}&{k}&{10}\end{array}\right]\qquad\overrightarrow{-3R_{1}+R_{2}\to R_{2}}\qquad\left[\begin{array}{ccc}{1}&{2}&{3}\\{0}&{k-6}&{1}\end{array}\right] \nonumber \]. If a consistent linear system has more variables than leading 1s, then . b) For all square matrices A, det(A^T)=det(A). In this example, they intersect at the point \((1,1)\) that is, when \(x=1\) and \(y=1\), both equations are satisfied and we have a solution to our linear system. Similarly, a linear transformation which is onto is often called a surjection. This meant that \(x_1\) and \(x_2\) were not free variables; since there was not a leading 1 that corresponded to \(x_3\), it was a free variable. How can we tell if a system is inconsistent? Draw a vector with its tail at the point \(\left( 0,0,0\right)\) and its tip at the point \(\left( a,b,c\right)\). The first two examples in this section had infinite solutions, and the third had no solution. Consider the reduced row echelon form of an augmented matrix of a linear system of equations. Question 8. We will now take a look at an example of a one to one and onto linear transformation. Definition. 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. It turns out that every linear transformation can be expressed as a matrix transformation, and thus linear transformations are exactly the same as matrix transformations. So far, whenever we have solved a system of linear equations, we have always found exactly one solution. Definition 5.5.2: Onto. How will we recognize that a system is inconsistent? If \(T\) is onto, then \(\mathrm{im}\left( T\right) =W\) and so \(\mathrm{rank}\left( T\right)\) which is defined as the dimension of \(\mathrm{im}\left( T\right)\) is \(m\). Hence \(\mathbb{F}^n\) is finite-dimensional. 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. For Property~3, note that a subspace \(U\) of a vector space \(V\) is closed under addition and scalar multiplication. The constants and coefficients of a matrix work together to determine whether a given system of linear equations has one, infinite, or no solution. A vector ~v2Rnis an n-tuple of real numbers. You can verify that \(T\) represents a linear transformation. Is it one to one? Let \(T: \mathbb{R}^4 \mapsto \mathbb{R}^2\) be a linear transformation defined by \[T \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] = \left [ \begin{array}{c} a + d \\ b + c \end{array} \right ] \mbox{ for all } \left [ \begin{array}{c} a \\ b \\ c \\ d \end{array} \right ] \in \mathbb{R}^4\nonumber \] Prove that \(T\) is onto but not one to one. Our first example explores officially a quick example used in the introduction of this section. The set of all linear combinations of some vectors v1,,vn is called the span of these vectors and contains always the origin. Consider now the general definition for a vector in \(\mathbb{R}^n\). Consider a linear system of equations with infinite solutions. 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. Recall that to find the matrix \(A\) of \(T\), we apply \(T\) to each of the standard basis vectors \(\vec{e}_i\) of \(\mathbb{R}^4\). It follows that \(T\) is not one to one. This vector it is obtained by starting at \(\left( 0,0,0\right)\), moving parallel to the \(x\) axis to \(\left( a,0,0\right)\) and then from here, moving parallel to the \(y\) axis to \(\left( a,b,0\right)\) and finally parallel to the \(z\) axis to \(\left( a,b,c\right).\) Observe that the same vector would result if you began at the point \(\left( d,e,f \right)\), moved parallel to the \(x\) axis to \(\left( d+a,e,f\right) ,\) then parallel to the \(y\) axis to \(\left( d+a,e+b,f\right) ,\) and finally parallel to the \(z\) axis to \(\left( d+a,e+b,f+c\right)\). 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. 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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