what does c mean in linear algebra

CLAPACK is the library which uder the hood uses very high-performance BLAS library, as do other libraries, like ATLAS. Which one of the following statements is TRUE about every. This gives us a new vector with dimensions (lx1). What exactly is a free variable? There is no right way of doing this; we are free to choose whatever we wish. 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. And linear algebra, as a branch of math, is used in everything from machine learning to organic chemistry. 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. A linear system will be inconsistent only when it implies that 0 equals 1. Rank (linear algebra) - Wikipedia In practical terms, we could respond by removing the corresponding column from the matrix and just keep in mind that that variable is free. Legal. 9.8: The Kernel and Image of a Linear Map Theorem 5.1.1: Matrix Transformations are Linear Transformations. Two F-vector spaces are called isomorphic if there exists an invertible linear map between them. Lets try another example, one that uses more variables. We denote the degree of \(p(z)\) by \(\deg(p(z))\). What does it mean for matrices to commute? | Linear algebra worked Here we dont differentiate between having one solution and infinite solutions, but rather just whether or not a solution exists. Lets summarize what we have learned up to this point. You can prove that \(T\) is in fact linear. Intro to linear equation standard form | Algebra (video) | Khan Academy \end{aligned}\end{align} \nonumber \]. 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}. 5.1: Linear Span - Mathematics LibreTexts 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. This definition is illustrated in the following picture for the special case of \(\mathbb{R}^{3}\). via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. In the two previous examples we have used the word free to describe certain variables. 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 \(T(\vec{x}) = \vec{0}\). 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 \(\left( d+a,e+b,f+c\right)\). First, we will prove that if \(T\) is one to one, then \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x}=\vec{0}\). Using this notation, we may use \(\vec{p}\) to denote the position vector of point \(P\). Thus \(T\) is onto. Using Theorem \(\PageIndex{1}\) we can show that \(T\) is onto but not one to one from the matrix of \(T\). 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 in fact, both \(\mathrm{im}\left( T\right)\) and \(\ker \left( T\right)\) are subspaces of \(W\) and \(V\) respectively. There are linear equations in one variable and linear equations in two variables. This leads to a homogeneous system of four equations in three variables. Accessibility StatementFor more information contact us atinfo@libretexts.org. Similarly, t and t 2 are linearly independent functions on the whole of the real line, more so [ 0, 1]. The answer to this question lies with properly understanding the reduced row echelon form of a matrix. Therefore, the reader is encouraged to employ some form of technology to find the reduced row echelon form. Let \(\vec{z}\in \mathbb{R}^m\). 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 \(\mathrm{rank}\left( T\right) =m\). linear independence for every finite subset {, ,} of B, if + + = for some , , in F, then = = =; spanning property for every vector v in V . \[\left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\nonumber \]. Let \(V,W\) be vector spaces and let \(T:V\rightarrow W\) be a linear transformation. We further visualize similar situations with, say, 20 equations with two variables. To discover what the solution is to a linear system, we first put the matrix into reduced row echelon form and then interpret that form properly. Thus every point \(P\) in \(\mathbb{R}^{n}\) determines its position vector \(\overrightarrow{0P}\). 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}. 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 \(\mathbb{R}^n\). 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. Vectors have both size (magnitude) and direction. Look also at the reduced matrix in Example \(\PageIndex{2}\). We generally write our solution with the dependent variables on the left and independent variables and constants on the right. Therefore, we have shown that for any \(a, b\), there is a \(\left [ \begin{array}{c} x \\ y \end{array} \right ]\) such that \(T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\). By definition, \[\ker(S)=\{ax^2+bx+c\in \mathbb{P}_2 ~|~ a+b=0, a+c=0, b-c=0, b+c=0\}.\nonumber \]. To find particular solutions, choose values for our free variables. 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. 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. A particular solution is one solution out of the infinite set of possible solutions. Introduction to linear independence (video) | Khan Academy A vector space that is not finite-dimensional is called infinite-dimensional. From here on out, in our examples, when we need the reduced row echelon form of a matrix, we will not show the steps involved. You may recall this example from earlier in Example 9.7.1. Let \(V\) be a vector space of dimension \(n\) and let \(W\) be a subspace. \[\begin{array}{c} x+y=a \\ x+2y=b \end{array}\nonumber \] Set up the augmented matrix and row reduce. 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 \(\vec{v}\in V\). M is the slope and b is the Y-Intercept. In previous sections we have only encountered linear systems with unique solutions (exactly one solution). 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. 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). 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. \nonumber \]. 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. 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. 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. a variable that does not correspond to a leading 1 is a free, or independent, variable. \\ \end{aligned}\end{align} \nonumber \] Notice how the variables \(x_1\) and \(x_3\) correspond to the leading 1s of the given matrix. The reduced row echelon form of the corresponding augmented matrix is, \[\left[\begin{array}{ccc}{1}&{1}&{0}\\{0}&{0}&{1}\end{array}\right] \nonumber \]. However, the second equation of our system says that \(2x+2y= 4\). Hence by Definition \(\PageIndex{1}\), \(T\) is one to one. We start by putting the corresponding matrix into reduced row echelon form. \[T(\vec{0})=T\left( \vec{0}+\vec{0}\right) =T(\vec{0})+T(\vec{0})\nonumber \] and so, adding the additive inverse of \(T(\vec{0})\) to both sides, one sees that \(T(\vec{0})=\vec{0}\). Lets continue this visual aspect of considering solutions to linear systems. The result is the \(2 \times 4\) matrix A given by \[A = \left [ \begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{array} \right ]\nonumber \] Fortunately, this matrix is already in reduced row-echelon form. \[\overrightarrow{PQ} = \left [ \begin{array}{c} q_{1}-p_{1}\\ \vdots \\ q_{n}-p_{n} \end{array} \right ] = \overrightarrow{0Q} - \overrightarrow{0P}\nonumber \]. As a general rule, when we are learning a new technique, it is best to not use technology to aid us. Linear Algebra - Definition, Topics, Formulas, Examples - Cuemath However, it boils down to look at the reduced form of the usual matrix.. These two equations tell us that the values of \(x_1\) and \(x_2\) depend on what \(x_3\) is. First, we will consider what \(\mathbb{R}^n\) looks like in more detail. Question 8. Here, the two vectors are dependent because (3,6) is a multiple of the (1,2) (or vice versa): . This form is also very useful when solving systems of two linear equations. Taking the vector \(\left [ \begin{array}{c} x \\ y \\ 0 \\ 0 \end{array} \right ] \in \mathbb{R}^4\) we have \[T \left [ \begin{array}{c} x \\ y \\ 0 \\ 0 \end{array} \right ] = \left [ \begin{array}{c} x + 0 \\ y + 0 \end{array} \right ] = \left [ \begin{array}{c} x \\ y \end{array} \right ]\nonumber \] This shows that \(T\) is onto. 1. 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 \(c_{k}=0\). Consider \(n=3\). 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 \(\ker \left( T\right)\). Use the kernel and image to determine if a linear transformation is one to one or onto. Our first example explores officially a quick example used in the introduction of this section. Consider now the general definition for a vector in \(\mathbb{R}^n\). 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. How do we recognize which variables are free and which are not? It is one of the most central topics of mathematics. 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. 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. That is, \[\ker \left( T\right) =\left\{ \vec{v}\in V:T(\vec{v})=\vec{0}\right\}\nonumber \]. In other words, \(A\vec{x}=0\) implies that \(\vec{x}=0\). Obviously, this is not true; we have reached a contradiction. Thus, \(T\) is one to one if it never takes two different vectors to the same vector. 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. This is the reason why it is named as a 'linear' equation. So far, whenever we have solved a system of linear equations, we have always found exactly one solution. Then. Recall that because \(T\) can be expressed as matrix multiplication, we know that \(T\) is a linear transformation. We can describe \(\mathrm{ker}(T)\) as follows. We define the range or image of \(T\) as the set of vectors of \(\mathbb{R}^{m}\) which are of the form \(T \left(\vec{x}\right)\) (equivalently, \(A\vec{x}\)) for some \(\vec{x}\in \mathbb{R}^{n}\). After moving it around, it is regarded as the same vector. ( 6 votes) Show more. as a standard basis, and therefore = More generally, =, and even more generally, = for any field. We formally define this and a few other terms in this following definition. 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 \(T\left( \vec{x}_1\right) = \vec{x}_2.\). Linear Algebra Book: Linear Algebra (Schilling, Nachtergaele and Lankham) 5: Span and Bases 5.1: Linear Span Expand/collapse global location . We will first find the kernel of \(T\). Definition 5.5.2: Onto. In previous sections, we have written vectors as columns, or \(n \times 1\) matrices. \nonumber \]. 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 \(n\). We can think as above that the first two coordinates determine a point in a plane. First, a definition: if there are infinite solutions, what do we call one of those infinite solutions? The numbers \(x_{j}\) are called the components of \(\vec{x}\). It is also widely applied in fields like physics, chemistry, economics, psychology, and engineering. Putting the augmented matrix in reduced row-echelon form: \[\left [\begin{array}{rrr|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right ] \rightarrow \cdots \rightarrow \left [\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right ].\nonumber \]. Suppose \(\vec{x}_1\) and \(\vec{x}_2\) are vectors in \(\mathbb{R}^n\). By convention, the degree of the zero polynomial \(p(z)=0\) is \(-\infty\). The two vectors would be linearly independent. Is \(T\) onto? Linear Algebra - GeeksforGeeks We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Linear Algebra finds applications in virtually every area of mathematics, including Multivariate Calculus, Differential Equations, and Probability Theory. Therefore, recognize that \[\left [ \begin{array}{r} 2 \\ 3 \end{array} \right ] = \left [ \begin{array}{rr} 2 & 3 \end{array} \right ]^T\nonumber \]. We conclude this section with a brief discussion regarding notation. Group all constants on the right side of the inequality. Compositions of linear transformations 1 (video) | Khan Academy 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. Let \(V\) and \(W\) be vector spaces and let \(T:V\rightarrow W\) be a linear transformation. For Property~3, note that a subspace \(U\) of a vector space \(V\) is closed under addition and scalar multiplication. Lets find out through an example. \[\begin{array}{ccccc}x_1&+&2x_2&=&3\\ 3x_1&+&kx_2&=&9\end{array} \nonumber \]. 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 \(y\). If \(\Span(v_1,\ldots,v_m)=V\), then we say that \((v_1,\ldots,v_m)\) spans \(V\) and we call \(V\) finite-dimensional. 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 \(\vec{v}-\vec{u}=0\). Consider the following linear system: \[x-y=0. 1.4: Existence and Uniqueness of Solutions - Mathematics LibreTexts Equivalently, if \(T\left( \vec{x}_1 \right) =T\left( \vec{x}_2\right) ,\) then \(\vec{x}_1 = \vec{x}_2\). 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. PDF LINEAR ALGEBRA. Part 0 Definitions. F R C Fn F A F linear, if for all A Dimension (vector space) - Wikipedia Therefore, they are equal. Legal. Therefore the dimension of \(\mathrm{im}(S)\), also called \(\mathrm{rank}(S)\), is equal to \(3\). In this case, we have an infinite solution set, just as if we only had the one equation \(x+y=1\). 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 \(v_1,v_2,\ldots,v_m\in V\). 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. 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. To express a plane, you would use a basis (minimum number of vectors in a set required to fill the subspace) of two vectors. 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. Yes, if the system includes other degrees (exponents) of the variables, but if you are talking about a system of linear equations, the lines can either cross, run parallel or coincide because linear equations represent lines. Therefore \(x_1\) and \(x_3\) are dependent variables; all other variables (in this case, \(x_2\) and \(x_4\)) are free variables. Notice that these vectors have the same span as the set above but are now linearly independent. \end{aligned}\end{align} \nonumber \], \[\begin{align}\begin{aligned} x_1 &= 3\\ x_2 &=1 \\ x_3 &= 1 . Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. T/F: A variable that corresponds to a leading 1 is free.. This follows from the definition of matrix multiplication. Once again, we get a bit of an unusual solution; while \(x_2\) is a dependent variable, it does not depend on any free variable; instead, it is always 1. Now consider the image. 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. Therefore, \(S \circ T\) is onto. We have just introduced a new term, the word free. Describe the kernel and image of a linear transformation. Now let us take the reduced matrix and write out the corresponding equations. 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. Since \(S\) is one to one, it follows that \(T (\vec{v}) = \vec{0}\). 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. The above examples demonstrate a method to determine if a linear transformation \(T\) is one to one or onto. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. If you are graphing a system with a quadratic and a linear equation, these will cross at either two points, one point or zero points. It is also a good practice to acknowledge the fact that our free variables are, in fact, free. Let T: Rn Rm be a linear transformation. Then \(\ker \left( T\right) \subseteq V\) and \(\mathrm{im}\left( T\right) \subseteq W\). 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. To express where it is in 3 dimensions, you would need a minimum, basis, of 3 independently linear vectors, span (V1,V2,V3). 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. Then, from the definition, \[\mathbb{R}^{2}= \left\{ \left(x_{1}, x_{2}\right) :x_{j}\in \mathbb{R}\text{ for }j=1,2 \right\}\nonumber \] Consider the familiar coordinate plane, with an \(x\) axis and a \(y\) axis. 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. We have infinite choices for the value of \(x_2\), so therefore we have infinite solutions. Notice that in this context, \(\vec{p} = \overrightarrow{0P}\). Again, more practice is called for. Try plugging these values back into the original equations to verify that these indeed are solutions. Some of the examples of the kinds of vectors that can be rephrased in terms of the function of vectors. Figure \(\PageIndex{1}\): The three possibilities for two linear equations with two unknowns. If \(x+y=0\), then it stands to reason, by multiplying both sides of this equation by 2, that \(2x+2y = 0\). Rows of zeros sometimes appear unexpectedly in matrices after they have been put in reduced row echelon form.

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