For this reason we may write both \(P=\left( p_{1},\cdots ,p_{n}\right) \in \mathbb{R}^{n}\) and \(\overrightarrow{0P} = \left [ p_{1} \cdots p_{n} \right ]^T \in \mathbb{R}^{n}\). By convention, the degree of the zero polynomial \(p(z)=0\) is \(-\infty\). Now, consider the case of \(\mathbb{R}^n\) for \(n=1.\) Then from the definition we can identify \(\mathbb{R}\) with points in \(\mathbb{R}^{1}\) as follows: \[\mathbb{R} = \mathbb{R}^{1}= \left\{ \left( x_{1}\right) :x_{1}\in \mathbb{R} \right\}\nonumber \] Hence, \(\mathbb{R}\) is defined as the set of all real numbers and geometrically, we can describe this as all the points on a line. We can describe \(\mathrm{ker}(T)\) as follows. Definition. a variable that does not correspond to a leading 1 is a free, or independent, variable. Let \(T:V\rightarrow W\) be a linear transformation where \(V,W\) are vector spaces. \[\left[\begin{array}{cccc}{0}&{1}&{-1}&{3}\\{1}&{0}&{2}&{2}\\{0}&{-3}&{3}&{-9}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{2}&{2}\\{0}&{1}&{-1}&{3}\\{0}&{0}&{0}&{0}\end{array}\right] \nonumber \], Now convert this reduced matrix back into equations. It is also a good practice to acknowledge the fact that our free variables are, in fact, free. 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. 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. row number of B and column number of A. 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. The kernel, \(\ker \left( T\right)\), consists of all \(\vec{v}\in V\) such that \(T(\vec{v})=\vec{0}\). Is it one to one? 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. Create the corresponding augmented matrix, and then put the matrix into reduced row echelon form. Then: a variable that corresponds to a leading 1 is a basic, or dependent, variable, and. Thus \(T\) is onto. 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 \(\mathrm{rank}\left( T\right) =m\). However, if \(k=6\), then our last row is \([0\ 0\ 1]\), meaning we have no solution. We often call a linear transformation which is one-to-one an injection. Accessibility StatementFor more information contact us [email protected]. lgebra is a subfield of mathematics pertaining to the manipulation of symbols and their governing rules. Determine if a linear transformation is onto or one to one. You can prove that \(T\) is in fact linear. 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. 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 not always the case; we will find in this section that some systems do not have a solution, and others have more than one. 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. This notation will be used throughout this chapter. We can also determine the position vector from \(P\) to \(Q\) (also called the vector from \(P\) to \(Q\)) defined as follows. GSL is a standalone C library, not as fast as any based on BLAS. We need to know how to do this; understanding the process has benefits. Legal. In previous sections we have only encountered linear systems with unique solutions (exactly one solution). \nonumber \]. If the trace of the matrix is positive, all its eigenvalues are positive. 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. Accessibility StatementFor more information contact us [email protected]. Use the kernel and image to determine if a linear transformation is one to one or onto. We can visualize this situation in Figure \(\PageIndex{1}\) (c); the two lines are parallel and never intersect. First, lets just think about it. So our final solution would look something like \[\begin{align}\begin{aligned} x_1 &= 4 +x_2 - 2x_4 \\ x_2 & \text{ is free} \\ x_3 &= 7+3x_4 \\ x_4 & \text{ is free}.\end{aligned}\end{align} \nonumber \]. And linear algebra, as a branch of math, is used in everything from machine learning to organic chemistry. 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. Learn linear algebra for freevectors, matrices, transformations, and more. Consider as an example the following diagram. [2] Then why include it? 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. 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 \(\mathrm{rank}\left( T\right) +\dim \left( \ker \left( T\right) \right) =\dim \left( V\right) .\). \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.). Now we have seen three more examples with different solution types. 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 \((ST)(\vec{x})=\vec{z}\). 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. Give the solution to a linear system whose augmented matrix in reduced row echelon form is, \[\left[\begin{array}{ccccc}{1}&{-1}&{0}&{2}&{4}\\{0}&{0}&{1}&{-3}&{7}\\{0}&{0}&{0}&{0}&{0}\end{array}\right] \nonumber \]. First here is a definition of what is meant by the image and kernel of a linear transformation. Then T is a linear transformation. Thus \(\ker \left( T\right)\) is a subspace of \(V\). 1. In linear algebra, the rank of a matrix A is the dimension of the vector space generated (or spanned) by its columns. 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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. Hence, every element in \(\mathbb{R}^2\) is identified by two components, \(x\) and \(y\), in the usual manner. We further visualize similar situations with, say, 20 equations with two variables. To express a plane, you would use a basis (minimum number of vectors in a set required to fill the subspace) of two vectors. 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. Lets continue this visual aspect of considering solutions to linear systems. Linear Algebra finds applications in virtually every area of mathematics, including Multivariate Calculus, Differential Equations, and Probability Theory. A consistent linear system with more variables than equations will always have infinite solutions. Therefore by the above theorem \(T\) is onto but not one to one. 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. for a finite set of \(k\) polynomials \(p_1(z),\ldots,p_k(z)\). 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. Give an example (different from those given in the text) of a 2 equation, 2 unknown linear system that is not consistent. It is also widely applied in fields like physics, chemistry, economics, psychology, and engineering. Find a basis for \(\mathrm{ker} (T)\) and \(\mathrm{im}(T)\). 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. \end{aligned}\end{align} \nonumber \]. Consider \(n=3\). \[\begin{array}{c} x+y=a \\ x+2y=b \end{array}\nonumber \] Set up the augmented matrix and row reduce. This section is devoted to studying two important characterizations of linear transformations, called one to one and onto. If \(T\) is onto, then \(\mathrm{im}\left( T\right) =W\) and so \(\mathrm{rank}\left( T\right)\) which is defined as the dimension of \(\mathrm{im}\left( T\right)\) is \(m\). [3] What kind of situation would lead to a column of all zeros? \[\left[\begin{array}{ccc}{1}&{1}&{1}\\{2}&{2}&{2}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{ccc}{1}&{1}&{1}\\{0}&{0}&{0}\end{array}\right] \nonumber \], Now convert the reduced matrix back into equations. 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 \(c_{k}=0\). 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. c) If a 3x3 matrix A is invertible, then rank(A)=3. Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. There are linear equations in one variable and linear equations in two variables. The above examples demonstrate a method to determine if a linear transformation \(T\) is one to one or onto. We have \[\begin{align}\begin{aligned} x_1 + 2x_3 &= 2 \\ x_2-x_3&=3 \end{aligned}\end{align} \nonumber \] or, equivalently, \[\begin{align}\begin{aligned} x_1 &= 2-2x_3 \\ x_2&=3+x_3\\x_3&\text{ is free.} 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 \[T\left( a\vec{v}_{1}+b\vec{v}_{2}\right) =aT(\vec{v}_{1})+bT(\vec{v}_{2})=a\vec{0}+b\vec{0}=\vec{0}\nonumber \]. 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. Linear algebra is a branch of mathematics that deals with linear equations and their representations in the vector space using matrices. If a consistent linear system of equations has a free variable, it has infinite solutions. A linear function is an algebraic equation in which each term is either a constant or the product of a constant and a single independent variable of power 1. Hence by Definition \(\PageIndex{1}\), \(T\) is one to one. In other words, \(\vec{v}=\vec{u}\), and \(T\) is one to one. The image of \(S\) is given by, \[\mathrm{im}(S) = \left\{ \left [\begin{array}{cc} a+b & a+c \\ b-c & b+c \end{array}\right ] \right\} = \mathrm{span} \left\{ \left [\begin{array}{rr} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [\begin{array}{rr} 1 & 0 \\ 1 & 1 \end{array} \right ], \left [\begin{array}{rr} 0 & 1 \\ -1 & 1 \end{array} \right ] \right\}\nonumber \]. Key Idea 1.4.1: Consistent Solution Types. If \(x+y=0\), then it stands to reason, by multiplying both sides of this equation by 2, that \(2x+2y = 0\). More succinctly, if we have a leading 1 in the last column of an augmented matrix, then the linear system has no solution. This gives us a new vector with dimensions (lx1). Now we want to know if \(T\) is one to one. In the two previous examples we have used the word free to describe certain variables. \end{aligned}\end{align} \nonumber \], \[\begin{align}\begin{aligned} x_1 &= 15\\ x_2 &=1 \\ x_3 &= -8 \\ x_4 &= -5. Suppose first that \(T\) is one to one and consider \(T(\vec{0})\). The answer to this question lies with properly understanding the reduced row echelon form of a matrix. \[\left [ \begin{array}{rr|r} 1 & 1 & a \\ 1 & 2 & b \end{array} \right ] \rightarrow \left [ \begin{array}{rr|r} 1 & 0 & 2a-b \\ 0 & 1 & b-a \end{array} \right ] \label{ontomatrix}\] You can see from this point that the system has a solution. Draw a vector with its tail at the point \(\left( 0,0,0\right)\) and its tip at the point \(\left( a,b,c\right)\). Here we consider the case where the linear map is not necessarily an isomorphism. Let us learn how to . A vector ~v2Rnis an n-tuple of real numbers. \], At the same time, though, note that \(\mathbb{F}[z]\) itself is infinite-dimensional. 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. It turns out that the matrix \(A\) of \(T\) can provide this information. We can verify that this system has no solution in two ways. We formally define this and a few other terms in this following definition. These definitions help us understand when a consistent system of linear equations will have infinite solutions. Every linear system of equations has exactly one solution, infinite solutions, or no solution. Hence \(\mathbb{F}^n\) is finite-dimensional. Figure \(\PageIndex{1}\): The three possibilities for two linear equations with two unknowns. It is used to stress that idea that \(x_2\) can take on any value; we are free to choose any value for \(x_2\). 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. 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 \(\left( d+a,e+b,f+c\right)\). In this case, we have an infinite solution set, just as if we only had the one equation \(x+y=1\). Recall that if \(S\) and \(T\) are linear transformations, we can discuss their composite denoted \(S \circ T\). Suppose \(\vec{x}_1\) and \(\vec{x}_2\) are vectors in \(\mathbb{R}^n\). You may have previously encountered the \(3\)-dimensional coordinate system, given by \[\mathbb{R}^{3}= \left\{ \left( x_{1}, x_{2}, x_{3}\right) :x_{j}\in \mathbb{R}\text{ for }j=1,2,3 \right\}\nonumber \]. Recall that to find the matrix \(A\) of \(T\), we apply \(T\) to each of the standard basis vectors \(\vec{e}_i\) of \(\mathbb{R}^4\). Then \(z^{m+1}\in\mathbb{F}[z]\), but \(z^{m+1}\notin \Span(p_1(z),\ldots,p_k(z))\). There is no solution to such a problem; this linear system has no solution. By Proposition \(\PageIndex{1}\) it is enough to show that \(A\vec{x}=0\) implies \(\vec{x}=0\). Then in fact, both \(\mathrm{im}\left( T\right)\) and \(\ker \left( T\right)\) are subspaces of \(W\) and \(V\) respectively. 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. Similarly, since \(T\) is one to one, it follows that \(\vec{v} = \vec{0}\). By Proposition \(\PageIndex{1}\) \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x} = \vec{0}\). Then \(\ker \left( T\right) \subseteq V\) and \(\mathrm{im}\left( T\right) \subseteq W\). Each vector, \(\overrightarrow{0P}\) and \(\overrightarrow{AB}\) has the same length (or magnitude) and direction. If there are no free variables, then there is exactly one solution; if there are any free variables, there are 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. Look also at the reduced matrix in Example \(\PageIndex{2}\). 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. Once this value is chosen, the value of \(x_1\) is determined. This is the reason why it is named as a 'linear' equation. If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive. Recall that a linear transformation has the property that \(T(\vec{0}) = \vec{0}\). Legal. As an extension of the previous example, consider the similar augmented matrix where the constant 9 is replaced with a 10. Since the unique solution is \(a=b=c=0\), \(\ker(S)=\{\vec{0}\}\), and thus \(S\) is one-to-one by Corollary \(\PageIndex{1}\). Let \(T:\mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. In other words, \(A\vec{x}=0\) implies that \(\vec{x}=0\). A map A : Fn Fm is called linear, if for all x,y Fn and all , F, we have A(x+y) = Ax+Ay. 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 \(\vec{u} = u_1 \vec{i} + u_2 \vec{j} + u_3 \vec{k}\). These two equations tell us that the values of \(x_1\) and \(x_2\) depend on what \(x_3\) is. 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 \(\mathbb{R}^n\), \[A\vec{x}=\sum_{k=1}^{n}x_{k}A_{k}\nonumber \]. That gives you linear independence. Before we start with a simple example, let us make a note about finding the reduced row echelon form of a matrix. When we learn about s and s, we will see that under certain circumstances this situation arises. 3 Answers. When this happens, we do learn something; it means that at least one equation was a combination of some of the others. Recall that because \(T\) can be expressed as matrix multiplication, we know that \(T\) is a linear transformation. However its performance is still quite good (not extremely good though) and is used quite often; mostly because of its portability. Let \(T: \mathbb{R}^k \mapsto \mathbb{R}^n\) and \(S: \mathbb{R}^n \mapsto \mathbb{R}^m\) be linear transformations. Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. 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. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. In practical terms, we could respond by removing the corresponding column from the matrix and just keep in mind that that variable is free. We can write the image of \(T\) as \[\mathrm{im}(T) = \left\{ \left [ \begin{array}{c} a - b \\ c + d \end{array} \right ] \right\}\nonumber \] Notice that this can be written as \[\mathrm{span} \left\{ \left [ \begin{array}{c} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} -1 \\ 0 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ], \left [ \begin{array}{c} 0 \\ 1 \end{array}\right ] \right\}\nonumber \], However this is clearly not linearly independent. We start with a very simple example. However, it boils down to look at the reduced form of the usual matrix.. First, a definition: if there are infinite solutions, what do we call one of those infinite solutions? ( 6 votes) Show more. Since \(0\neq 4\), we have a contradiction and hence our system has no solution. 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. To find particular solutions, choose values for our free variables. Therefore, recognize that \[\left [ \begin{array}{r} 2 \\ 3 \end{array} \right ] = \left [ \begin{array}{rr} 2 & 3 \end{array} \right ]^T\nonumber \]. The vectors \(v_1=(1,1,0)\) and \(v_2=(1,-1,0)\) span a subspace of \(\mathbb{R}^3\). \[\left[\begin{array}{cccc}{1}&{1}&{1}&{5}\\{1}&{-1}&{1}&{3}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{1}&{4}\\{0}&{1}&{0}&{1}\end{array}\right] \nonumber \], 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 \[\begin{align}\begin{aligned} x_1&= 4-x_3\\x_2&=1\\x_3 &\text{ is free}.\\ \end{aligned}\end{align} \nonumber \]. 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. 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. Then \(T\) is one to one if and only if the rank of \(A\) is \(n\). Therefore, well do a little more practice. Then. In other words, linear algebra is the study of linear functions and vectors. Then \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies \(\vec{x}=\vec{0}\). We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. Since \(S\) is onto, there exists a vector \(\vec{y}\in \mathbb{R}^n\) such that \(S(\vec{y})=\vec{z}\). 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. We now wish to find a basis for \(\mathrm{im}(T)\). 3.Now multiply the resulting matrix in 2 with the vector x we want to transform. \[\mathrm{ker}(T) = \left\{ \left [ \begin{array}{cc} s & s \\ t & -t \end{array} \right ] \right\} = \mathrm{span} \left\{ \left [ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right ] \right\}\nonumber \] It is clear that this set is linearly independent and therefore forms a basis for \(\mathrm{ker}(T)\).

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