Let \(T:\mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. Similarly, a linear transformation which is onto is often called a surjection. Recall that the point given by 0 = (0, , 0) is called the origin. 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 \(T\). So the span of the plane would be span (V1,V2). Rows of zeros sometimes appear unexpectedly in matrices after they have been put in reduced row echelon form. Recall that a linear transformation has the property that \(T(\vec{0}) = \vec{0}\). More succinctly, if we have a leading 1 in the last column of an augmented matrix, then the linear system has no solution. Thus \(\ker \left( T\right)\) is a subspace of \(V\). 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. As before, let \(V\) denote a vector space over \(\mathbb{F}\). \[\begin{array}{ccccc} x_1 & +& x_2 & = & 1\\ 2x_1 & + & 2x_2 & = &2\end{array} . In those cases we leave the variable in the system just to remind ourselves that it is there. It is like you took an actual arrow, and moved it from one location to another keeping it pointing the same direction. It is also a good practice to acknowledge the fact that our free variables are, in fact, free. \[\begin{array}{ccccc}x_1&+&2x_2&=&3\\ 3x_1&+&kx_2&=&9\end{array} \nonumber \]. \\ \end{aligned}\end{align} \nonumber \] Notice how the variables \(x_1\) and \(x_3\) correspond to the leading 1s of the given matrix. First, we will consider what \(\mathbb{R}^n\) looks like in more detail. We also could have seen that \(T\) is one to one from our above solution for onto. If a consistent linear system of equations has a free variable, it has infinite solutions. (We cannot possibly pick values for \(x\) and \(y\) so that \(2x+2y\) equals both 0 and 4. We can describe \(\mathrm{ker}(T)\) as follows. 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. 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 \(P\). Here we consider the case where the linear map is not necessarily an isomorphism. Let \(m=\max(\deg p_1(z),\ldots,\deg p_k(z))\). We generally write our solution with the dependent variables on the left and independent variables and constants on the right. We start with a very simple example. Therefore, well do a little more practice. For example, if we set \(x_2 = 0\), then \(x_1 = 1\); if we set \(x_2 = 5\), then \(x_1 = -4\). If \(k\neq 6\), then our next step would be to make that second row, second column entry a leading one. \end{aligned}\end{align} \nonumber \]. In looking at the second row, we see that if \(k=6\), then that row contains only zeros and \(x_2\) is a free variable; we have infinite solutions. 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. Give an example (different from those given in the text) of a 2 equation, 2 unknown linear system that is not consistent. Recall that the point given by \(0=\left( 0, \cdots, 0 \right)\) is called the origin. Again, more practice is called for. Notice that these vectors have the same span as the set above but are now linearly independent. Figure \(\PageIndex{1}\): The three possibilities for two linear equations with two unknowns. Accessibility StatementFor more information contact us atinfo@libretexts.org. 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. One can probably see that free and independent are relatively synonymous. Furthermore, since \(T\) is onto, there exists a vector \(\vec{x}\in \mathbb{R}^k\) such that \(T(\vec{x})=\vec{y}\). 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. Give the solution to a linear system whose augmented matrix in reduced row echelon form is, \[\left[\begin{array}{ccccc}{1}&{-1}&{0}&{2}&{4}\\{0}&{0}&{1}&{-3}&{7}\\{0}&{0}&{0}&{0}&{0}\end{array}\right] \nonumber \]. Some of the examples of the kinds of vectors that can be rephrased in terms of the function of vectors. 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}. Precisely, \[\begin{array}{c} \vec{u}=\vec{v} \; \mbox{if and only if}\\ u_{j}=v_{j} \; \mbox{for all}\; j=1,\cdots ,n \end{array}\nonumber \] Thus \(\left [ \begin{array}{rrr} 1 & 2 & 4 \end{array} \right ]^T \in \mathbb{R}^{3}\) and \(\left [ \begin{array}{rrr} 2 & 1 & 4 \end{array} \right ]^T \in \mathbb{R}^{3}\) but \(\left [ \begin{array}{rrr} 1 & 2 & 4 \end{array} \right ]^T \neq \left [ \begin{array}{rrr} 2 & 1 & 4 \end{array} \right ]^T\) because, even though the same numbers are involved, the order of the numbers is different. \[\begin{align}\begin{aligned} x_1 &= 4\\ x_2 &=1 \\ x_3 &= 0 . The two vectors would be linearly independent. 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. These notations may be used interchangeably. - Sarvesh Ravichandran Iyer If there are no free variables, then there is exactly one solution; if there are any free variables, there are infinite solutions. 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 \(T\left( \vec{x}_1\right) = \vec{x}_2.\). To see this, assume the contrary, namely that, \[ \mathbb{F}[z] = \Span(p_1(z),\ldots,p_k(z))\]. When we learn about s and s, we will see that under certain circumstances this situation arises. You may recall this example from earlier in Example 9.7.1. Below we see the augmented matrix and one elementary row operation that starts the Gaussian elimination process. INTRODUCTION Linear algebra is the math of vectors and matrices. Computer programs such as Mathematica, MATLAB, Maple, and Derive can be used; many handheld calculators (such as Texas Instruments calculators) will perform these calculations very quickly. A consistent linear system with more variables than equations will always have infinite solutions. \nonumber \]. Consider the reduced row echelon form of an augmented matrix of a linear system of equations. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Now we want to find a way to describe all matrices \(A\) such that \(T(A) = \vec{0}\), that is the matrices in \(\mathrm{ker}(T)\). 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. \], At the same time, though, note that \(\mathbb{F}[z]\) itself is infinite-dimensional. Lets continue this visual aspect of considering solutions to linear systems. Discuss it. Actually, the correct formula for slope intercept form is . We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. By looking at the matrix given by \(\eqref{ontomatrix}\), you can see that there is a unique solution given by \(x=2a-b\) and \(y=b-a\). How can one tell what kind of solution a linear system of equations has? 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 \[T\left( a\vec{v}_{1}+b\vec{v}_{2}\right) =aT(\vec{v}_{1})+bT(\vec{v}_{2})=a\vec{0}+b\vec{0}=\vec{0}\nonumber \]. Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. Since we have infinite choices for the value of \(x_3\), we have infinite solutions. They are given by \[\vec{i} = \left [ \begin{array}{rrr} 1 & 0 & 0 \end{array} \right ]^T\nonumber \] \[\vec{j} = \left [ \begin{array}{rrr} 0 & 1 & 0 \end{array} \right ]^T\nonumber \] \[\vec{k} = \left [ \begin{array}{rrr} 0 & 0 & 1 \end{array} \right ]^T\nonumber \] We can write any vector \(\vec{u} = \left [ \begin{array}{rrr} u_1 & u_2 & u_3 \end{array} \right ]^T\) as a linear combination of these vectors, written as \(\vec{u} = u_1 \vec{i} + u_2 \vec{j} + u_3 \vec{k}\). Systems with exactly one solution or no solution are the easiest to deal with; systems with infinite solutions are a bit harder to deal with. Linear Algebra Book: Linear Algebra (Schilling, Nachtergaele and Lankham) 5: Span and Bases 5.1: Linear Span Expand/collapse global location . Now suppose \(n=3\). Linear Algebra finds applications in virtually every area of mathematics, including Multivariate Calculus, Differential Equations, and Probability Theory. 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. It consists of all polynomials in \(\mathbb{P}_1\) that have \(1\) for a root. CLAPACK is the library which uder the hood uses very high-performance BLAS library, as do other libraries, like ATLAS. ( 6 votes) Show more. Let \(T:V\rightarrow W\) be a linear map where the dimension of \(V\) is \(n\) and the dimension of \(W\) is \(m\). The linear span of a set of vectors is therefore a vector space. How can we tell if a system is inconsistent? a variable that does not correspond to a leading 1 is a free, or independent, variable. We conclude this section with a brief discussion regarding notation. In other words, linear algebra is the study of linear functions and vectors. Most modern geometrical concepts are based on linear algebra. 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 write \[\overrightarrow{0P} = \left [ \begin{array}{c} p_{1} \\ \vdots \\ p_{n} \end{array} \right ]\nonumber \]. The following proposition is an important result. These two equations tell us that the values of \(x_1\) and \(x_2\) depend on what \(x_3\) is. Hence \(\mathbb{F}^n\) is finite-dimensional. Confirm that the linear system \[\begin{array}{ccccc} x&+&y&=&0 \\2x&+&2y&=&4 \end{array} \nonumber \] has no solution. However the last row gives us the equation \[0x_1+0x_2+0x_3 = 1 \nonumber \] or, more concisely, \(0=1\). 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. For convenience in this chapter we may write vectors as the transpose of row vectors, or \(1 \times n\) matrices. Try plugging these values back into the original equations to verify that these indeed are solutions. Since this is the only place the two lines intersect, this is the only solution. \(T\) is onto if and only if the rank of \(A\) is \(m\). Let \(S:\mathbb{P}_2\to\mathbb{M}_{22}\) be a linear transformation defined by \[S(ax^2+bx+c) = \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \mbox{ for all } ax^2+bx+c\in \mathbb{P}_2.\nonumber \] Prove that \(S\) is one to one but not onto. Now suppose \(n=2\). AboutTranscript. The coordinates \(x, y\) (or \(x_1\),\(x_2\)) uniquely determine a point in the plan. The rank of \(A\) is \(2\). \nonumber \]. 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. 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. 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By Proposition \(\PageIndex{1}\) \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x} = \vec{0}\). When this happens, we do learn something; it means that at least one equation was a combination of some of the others. [1] That sure seems like a mouthful in and of itself. Example: Let V = Span { [0, 0, 1], [2, 0, 1], [4, 1, 2]}. If \(x+y=0\), then it stands to reason, by multiplying both sides of this equation by 2, that \(2x+2y = 0\). It turns out that the matrix \(A\) of \(T\) can provide this information. 3 Answers. Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. 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. Definition. To express a plane, you would use a basis (minimum number of vectors in a set required to fill the subspace) of two vectors. If a consistent linear system of equations has a free variable, it has infinite solutions. 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. Therefore, we have shown that for any \(a, b\), there is a \(\left [ \begin{array}{c} x \\ y \end{array} \right ]\) such that \(T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\). Accessibility StatementFor more information contact us atinfo@libretexts.org. Thus every point \(P\) in \(\mathbb{R}^{n}\) determines its position vector \(\overrightarrow{0P}\). The vectors \(v_1=(1,1,0)\) and \(v_2=(1,-1,0)\) span a subspace of \(\mathbb{R}^3\). Each vector, \(\overrightarrow{0P}\) and \(\overrightarrow{AB}\) has the same length (or magnitude) and direction. From this theorem follows the next corollary. 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? 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 \(T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\). This section is devoted to studying two important characterizations of linear transformations, called one to one and onto. Lets find out through an example. 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. Then \(n=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im} \left( T\right) \right)\). This is as far as we need to go. 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). M is the slope and b is the Y-Intercept. A linear system is inconsistent if it does not have a solution. This question is familiar to you. Let T: Rn Rm be a transformation defined by T(x) = Ax. Let \(T: \mathbb{R}^k \mapsto \mathbb{R}^n\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^m\) be linear transformations. In this example, it is not possible to have no solutions. The following is a compilation of symbols from the different branches of algebra, which . Then \(T\) is one to one if and only if the rank of \(A\) is \(n\). Then \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies \(\vec{x}=\vec{0}\). 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.
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