what does c mean in linear algebra

Linear algebra is a branch of mathematics that deals with linear equations and their representations in the vector space using matrices. Now let us take the reduced matrix and write out the corresponding equations. We start with a very simple example. Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. 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). A comprehensive collection of 225+ symbols used in algebra, categorized by subject and type into tables along with each symbol's name, usage and example. If a consistent linear system of equations has a free variable, it has infinite solutions. Lets summarize what we have learned up to this point. It consists of all numbers which can be obtained by evaluating all polynomials in \(\mathbb{P}_1\) at \(1\). Let T: Rn Rm be a transformation defined by T(x) = Ax. Before we start with a simple example, let us make a note about finding the reduced row echelon form of a matrix. If a consistent linear system has more variables than leading 1s, then . 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. lgebra is a subfield of mathematics pertaining to the manipulation of symbols and their governing rules. INTRODUCTION Linear algebra is the math of vectors and matrices. Notice that in this context, \(\vec{p} = \overrightarrow{0P}\). First, lets just think about it. By convention, the degree of the zero polynomial \(p(z)=0\) is \(-\infty\). If the trace of the matrix is positive, all its eigenvalues are positive. Define \( \mathbb{F}_m[z] = \) set of all polynomials in \( \mathbb{F}[z] \) of degree at most m. Then \(\mathbb{F}_m[z]\subset \mathbb{F}[z]\) is a subspace since \(\mathbb{F}_m[z]\) contains the zero polynomial and is closed under addition and scalar multiplication. For convenience in this chapter we may write vectors as the transpose of row vectors, or \(1 \times n\) matrices. Let \(V,W\) be vector spaces and let \(T:V\rightarrow W\) be a linear transformation. Confirm that the linear system \[\begin{array}{ccccc} x&+&y&=&0 \\2x&+&2y&=&4 \end{array} \nonumber \] has no solution. \[\left[\begin{array}{ccc}{1}&{1}&{1}\\{2}&{2}&{2}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{ccc}{1}&{1}&{1}\\{0}&{0}&{0}\end{array}\right] \nonumber \], Now convert the reduced matrix back into equations. 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. This is as far as we need to go. 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. 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. By looking at the matrix given by \(\eqref{ontomatrix}\), you can see that there is a unique solution given by \(x=2a-b\) and \(y=b-a\). Legal. We dont particularly care about the solution, only that we would have exactly one as both \(x_1\) and \(x_2\) would correspond to a leading one and hence be dependent variables. A system of linear equations is consistent if it has a solution (perhaps more than one). 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. The linear span (or just span) of a set of vectors in a vector space is the intersection of all subspaces containing that set. Then \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies \(\vec{x}=\vec{0}\). Recall that the point given by \(0=\left( 0, \cdots, 0 \right)\) is called the origin. Our first example explores officially a quick example used in the introduction of this section. Thus by Lemma 9.7.1 \(T\) is one to one. Similarly, since \(T\) is one to one, it follows that \(\vec{v} = \vec{0}\). A consistent linear system with more variables than equations will always have infinite solutions. From this theorem follows the next corollary. 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. This corresponds to the maximal number of linearly independent columns of A.This, in turn, is identical to the dimension of the vector space spanned by its rows. In those cases we leave the variable in the system just to remind ourselves that it is there. In linear algebra, vectors are taken while forming linear functions. These are of course equivalent and we may move between both notations. Accessibility StatementFor more information contact us atinfo@libretexts.org. 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. \nonumber \] There are obviously infinite solutions to this system; as long as \(x=y\), we have a solution. So suppose \(\left [ \begin{array}{c} a \\ b \end{array} \right ] \in \mathbb{R}^{2}.\) Does there exist \(\left [ \begin{array}{c} x \\ y \end{array} \right ] \in \mathbb{R}^2\) such that \(T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ] ?\) If so, then since \(\left [ \begin{array}{c} a \\ b \end{array} \right ]\) is an arbitrary vector in \(\mathbb{R}^{2},\) it will follow that \(T\) is onto. If \(k\neq 6\), then our next step would be to make that second row, second column entry a leading one. These matrices are linearly independent which means this set forms a basis for \(\mathrm{im}(S)\). The idea behind the more general \(\mathbb{R}^n\) is that we can extend these ideas beyond \(n = 3.\) This discussion regarding points in \(\mathbb{R}^n\) leads into a study of vectors in \(\mathbb{R}^n\). \nonumber \]. It is also a good practice to acknowledge the fact that our free variables are, in fact, free. Find the solution to the linear system \[\begin{array}{ccccccc}x_1&+&x_2&+&x_3&=&5\\x_1&-&x_2&+&x_3&=&3\\ \end{array} \nonumber \] and give two particular solutions. By setting up the augmented matrix and row reducing, we end up with \[\left [ \begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right ]\nonumber \], This tells us that \(x = 0\) and \(y = 0\). Now, consider the case of Rn . The notation "2S" is read "element of S." For example, consider a vector that has three components: ~v= (v 1;v 2;v Since we have infinite choices for the value of \(x_3\), we have infinite solutions. Thus \(T\) is onto. Then \(\ker \left( T\right) \subseteq V\) and \(\mathrm{im}\left( T\right) \subseteq W\). Let \(T: \mathbb{R}^k \mapsto \mathbb{R}^n\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^m\) be linear transformations. Then \(n=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im} \left( T\right) \right)\). Let \(T:V\rightarrow W\) be a linear transformation where \(V,W\) are vector spaces. 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}. Notice that there is only one leading 1 in that matrix, and that leading 1 corresponded to the \(x_1\) variable. Consider a linear system of equations with infinite solutions. For example, 2x+3y=5 is a linear equation in standard form. Hence there are scalars \(a_{i}\) such that \[\vec{v}-\sum_{i=1}^{r}c_{i}\vec{v}_{i}=\sum_{j=1}^{s}a_{j}\vec{u}_{j}\nonumber \] Hence \(\vec{v}=\sum_{i=1}^{r}c_{i}\vec{v}_{i}+\sum_{j=1}^{s}a_{j}\vec{u} _{j}.\) Since \(\vec{v}\) is arbitrary, it follows that \[V=\mathrm{span}\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots , \vec{v}_{r}\right\}\nonumber \] If the vectors \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots , \vec{v}_{r}\right\}\) are linearly independent, then it will follow that this set is a basis. \[\left[\begin{array}{cccc}{1}&{1}&{1}&{5}\\{1}&{-1}&{1}&{3}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{1}&{4}\\{0}&{1}&{0}&{1}\end{array}\right] \nonumber \], 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 \[\begin{align}\begin{aligned} x_1&= 4-x_3\\x_2&=1\\x_3 &\text{ is free}.\\ \end{aligned}\end{align} \nonumber \]. Key Idea \(\PageIndex{1}\) applies only to consistent systems. However its performance is still quite good (not extremely good though) and is used quite often; mostly because of its portability. Recall that because \(T\) can be expressed as matrix multiplication, we know that \(T\) is a linear transformation. T/F: It is possible for a linear system to have exactly 5 solutions. \[\left[\begin{array}{ccc}{1}&{2}&{3}\\{3}&{k}&{9}\end{array}\right]\qquad\overrightarrow{-3R_{1}+R_{2}\to R_{2}}\qquad\left[\begin{array}{ccc}{1}&{2}&{3}\\{0}&{k-6}&{0}\end{array}\right] \nonumber \]. Definition 5.5.2: Onto. 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. In linear algebra, the rank of a matrix A is the dimension of the vector space generated (or spanned) by its columns. To find the solution, put the corresponding matrix into reduced row echelon form. 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. The standard form for linear equations in two variables is Ax+By=C. If \(T\) and \(S\) are onto, then \(S \circ T\) is onto. This page titled 4.1: Vectors in R 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. Since the unique solution is \(a=b=c=0\), \(\ker(S)=\{\vec{0}\}\), and thus \(S\) is one-to-one by Corollary \(\PageIndex{1}\). Therefore, recognize that \[\left [ \begin{array}{r} 2 \\ 3 \end{array} \right ] = \left [ \begin{array}{rr} 2 & 3 \end{array} \right ]^T\nonumber \]. Conversely, every such position vector \(\overrightarrow{0P}\) which has its tail at \(0\) and point at \(P\) determines the point \(P\) of \(\mathbb{R}^{n}\). In the previous section, we learned how to find the reduced row echelon form of a matrix using Gaussian elimination by hand. While it becomes harder to visualize when we add variables, no matter how many equations and variables we have, solutions to linear equations always come in one of three forms: exactly one solution, infinite solutions, or no solution. 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 \(T(\vec{x}) = \vec{0}\). However, if \(k=6\), then our last row is \([0\ 0\ 1]\), meaning we have no solution. Consider the system \[\begin{align}\begin{aligned} x+y&=2\\ x-y&=0. You may recall this example from earlier in Example 9.7.1. \[\overrightarrow{PQ} = \left [ \begin{array}{c} q_{1}-p_{1}\\ \vdots \\ q_{n}-p_{n} \end{array} \right ] = \overrightarrow{0Q} - \overrightarrow{0P}\nonumber \]. 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 \(\mathbb{R}^n\), \[A\vec{x}=\sum_{k=1}^{n}x_{k}A_{k}\nonumber \]. To find two particular solutions, we pick values for our free variables. (lxm) and (mxn) matrices give us (lxn) matrix. We have infinite choices for the value of \(x_2\), so therefore we have infinite solutions. This is the reason why it is named as a 'linear' equation. 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 \(\ker \left( T\right) ,\left\{ \vec{u} _{1},\cdots ,\vec{u}_{s}\right\}\). Step-by-step solution. 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Notice that these vectors have the same span as the set above but are now linearly independent. 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. Therefore, \(A \left( \mathbb{R}^n \right)\) is the collection of all linear combinations of these products. (We can think of it as depending on the value of 1.) Two F-vector spaces are called isomorphic if there exists an invertible linear map between them. Then the rank of \(T\) denoted as \(\mathrm{rank}\left( T\right)\) is defined as the dimension of \(\mathrm{im}\left( T\right) .\) The nullity of \(T\) is the dimension of \(\ker \left( T\right) .\) Thus the above theorem says that \(\mathrm{rank}\left( T\right) +\dim \left( \ker \left( T\right) \right) =\dim \left( V\right) .\). We will start by looking at onto. By setting \(x_2 = 0 = x_4\), we have the solution \(x_1 = 4\), \(x_2 = 0\), \(x_3 = 7\), \(x_4 = 0\). One can probably see that free and independent are relatively synonymous. How will we recognize that a system is inconsistent? We answer this question by forming the augmented matrix and starting the process of putting it into reduced row echelon form. Therefore, no solution exists; this system is inconsistent. The vectors \(v_1=(1,1,0)\) and \(v_2=(1,-1,0)\) span a subspace of \(\mathbb{R}^3\). So our final solution would look something like \[\begin{align}\begin{aligned} x_1 &= 4 +x_2 - 2x_4 \\ x_2 & \text{ is free} \\ x_3 &= 7+3x_4 \\ x_4 & \text{ is free}.\end{aligned}\end{align} \nonumber \]. How can we tell what kind of solution (if one exists) a given system of linear equations has? By removing vectors from the set to create an independent set gives a basis of \(\mathrm{im}(T)\). What exactly is a free variable? 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 : \[T\left( \vec{x}_1 \right) \neq T \left(\vec{x}_2\right)\nonumber \]. Most modern geometrical concepts are based on linear algebra. Let \(V\) be a vector space of dimension \(n\) and let \(W\) be a subspace. 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. When we learn about s and s, we will see that under certain circumstances this situation arises. If \(k\neq 6\), there is exactly one solution; if \(k=6\), there are infinite solutions. 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 \(\left( d+a,e+b,f+c\right)\). Vectors have both size (magnitude) and direction. 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 this is the only place the two lines intersect, this is the only solution. 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. B. First consider \(\ker \left( T\right) .\) It is necessary to show that if \(\vec{v}_{1},\vec{v}_{2}\) are vectors in \(\ker \left( T\right)\) and if \(a,b\) are scalars, then \(a\vec{v}_{1}+b\vec{v}_{2}\) is also in \(\ker \left( T\right) .\) But \[T\left( a\vec{v}_{1}+b\vec{v}_{2}\right) =aT(\vec{v}_{1})+bT(\vec{v}_{2})=a\vec{0}+b\vec{0}=\vec{0}\nonumber \]. Then. \end{aligned}\end{align} \nonumber \]. The above examples demonstrate a method to determine if a linear transformation \(T\) is one to one or onto. Some of the examples of the kinds of vectors that can be rephrased in terms of the function of vectors. A. For example, if we set \(x_2 = 0\), then \(x_1 = 1\); if we set \(x_2 = 5\), then \(x_1 = -4\). To show that \(T\) is onto, let \(\left [ \begin{array}{c} x \\ y \end{array} \right ]\) be an arbitrary vector in \(\mathbb{R}^2\). \[\left [ \begin{array}{rr|r} 1 & 1 & a \\ 1 & 2 & b \end{array} \right ] \rightarrow \left [ \begin{array}{rr|r} 1 & 0 & 2a-b \\ 0 & 1 & b-a \end{array} \right ] \label{ontomatrix}\] You can see from this point that the system has a solution. We can also determine the position vector from \(P\) to \(Q\) (also called the vector from \(P\) to \(Q\)) defined as follows. First, we will prove that if \(T\) is one to one, then \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x}=\vec{0}\). . 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 \(c_{k}=0\). A linear system is inconsistent if it does not have a solution. 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. 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. 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\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), A One to One and Onto Linear Transformation, 5.4: Special Linear Transformations in R, Lemma \(\PageIndex{1}\): Range of a Matrix Transformation, Definition \(\PageIndex{1}\): One to One, Proposition \(\PageIndex{1}\): One to One, Example \(\PageIndex{1}\): A One to One and Onto Linear Transformation, Example \(\PageIndex{2}\): An Onto Transformation, Theorem \(\PageIndex{1}\): Matrix of a One to One or Onto Transformation, Example \(\PageIndex{3}\): An Onto Transformation, Example \(\PageIndex{4}\): Composite of Onto Transformations, Example \(\PageIndex{5}\): Composite of One to One Transformations, source@https://lyryx.com/first-course-linear-algebra. Thus \[\vec{z} = S(\vec{y}) = S(T(\vec{x})) = (ST)(\vec{x}),\nonumber \] showing that for each \(\vec{z}\in \mathbb{R}^m\) there exists and \(\vec{x}\in \mathbb{R}^k\) such that \((ST)(\vec{x})=\vec{z}\). Therefore the dimension of \(\mathrm{im}(S)\), also called \(\mathrm{rank}(S)\), is equal to \(3\). The first two examples in this section had infinite solutions, and the third had no solution. Rather, we will give the initial matrix, then immediately give the reduced row echelon form of the matrix. As before, let \(V\) denote a vector space over \(\mathbb{F}\). Accessibility StatementFor more information contact us atinfo@libretexts.org. Our final analysis is then this. Hence \(\mathbb{F}^n\) is finite-dimensional. The two vectors would be linearly independent. Now we have seen three more examples with different solution types. (So if a given linear system has exactly one solution, it will always have exactly one solution even if the constants are changed.)

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