Legal. Legal. More precisely, if we write the vectors in \(\mathbb{R}^3\) as 3-tuples of the form \((x,y,z)\), then \(\Span(v_1,v_2)\) is the \(xy\)-plane in \(\mathbb{R}^3\). 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}. A consistent linear system of equations will have exactly one solution if and only if there is a leading 1 for each variable in the system. \[\begin{aligned} \mathrm{im}(T) & = \{ p(1) ~|~ p(x)\in \mathbb{P}_1 \} \\ & = \{ a+b ~|~ ax+b\in \mathbb{P}_1 \} \\ & = \{ a+b ~|~ a,b\in\mathbb{R} \}\\ & = \mathbb{R}\end{aligned}\] 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. Then \(T\) is one to one if and only if \(\ker \left( T\right) =\left\{ \vec{0}\right\}\) and \(T\) is onto if and only if \(\mathrm{rank}\left( T\right) =m\). This leads us to a definition. 3 Answers. One can probably see that free and independent are relatively synonymous. \nonumber \] There are obviously infinite solutions to this system; as long as \(x=y\), we have a solution. The next example shows the same concept with regards to one-to-one transformations. Then \(n=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im} \left( T\right) \right)\). 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. \[\begin{array}{c} x+y=a \\ x+2y=b \end{array}\nonumber \] Set up the augmented matrix and row reduce. 3.Now multiply the resulting matrix in 2 with the vector x we want to transform. Once this value is chosen, the value of \(x_1\) is determined. It consists of all numbers which can be obtained by evaluating all polynomials in \(\mathbb{P}_1\) at \(1\). There are linear equations in one variable and linear equations in two variables. \[\begin{aligned} \mathrm{ker}(T) & = \{ p(x)\in \mathbb{P}_1 ~|~ p(1)=0\} \\ & = \{ ax+b ~|~ a,b\in\mathbb{R} \mbox{ and }a+b=0\} \\ & = \{ ax-a ~|~ a\in\mathbb{R} \}\end{aligned}\] Therefore a basis for \(\mathrm{ker}(T)\) is \[\left\{ x-1 \right\}\nonumber \] Notice that this is a subspace of \(\mathbb{P}_1\). If we were to consider a linear system with three equations and two unknowns, we could visualize the solution by graphing the corresponding three lines. The notation Rn refers to the collection of ordered lists of n real numbers, that is Rn = {(x1xn): xj R for j = 1, , n} In this chapter, we take a closer look at vectors in Rn. Notice that in this context, \(\vec{p} = \overrightarrow{0P}\). 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. Then. a variable that does not correspond to a leading 1 is a free, or independent, variable. Question 4227: what does m+c mean in a linear graph when y=mx+c. Lets continue this visual aspect of considering solutions to linear systems. A linear transformation \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) is called one to one (often written as \(1-1)\) if whenever \(\vec{x}_1 \neq \vec{x}_2\) it follows that : \[T\left( \vec{x}_1 \right) \neq T \left(\vec{x}_2\right)\nonumber \]. B. \\ \end{aligned}\end{align} \nonumber \]. Here we consider the case where the linear map is not necessarily an isomorphism. 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. Accessibility StatementFor more information contact us atinfo@libretexts.org. Now consider the linear system \[\begin{align}\begin{aligned} x+y&=1\\2x+2y&=2.\end{aligned}\end{align} \nonumber \] It is clear that while we have two equations, they are essentially the same equation; the second is just a multiple of the first. We can describe \(\mathrm{ker}(T)\) as follows. Here, the vector would have its tail sitting at the point determined by \(A= \left( d,e,f\right)\) and its point at \(B=\left( d+a,e+b,f+c\right) .\) It is the same vector because it will point in the same direction and have the same length. You may have previously encountered the \(3\)-dimensional coordinate system, given by \[\mathbb{R}^{3}= \left\{ \left( x_{1}, x_{2}, x_{3}\right) :x_{j}\in \mathbb{R}\text{ for }j=1,2,3 \right\}\nonumber \]. In those cases we leave the variable in the system just to remind ourselves that it is there. Therefore \(x_1\) and \(x_3\) are dependent variables; all other variables (in this case, \(x_2\) and \(x_4\)) are free variables. Before we start with a simple example, let us make a note about finding the reduced row echelon form of a matrix. Let \(V\) and \(W\) be vector spaces and let \(T:V\rightarrow W\) be a linear transformation. Therefore, there is only one vector, specifically \(\left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} 2a-b\\ b-a \end{array} \right ]\) such that \(T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\). \[\left [ \begin{array}{rr|r} 1 & 1 & a \\ 1 & 2 & b \end{array} \right ] \rightarrow \left [ \begin{array}{rr|r} 1 & 0 & 2a-b \\ 0 & 1 & b-a \end{array} \right ] \label{ontomatrix}\] You can see from this point that the system has a solution. We can picture that perhaps all three lines would meet at one point, giving exactly 1 solution; perhaps all three equations describe the same line, giving an infinite number of solutions; perhaps we have different lines, but they do not all meet at the same point, giving no solution. We trust that the reader can verify the accuracy of this form by both performing the necessary steps by hand or utilizing some technology to do it for them. Then in fact, both \(\mathrm{im}\left( T\right)\) and \(\ker \left( T\right)\) are subspaces of \(W\) and \(V\) respectively. Then \(T\) is called onto if whenever \(\vec{x}_2 \in \mathbb{R}^{m}\) there exists \(\vec{x}_1 \in \mathbb{R}^{n}\) such that \(T\left( \vec{x}_1\right) = \vec{x}_2.\). It is easier to read this when are variables are listed vertically, so we repeat these solutions: \[\begin{align}\begin{aligned} x_1 &= 4\\ x_2 &=0 \\ x_3 &= 7 \\ x_4 &= 0. We can essentially ignore the third row; it does not divulge any information about the solution.\(^{2}\) The first and second rows can be rewritten as the following equations: \[\begin{align}\begin{aligned} x_1 - x_2 + 2x_4 &=4 \\ x_3 - 3x_4 &= 7. T/F: It is possible for a linear system to have exactly 5 solutions. 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}. Recall that the point given by \(0=\left( 0, \cdots, 0 \right)\) is called the origin. This notation will be used throughout this chapter. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . This situation feels a little unusual,\(^{3}\) for \(x_3\) doesnt appear in any of the equations above, but cannot overlook it; it is still a free variable since there is not a leading 1 that corresponds to it. How can we tell what kind of solution (if one exists) a given system of linear equations has? And linear algebra, as a branch of math, is used in everything from machine learning to organic chemistry. \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.). We define them now. We often call a linear transformation which is one-to-one an injection. We can picture all of these solutions by thinking of the graph of the equation \(y=x\) on the traditional \(x,y\) coordinate plane. Use the kernel and image to determine if a linear transformation is one to one or onto. Use the kernel and image to determine if a linear transformation is one to one or onto. Accessibility StatementFor more information contact us atinfo@libretexts.org. Let \(A\) be an \(m\times n\) matrix where \(A_{1},\cdots , A_{n}\) denote the columns of \(A.\) Then, for a vector \(\vec{x}=\left [ \begin{array}{c} x_{1} \\ \vdots \\ x_{n} \end{array} \right ]\) in \(\mathbb{R}^n\), \[A\vec{x}=\sum_{k=1}^{n}x_{k}A_{k}\nonumber \]. In very large systems, it might be hard to determine whether or not a variable is actually used and one would not worry about it. \[\begin{align}\begin{aligned} x_1 &= 3\\ x_2 &=5 \\ x_3 &= 1000 \\ x_4 &= 0. 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. The rank of \(A\) is \(2\). \\ \end{aligned}\end{align} \nonumber \] Notice how the variables \(x_1\) and \(x_3\) correspond to the leading 1s of the given matrix. Therefore, when we graph the two equations, we are graphing the same line twice (see Figure \(\PageIndex{1}\)(b); the thicker line is used to represent drawing the line twice). Otherwise, if there is a leading 1 for each variable, then there is exactly one solution; otherwise (i.e., there are free variables) there are infinite solutions. INTRODUCTION Linear algebra is the math of vectors and matrices. A system of linear equations is inconsistent if the reduced row echelon form of its corresponding augmented matrix has a leading 1 in the last column. We write \[\overrightarrow{0P} = \left [ \begin{array}{c} p_{1} \\ \vdots \\ p_{n} \end{array} \right ]\nonumber \]. Recall that a linear transformation has the property that \(T(\vec{0}) = \vec{0}\). 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. 7. To have such a column, the original matrix needed to have a column of all zeros, meaning that while we acknowledged the existence of a certain variable, we never actually used it in any equation. However, the second equation of our system says that \(2x+2y= 4\). Rank is thus a measure of the "nondegenerateness" of the system of linear equations and linear transformation . Hence by Definition \(\PageIndex{1}\), \(T\) is one to one. Now consider the image. There is no solution to such a problem; this linear system has no solution. 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How can one tell what kind of solution a linear system of equations has? 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. Here we will determine that \(S\) is one to one, but not onto, using the method provided in Corollary \(\PageIndex{1}\). In other words, linear algebra is the study of linear functions and vectors. 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 \(\ker \left( T\right)\). Each vector, \(\overrightarrow{0P}\) and \(\overrightarrow{AB}\) has the same length (or magnitude) and direction. [1] That sure seems like a mouthful in and of itself. Legal. Vectors have both size (magnitude) and direction. For convenience in this chapter we may write vectors as the transpose of row vectors, or \(1 \times n\) matrices. This gives us a new vector with dimensions (lx1). Suppose first that \(T\) is one to one and consider \(T(\vec{0})\). (By the way, since infinite solutions exist, this system of equations is consistent.). It follows that \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots ,\vec{v} _{r}\right\}\) is a basis for \(V\) and so \[n=s+r=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right)\nonumber \], Let \(T:V\rightarrow W\) be a linear transformation and suppose \(V,W\) are finite dimensional vector spaces. Suppose the dimension of \(V\) is \(n\). Consider \(n=3\). 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. However, actually executing the process by hand for every problem is not usually beneficial. However, it boils down to look at the reduced form of the usual matrix.. Definition. (lxn) matrix and (nx1) vector multiplication. It follows that \(T\) is not one to one. We often write the solution as \(x=1-y\) to demonstrate that \(y\) can be any real number, and \(x\) is determined once we pick a value for \(y\). As an extension of the previous example, consider the similar augmented matrix where the constant 9 is replaced with a 10. CLAPACK is the library which uder the hood uses very high-performance BLAS library, as do other libraries, like ATLAS. 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 \(P\). \[\left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\nonumber \]. Notice that these vectors have the same span as the set above but are now linearly independent. (We can think of it as depending on the value of 1.) Now assume that if \(T(\vec{x})=\vec{0},\) then it follows that \(\vec{x}=\vec{0}.\) If \(T(\vec{v})=T(\vec{u}),\) then \[T(\vec{v})-T(\vec{u})=T\left( \vec{v}-\vec{u}\right) =\vec{0}\nonumber \] which shows that \(\vec{v}-\vec{u}=0\). The vectors \(e_1=(1,0,\ldots,0)\), \(e_2=(0,1,0,\ldots,0), \ldots, e_n=(0,\ldots,0,1)\) span \(\mathbb{F}^n\). Let us learn how to . For what values of \(k\) will the given system have exactly one solution, infinite solutions, or no solution? Since we have infinite choices for the value of \(x_3\), we have infinite solutions. So far, whenever we have solved a system of linear equations, we have always found exactly one solution. By setting \(x_2 = 1\) and \(x_4 = -5\), we have the solution \(x_1 = 15\), \(x_2 = 1\), \(x_3 = -8\), \(x_4 = -5\). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Similarly, a linear transformation which is onto is often called a surjection. \[\left[\begin{array}{cccc}{1}&{1}&{1}&{5}\\{1}&{-1}&{1}&{3}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{1}&{4}\\{0}&{1}&{0}&{1}\end{array}\right] \nonumber \], 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 \[\begin{align}\begin{aligned} x_1&= 4-x_3\\x_2&=1\\x_3 &\text{ is free}.\\ \end{aligned}\end{align} \nonumber \]. Figure \(\PageIndex{1}\): The three possibilities for two linear equations with two unknowns. 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. First here is a definition of what is meant by the image and kernel of a linear transformation. Hence, every element in \(\mathbb{R}^2\) is identified by two components, \(x\) and \(y\), in the usual manner. 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. We now wish to find a basis for \(\mathrm{im}(T)\). Consider a linear system of equations with infinite solutions. We further visualize similar situations with, say, 20 equations with two variables. Equivalently, if \(T\left( \vec{x}_1 \right) =T\left( \vec{x}_2\right) ,\) then \(\vec{x}_1 = \vec{x}_2\). 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? 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. Accessibility StatementFor more information contact us atinfo@libretexts.org. To express a plane, you would use a basis (minimum number of vectors in a set required to fill the subspace) of two vectors. It turns out that the matrix \(A\) of \(T\) can provide this information. Group all constants on the right side of the inequality. After moving it around, it is regarded as the same vector. Let \(T: \mathbb{R}^k \mapsto \mathbb{R}^n\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^m\) be linear transformations. Key Idea 1.4.1: Consistent Solution Types. Similarly, by Corollary \(\PageIndex{1}\), if \(S\) is onto it will have \(\mathrm{rank}(S) = \mathrm{dim}(\mathbb{M}_{22}) = 4\). First, we will consider what Rn looks like in more detail. Let \(T:V\rightarrow W\) be a linear transformation where \(V,W\) are vector spaces. As examples, \(x_1 = 2\), \(x_2 = 3\), \(x_3 = 0\) is one solution; \(x_1 = -2\), \(x_2 = 5\), \(x_3 = 2\) is another solution. This is the reason why it is named as a 'linear' equation. For example, 2x+3y=5 is a linear equation in standard form. [2] Then why include it? If you're seeing this message, it means we're having trouble loading external resources on our website. First, a definition: if there are infinite solutions, what do we call one of those infinite solutions? \end{aligned}\end{align} \nonumber \]. \[\mathrm{ker}(T) = \left\{ \left [ \begin{array}{cc} s & s \\ t & -t \end{array} \right ] \right\} = \mathrm{span} \left\{ \left [ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right ] \right\}\nonumber \] It is clear that this set is linearly independent and therefore forms a basis for \(\mathrm{ker}(T)\). When we learn about s and s, we will see that under certain circumstances this situation arises. We have just introduced a new term, the word free. \[\begin{align}\begin{aligned} x_1 &= 4\\ x_2 &=1 \\ x_3 &= 0 . In the next section, well look at situations which create linear systems that need solving (i.e., word problems). 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 \(T\). 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. Similarly, since \(T\) is one to one, it follows that \(\vec{v} = \vec{0}\). Recall that if \(S\) and \(T\) are linear transformations, we can discuss their composite denoted \(S \circ T\). 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}. In linear algebra, vectors are taken while forming linear functions. More succinctly, if we have a leading 1 in the last column of an augmented matrix, then the linear system has no solution. Thus every point \(P\) in \(\mathbb{R}^{n}\) determines its position vector \(\overrightarrow{0P}\). c) If a 3x3 matrix A is invertible, then rank(A)=3. By looking at the matrix given by \(\eqref{ontomatrix}\), you can see that there is a unique solution given by \(x=2a-b\) and \(y=b-a\).
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