\[\begin{align}\begin{aligned} x_1 &= 3\\ x_2 &=5 \\ x_3 &= 1000 \\ x_4 &= 0. However its performance is still quite good (not extremely good though) and is used quite often; mostly because of its portability. 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 \(\ker \left( T\right) ,\left\{ \vec{u} _{1},\cdots ,\vec{u}_{s}\right\}\). \[\mathrm{ker}(T) = \left\{ \left [ \begin{array}{cc} s & s \\ t & -t \end{array} \right ] \right\} = \mathrm{span} \left\{ \left [ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right ] \right\}\nonumber \] It is clear that this set is linearly independent and therefore forms a basis for \(\mathrm{ker}(T)\). B. 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. This notation will be used throughout this chapter. 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. \[\begin{aligned} \mathrm{im}(T) & = \{ p(1) ~|~ p(x)\in \mathbb{P}_1 \} \\ & = \{ a+b ~|~ ax+b\in \mathbb{P}_1 \} \\ & = \{ a+b ~|~ a,b\in\mathbb{R} \}\\ & = \mathbb{R}\end{aligned}\] 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. Consider Example \(\PageIndex{2}\). 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 have just introduced a new term, the word free. A vector belongs to V when you can write it as a linear combination of the generators of V. Related to Graph - Spanning ? You see that the ordered triples correspond to points in space just as the ordered pairs correspond to points in a plane and single real numbers correspond to points on a line. Similarly, since \(T\) is one to one, it follows that \(\vec{v} = \vec{0}\). If \(k\neq 6\), then our next step would be to make that second row, second column entry a leading one. T/F: A particular solution for a linear system with infinite solutions can be found by arbitrarily picking values for the free variables. Rather, we will give the initial matrix, then immediately give the reduced row echelon form of the matrix. 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. In the or not case, the constants determine whether or not infinite solutions or no solution exists. The complex numbers are both a real and complex vector space; we have = and = So the dimension depends on the base field. \\ \end{aligned}\end{align} \nonumber \] Notice how the variables \(x_1\) and \(x_3\) correspond to the leading 1s of the given matrix. The answer to this question lies with properly understanding the reduced row echelon form of a matrix. Try plugging these values back into the original equations to verify that these indeed are solutions. GATE-CS-2014- (Set-2) Linear Algebra. Consider the system \[\begin{align}\begin{aligned} x+y&=2\\ x-y&=0. We will start by looking at onto. Thus every point \(P\) in \(\mathbb{R}^{n}\) determines its position vector \(\overrightarrow{0P}\). By removing vectors from the set to create an independent set gives a basis of \(\mathrm{im}(T)\). Suppose first that \(T\) is one to one and consider \(T(\vec{0})\). We can now use this theorem to determine this fact about \(T\). A basis B of a vector space V over a field F (such as the real numbers R or the complex numbers C) is a linearly independent subset of V that spans V.This means that a subset B of V is a basis if it satisfies the two following conditions: . So far, whenever we have solved a system of linear equations, we have always found exactly one solution. Create the corresponding augmented matrix, and then put the matrix into reduced row echelon form. Definition 5.1.3: finite-dimensional and Infinite-dimensional vector spaces. Suppose \[T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array} \right ] \left [ \begin{array}{r} x \\ y \end{array} \right ]\nonumber \] Then, \(T:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}\) is a linear transformation. 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. Linear Equation Definition: A linear equation is an algebraic equation where each term has an exponent of 1 and when this equation is graphed, it always results in a straight line. T/F: It is possible for a linear system to have exactly 5 solutions. Let \(T:V\rightarrow W\) be a linear transformation where \(V,W\) are vector spaces. 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. 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. The constants and coefficients of a matrix work together to determine whether a given system of linear equations has one, infinite, or no solution. This leads to a homogeneous system of four equations in three variables. If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive. \\ \end{aligned}\end{align} \nonumber \]. A vector ~v2Rnis an n-tuple of real numbers. Some of the examples of the kinds of vectors that can be rephrased in terms of the function of vectors. Below we see the augmented matrix and one elementary row operation that starts the Gaussian elimination process. 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. The corresponding augmented matrix and its reduced row echelon form are given below. We start by putting the corresponding matrix into reduced row echelon form. It is one of the most central topics of mathematics. By Proposition \(\PageIndex{1}\), \(A\) is one to one, and so \(T\) is also one to one. 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. 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}. There is no solution to such a problem; this linear system has no solution. Then T is called onto if whenever x2 Rm there exists x1 Rn such that T(x1) = x2. 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}. Two F-vector spaces are called isomorphic if there exists an invertible linear map between them. Find a basis for \(\mathrm{ker} (T)\) and \(\mathrm{im}(T)\). How can we tell what kind of solution (if one exists) a given system of linear equations has? Legal. [3] What kind of situation would lead to a column of all zeros? 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 \(\mathrm{rank}\left( T\right) +\dim \left( \ker \left( T\right) \right) =\dim \left( V\right) .\). Consider now the general definition for a vector in \(\mathbb{R}^n\). It is used to stress that idea that \(x_2\) can take on any value; we are free to choose any value for \(x_2\). \end{aligned}\end{align} \nonumber \], \[\begin{align}\begin{aligned} x_1 &= 3\\ x_2 &=1 \\ x_3 &= 1 . . Hence, if \(v_1,\ldots,v_m\in U\), then any linear combination \(a_1v_1+\cdots +a_m v_m\) must also be an element of \(U\). Use the kernel and image to determine if a linear transformation is one to one or onto. By Proposition \(\PageIndex{1}\) it is enough to show that \(A\vec{x}=0\) implies \(\vec{x}=0\). Now suppose \(n=3\). Consider \(n=3\). We also could have seen that \(T\) is one to one from our above solution for onto. We have now seen examples of consistent systems with exactly one solution and others with infinite solutions. Therefore, \(A \left( \mathbb{R}^n \right)\) is the collection of all linear combinations of these products. Accessibility StatementFor more information contact us [email protected]. Definition 9.8.1: Kernel and Image The only vector space with dimension is {}, the vector space consisting only of its zero element.. Properties. 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 \(T(\vec{x}) = \vec{0}\). We need to know how to do this; understanding the process has benefits. You can prove that \(T\) is in fact linear. We now wish to find a basis for \(\mathrm{im}(T)\). Then \(z^{m+1}\in\mathbb{F}[z]\), but \(z^{m+1}\notin \Span(p_1(z),\ldots,p_k(z))\). 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. We will now take a look at an example of a one to one and onto linear transformation. Remember, dependent vectors mean that one vector is a linear combination of the other(s). 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. If we have any row where all entries are 0 except for the entry in the last column, then the system implies 0=1. \end{aligned}\end{align} \nonumber \]. In the previous section, we learned how to find the reduced row echelon form of a matrix using Gaussian elimination by hand. 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. A particular solution is one solution out of the infinite set of possible solutions. 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 \(y=b-a\). A consistent linear system with more variables than equations will always have infinite solutions. The vectors \(v_1=(1,1,0)\) and \(v_2=(1,-1,0)\) span a subspace of \(\mathbb{R}^3\). First, we will consider what Rn looks like in more detail. Recall that if \(p(z)=a_mz^m + a_{m-1} z^{m-1} + \cdots + a_1z + a_0\in \mathbb{F}[z]\) is a polynomial with coefficients in \(\mathbb{F}\) such that \(a_m\neq 0\), then we say that \(p(z)\) has degree \(m\). This is the reason why it is named as a 'linear' equation. 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. Discuss it. 3.Now multiply the resulting matrix in 2 with the vector x we want to transform. Therefore, \(S \circ T\) is onto. (We cannot possibly pick values for \(x\) and \(y\) so that \(2x+2y\) equals both 0 and 4. GSL is a standalone C library, not as fast as any based on BLAS. It follows that if a variable is not independent, it must be dependent; the word basic comes from connections to other areas of mathematics that we wont explore here. This follows from the definition of matrix multiplication. There is no right way of doing this; we are free to choose whatever we wish. The linear span (or just span) of a set of vectors in a vector space is the intersection of all subspaces containing that set. \end{aligned}\end{align} \nonumber \], \[\begin{align}\begin{aligned} x_1 &= 15\\ x_2 &=1 \\ x_3 &= -8 \\ x_4 &= -5. Let T: Rn Rm be a linear transformation. Definition. 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 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. Lets look at an example to get an idea of how the values of constants and coefficients work together to determine the solution type. \[\left [ \begin{array}{rr|r} 1 & 1 & a \\ 1 & 2 & b \end{array} \right ] \rightarrow \left [ \begin{array}{rr|r} 1 & 0 & 2a-b \\ 0 & 1 & b-a \end{array} \right ] \label{ontomatrix}\] You can see from this point that the system has a solution. 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. Prove that if \(T\) and \(S\) are one to one, then \(S \circ T\) is one-to-one. After moving it around, it is regarded as the same vector. As before, let \(V\) denote a vector space over \(\mathbb{F}\). The coordinates \(x, y\) (or \(x_1\),\(x_2\)) uniquely determine a point in the plan. [1] That sure seems like a mouthful in and of itself. However the last row gives us the equation \[0x_1+0x_2+0x_3 = 1 \nonumber \] or, more concisely, \(0=1\). 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 \(T\left( \vec{x}_1\right) = \vec{x}_2.\). 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. 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. To express where it is in 3 dimensions, you would need a minimum, basis, of 3 independently linear vectors, span (V1,V2,V3). Since \(S\) is onto, there exists a vector \(\vec{y}\in \mathbb{R}^n\) such that \(S(\vec{y})=\vec{z}\). You can think of the components of a vector as directions for obtaining the vector. To find two particular solutions, we pick values for our free variables. To see this, assume the contrary, namely that, \[ \mathbb{F}[z] = \Span(p_1(z),\ldots,p_k(z))\]. For Property~3, note that a subspace \(U\) of a vector space \(V\) is closed under addition and scalar multiplication. To express a plane, you would use a basis (minimum number of vectors in a set required to fill the subspace) of two vectors. Notice that two vectors \(\vec{u} = \left [ u_{1} \cdots u_{n}\right ]^T\) and \(\vec{v}=\left [ v_{1} \cdots v_{n}\right ]^T\) are equal if and only if all corresponding components are equal. Then \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies \(\vec{x}=\vec{0}\). In those cases we leave the variable in the system just to remind ourselves that it is there. Actually, the correct formula for slope intercept form is . Rank is thus a measure of the "nondegenerateness" of the system of linear equations and linear transformation . If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. However, if \(k=6\), then our last row is \([0\ 0\ 1]\), meaning we have no solution. This page titled 9.8: The Kernel and Image of a Linear Map 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. Then \(W=V\) if and only if the dimension of \(W\) is also \(n\). For Property~2, note that \(0\in\Span(v_1,v_2,\ldots,v_m)\) and that \(\Span(v_1,v_2,\ldots,v_m)\) is closed under addition and scalar multiplication. Accessibility StatementFor more information contact us [email protected]. Each vector, \(\overrightarrow{0P}\) and \(\overrightarrow{AB}\) has the same length (or magnitude) and direction. In the two previous examples we have used the word free to describe certain variables. These definitions help us understand when a consistent system of linear equations will have infinite solutions. From this theorem follows the next corollary. 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 : \[T\left( \vec{x}_1 \right) \neq T \left(\vec{x}_2\right)\nonumber \]. Our main concern is what the rref is, not what exact steps were used to arrive there. We write our solution as: \[\begin{align}\begin{aligned} x_1 &= 3-2x_4 \\ x_2 &=5-4x_4 \\ x_3 & \text{ is free} \\ x_4 & \text{ is free}. \nonumber \]. I'm having trouble with some true/false questions in my linear algebra class and was hoping someone could help me out. Vectors have both size (magnitude) and direction. 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 \(c_{k}=0\). However, actually executing the process by hand for every problem is not usually beneficial. Group all constants on the right side of the inequality. 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 \(\mathrm{rank}\left( T\right) =m\). It is also widely applied in fields like physics, chemistry, economics, psychology, and engineering. Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. The following is a compilation of symbols from the different branches of algebra, which . Therefore, recognize that \[\left [ \begin{array}{r} 2 \\ 3 \end{array} \right ] = \left [ \begin{array}{rr} 2 & 3 \end{array} \right ]^T\nonumber \]. 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. If there are no free variables, then there is exactly one solution; if there are any free variables, there are infinite solutions. Lets summarize what we have learned up to this point. ), Now let us confirm this using the prescribed technique from above. via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. We further visualize similar situations with, say, 20 equations with two variables. Definition First, a definition: if there are infinite solutions, what do we call one of those infinite solutions? We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Then \(n=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im} \left( T\right) \right)\). This definition is illustrated in the following picture for the special case of \(\mathbb{R}^{3}\). One can probably see that free and independent are relatively synonymous. \[\begin{array}{ccccc} x_1 & +& x_2 & = & 1\\ 2x_1 & + & 2x_2 & = &2\end{array} . The first two examples in this section had infinite solutions, and the third had no solution. The following examines what happens if both \(S\) and \(T\) are onto. (We can think of it as depending on the value of 1.) Let \(T:V\rightarrow W\) be a linear map where the dimension of \(V\) is \(n\) and the dimension of \(W\) is \(m\). Using Theorem \(\PageIndex{1}\) we can show that \(T\) is onto but not one to one from the matrix of \(T\). That told us that \(x_1\) was not a free variable; since \(x_2\) did not correspond to a leading 1, it was a free variable. Therefore, they are equal. Draw a vector with its tail at the point \(\left( 0,0,0\right)\) and its tip at the point \(\left( a,b,c\right)\). Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. The second important characterization is called onto. as a standard basis, and therefore = More generally, =, and even more generally, = for any field. Hence by Definition \(\PageIndex{1}\), \(T\) is one to one. We can also determine the position vector from \(P\) to \(Q\) (also called the vector from \(P\) to \(Q\)) defined as follows. The two vectors would be linearly independent. 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. b) For all square matrices A, det(A^T)=det(A). \[\begin{array}{ccccc}x_1&+&2x_2&=&3\\ 3x_1&+&kx_2&=&9\end{array} \nonumber \]. Consider the system \(A\vec{x}=0\) given by: \[\left [ \begin{array}{cc} 1 & 1 \\ 1 & 2\\ \end{array} \right ] \left [ \begin{array}{c} x\\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber \], \[\begin{array}{c} x + y = 0 \\ x + 2y = 0 \end{array}\nonumber \], We need to show that the solution to this system is \(x = 0\) and \(y = 0\). First, we will prove that if \(T\) is one to one, then \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x}=\vec{0}\). 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 \(\mathbb{R}^n\), \[A\vec{x}=\sum_{k=1}^{n}x_{k}A_{k}\nonumber \]. Lemma 5.1.2 implies that \(\Span(v_1,v_2,\ldots,v_m)\) is the smallest subspace of \(V\) containing each of \(v_1,v_2,\ldots,v_m\). 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\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. These two equations tell us that the values of \(x_1\) and \(x_2\) depend on what \(x_3\) is. 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. The textbook definition of linear is: "progressing from one stage to another in a single series of steps; sequential." Which makes sense because if we are transforming these matrices linearly they would follow a sequence based on how they are scaled up or down.
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