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[2] Then why include it? This follows from the definition of matrix multiplication. Similarly, by Corollary \(\PageIndex{1}\), if \(S\) is onto it will have \(\mathrm{rank}(S) = \mathrm{dim}(\mathbb{M}_{22}) = 4\). 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. It consists of all polynomials in \(\mathbb{P}_1\) that have \(1\) for a root. We can also determine the position vector from \(P\) to \(Q\) (also called the vector from \(P\) to \(Q\)) defined as follows. A consistent linear system with more variables than equations will always have infinite solutions. Now we want to know if \(T\) is one to one. First, we will consider what \(\mathbb{R}^n\) looks like in more detail. 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. Recall that a linear transformation has the property that \(T(\vec{0}) = \vec{0}\). Consider as an example the following diagram. The first two examples in this section had infinite solutions, and the third had no solution. . Then \(W=V\) if and only if the dimension of \(W\) is also \(n\). Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation. By Proposition \(\PageIndex{1}\), \(A\) is one to one, and so \(T\) is also one to one. Above we showed that \(T\) was onto but not one to one. 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. 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. Our first example explores officially a quick example used in the introduction of this section. Our final analysis is then this. By picking two values for \(x_3\), we get two particular solutions. Therefore \(x_1\) and \(x_3\) are dependent variables; all other variables (in this case, \(x_2\) and \(x_4\)) are free variables. Therefore, \(S \circ T\) is onto. Actually, the correct formula for slope intercept form is . If \(x+y=0\), then it stands to reason, by multiplying both sides of this equation by 2, that \(2x+2y = 0\). Since \(S\) is one to one, it follows that \(T (\vec{v}) = \vec{0}\). 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 \(\left( d+a,e+b,f+c\right)\). The vectors \(e_1=(1,0,\ldots,0)\), \(e_2=(0,1,0,\ldots,0), \ldots, e_n=(0,\ldots,0,1)\) span \(\mathbb{F}^n\). There is no solution to such a problem; this linear system has no solution. 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. M is the slope and b is the Y-Intercept. Let \(V\) be a vector space of dimension \(n\) and let \(W\) be a subspace. However, if \(k=6\), then our last row is \([0\ 0\ 1]\), meaning we have no solution. Similarly, t and t 2 are linearly independent functions on the whole of the real line, more so [ 0, 1]. \end{aligned}\end{align} \nonumber \]. I'm having trouble with some true/false questions in my linear algebra class and was hoping someone could help me out. We start with a very simple example. Recall that the point given by \(0=\left( 0, \cdots, 0 \right)\) is called the origin. To see this, assume the contrary, namely that, \[ \mathbb{F}[z] = \Span(p_1(z),\ldots,p_k(z))\]. If \(k\neq 6\), then our next step would be to make that second row, second column entry a leading one. Describe the kernel and image of a linear transformation. In the next section, well look at situations which create linear systems that need solving (i.e., word problems). We generally write our solution with the dependent variables on the left and independent variables and constants on the right. Consider a linear system of equations with infinite solutions. We can think as above that the first two coordinates determine a point in a plane. 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? Similarly, a linear transformation which is onto is often called a surjection. 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 \] Each of these equations can be viewed as lines in the coordinate plane, and since their slopes are different, we know they will intersect somewhere (see Figure \(\PageIndex{1}\)(a)). Again, there is no right way of doing this (in fact, there are \(\ldots\) infinite ways of doing this) so we give only an example here. Therefore, we have shown that for any \(a, b\), there is a \(\left [ \begin{array}{c} x \\ y \end{array} \right ]\) such that \(T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\). 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. 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. If a consistent linear system has more variables than leading 1s, then . Then. 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. 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. a variable that does not correspond to a leading 1 is a free, or independent, variable. We have infinite choices for the value of \(x_2\), so therefore we have infinite solutions. A vector space that is not finite-dimensional is called infinite-dimensional. Then T is a linear transformation. Suppose \(p(x)=ax^2+bx+c\in\ker(S)\). Again, more practice is called for. 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. Since the unique solution is \(a=b=c=0\), \(\ker(S)=\{\vec{0}\}\), and thus \(S\) is one-to-one by Corollary \(\PageIndex{1}\). However the last row gives us the equation \[0x_1+0x_2+0x_3 = 1 \nonumber \] or, more concisely, \(0=1\). These notations may be used interchangeably. 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. 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. Accessibility StatementFor more information contact us atinfo@libretexts.org. 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. \[\left[\begin{array}{ccc}{1}&{2}&{3}\\{3}&{k}&{9}\end{array}\right]\qquad\overrightarrow{-3R_{1}+R_{2}\to R_{2}}\qquad\left[\begin{array}{ccc}{1}&{2}&{3}\\{0}&{k-6}&{0}\end{array}\right] \nonumber \]. Then in fact, both \(\mathrm{im}\left( T\right)\) and \(\ker \left( T\right)\) are subspaces of \(W\) and \(V\) respectively. 7. linear independence for every finite subset {, ,} of B, if + + = for some , , in F, then = = =; spanning property for every vector v in V . We will now take a look at an example of a one to one and onto linear transformation. 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. Let T: Rn Rm be a transformation defined by T(x) = Ax. If you're seeing this message, it means we're having trouble loading external resources on our website. 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. This is a fact that we will not prove here, but it deserves to be stated. If a consistent linear system has more variables than leading 1s, then the system will have infinite solutions. \], At the same time, though, note that \(\mathbb{F}[z]\) itself is infinite-dimensional. 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. 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. Lets summarize what we have learned up to this point. Therefore, \(A \left( \mathbb{R}^n \right)\) is the collection of all linear combinations of these products. Therefore by the above theorem \(T\) is onto but not one to one. 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. Let T: Rn Rm be a linear transformation. Let \(T: \mathbb{R}^n \mapsto \mathbb{R}^m\) be a linear transformation induced by the \(m \times n\) matrix \(A\). 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. For example, in the following graph, the Y-Intercept is 4, which is where the line on the graph . 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 \[T\left( a\vec{v}_{1}+b\vec{v}_{2}\right) =aT(\vec{v}_{1})+bT(\vec{v}_{2})=a\vec{0}+b\vec{0}=\vec{0}\nonumber \]. Accessibility StatementFor more information contact us atinfo@libretexts.org. Here we consider the case where the linear map is not necessarily an isomorphism. By setting \(x_2 = 0 = x_4\), we have the solution \(x_1 = 4\), \(x_2 = 0\), \(x_3 = 7\), \(x_4 = 0\). 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 : \[T\left( \vec{x}_1 \right) \neq T \left(\vec{x}_2\right)\nonumber \]. 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. 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. That is, \[\ker \left( T\right) =\left\{ \vec{v}\in V:T(\vec{v})=\vec{0}\right\}\nonumber \]. - Sarvesh Ravichandran Iyer It is like you took an actual arrow, and moved it from one location to another keeping it pointing the same direction. Key Idea \(\PageIndex{1}\) applies only to consistent systems. We can now use this theorem to determine this fact about \(T\). This situation feels a little unusual,\(^{3}\) for \(x_3\) doesnt appear in any of the equations above, but cannot overlook it; it is still a free variable since there is not a leading 1 that corresponds to it. ( 6 votes) Show more. Determine if a linear transformation is onto or one to one. Therefore, they are equal. This is the reason why it is named as a 'linear' equation. \[\mathrm{ker}(T) = \left\{ \left [ \begin{array}{cc} s & s \\ t & -t \end{array} \right ] \right\} = \mathrm{span} \left\{ \left [ \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right ], \left [ \begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array} \right ] \right\}\nonumber \] It is clear that this set is linearly independent and therefore forms a basis for \(\mathrm{ker}(T)\). Thus, \(T\) is one to one if it never takes two different vectors to the same vector. (We can think of it as depending on the value of 1.) How do we recognize which variables are free and which are not? For example, if we set \(x_2 = 0\), then \(x_1 = 1\); if we set \(x_2 = 5\), then \(x_1 = -4\). Before we start with a simple example, let us make a note about finding the reduced row echelon form of a matrix. \[\left[\begin{array}{cccc}{1}&{1}&{1}&{1}\\{1}&{2}&{1}&{2}\\{2}&{3}&{2}&{0}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{1}&{0}\\{0}&{1}&{0}&{0}\\{0}&{0}&{0}&{1}\end{array}\right] \nonumber \]. \[\left[\begin{array}{cccc}{0}&{1}&{-1}&{3}\\{1}&{0}&{2}&{2}\\{0}&{-3}&{3}&{-9}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{2}&{2}\\{0}&{1}&{-1}&{3}\\{0}&{0}&{0}&{0}\end{array}\right] \nonumber \], Now convert this reduced matrix back into equations. The next example shows the same concept with regards to one-to-one transformations. Answer by ntnk (54) ( Show Source ): You can put this solution on YOUR website! A major result is the relation between the dimension of the kernel and dimension of the image of a linear transformation. A. \[\begin{array}{ccccc} x_1 & +& x_2 & = & 1\\ 2x_1 & + & 2x_2 & = &2\end{array} . 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 \(y = 0\). A vector belongs to V when you can write it as a linear combination of the generators of V. Related to Graph - Spanning ? Therefore, the reader is encouraged to employ some form of technology to find the reduced row echelon form. row number of B and column number of A. If \(\mathrm{ rank}\left( T\right) =m,\) then by Theorem \(\PageIndex{2}\), since \(\mathrm{im} \left( T\right)\) is a subspace of \(W,\) it follows that \(\mathrm{im}\left( T\right) =W\). Furthermore, since \(T\) is onto, there exists a vector \(\vec{x}\in \mathbb{R}^k\) such that \(T(\vec{x})=\vec{y}\). We conclude this section with a brief discussion regarding notation. This helps us learn not only the technique but some of its inner workings. We can then use technology once we have mastered the technique and are now learning how to use it to solve problems. This form is also very useful when solving systems of two linear equations. Putting the augmented matrix in reduced row-echelon form: \[\left [\begin{array}{rrr|c} 1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 1 & 1 & 0 \end{array}\right ] \rightarrow \cdots \rightarrow \left [\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right ].\nonumber \]. Notice that there is only one leading 1 in that matrix, and that leading 1 corresponded to the \(x_1\) variable. Once this value is chosen, the value of \(x_1\) is determined. Notice that in this context, \(\vec{p} = \overrightarrow{0P}\). 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. 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 \(T(\vec{x}) = \vec{0}\). \[T(\vec{0})=T\left( \vec{0}+\vec{0}\right) =T(\vec{0})+T(\vec{0})\nonumber \] and so, adding the additive inverse of \(T(\vec{0})\) to both sides, one sees that \(T(\vec{0})=\vec{0}\). By looking at the matrix given by \(\eqref{ontomatrix}\), you can see that there is a unique solution given by \(x=2a-b\) and \(y=b-a\). Using this notation, we may use \(\vec{p}\) to denote the position vector of point \(P\). The reduced row echelon form of the corresponding augmented matrix is, \[\left[\begin{array}{ccc}{1}&{1}&{0}\\{0}&{0}&{1}\end{array}\right] \nonumber \]. The corresponding augmented matrix and its reduced row echelon form are given below. To find the solution, put the corresponding matrix into reduced row echelon form. First, we will prove that if \(T\) is one to one, then \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x}=\vec{0}\). Let \(V,W\) be vector spaces and let \(T:V\rightarrow W\) be a linear transformation. Now consider the image. 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 \(\mathbb{R}^n\), \[A\vec{x}=\sum_{k=1}^{n}x_{k}A_{k}\nonumber \]. \end{aligned}\end{align} \nonumber \], \[\begin{align}\begin{aligned} x_1 &= 3\\ x_2 &=1 \\ x_3 &= 1 . The following proposition is an important result. Prove that if \(T\) and \(S\) are one to one, then \(S \circ T\) is one-to-one. 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: . (lxm) and (mxn) matrices give us (lxn) matrix. \[\left[\begin{array}{cccc}{1}&{1}&{1}&{5}\\{1}&{-1}&{1}&{3}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{cccc}{1}&{0}&{1}&{4}\\{0}&{1}&{0}&{1}\end{array}\right] \nonumber \], 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 \[\begin{align}\begin{aligned} x_1&= 4-x_3\\x_2&=1\\x_3 &\text{ is free}.\\ \end{aligned}\end{align} \nonumber \]. In linear algebra, the rank of a matrix A is the dimension of the vector space generated (or spanned) by its columns. Linear algebra is a branch of mathematics that deals with linear equations and their representations in the vector space using matrices. And linear algebra, as a branch of math, is used in everything from machine learning to organic chemistry. Definition 5.1.3: finite-dimensional and Infinite-dimensional vector spaces. We have now seen examples of consistent systems with exactly one solution and others with infinite solutions. We have just introduced a new term, the word free. More succinctly, if we have a leading 1 in the last column of an augmented matrix, then the linear system has no solution. 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. 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. It follows that \(\left\{ \vec{u}_{1},\cdots ,\vec{u}_{s},\vec{v}_{1},\cdots ,\vec{v} _{r}\right\}\) is a basis for \(V\) and so \[n=s+r=\dim \left( \ker \left( T\right) \right) +\dim \left( \mathrm{im}\left( T\right) \right)\nonumber \], Let \(T:V\rightarrow W\) be a linear transformation and suppose \(V,W\) are finite dimensional vector spaces. Here, the two vectors are dependent because (3,6) is a multiple of the (1,2) (or vice versa): . First here is a definition of what is meant by the image and kernel of a linear transformation. Discuss it. It turns out that the matrix \(A\) of \(T\) can provide this information. If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive. Now we have seen three more examples with different solution types. 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 \(T\left [ \begin{array}{c} x \\ y \end{array} \right ] =\left [ \begin{array}{c} a \\ b \end{array} \right ]\). 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 \(P\). \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.). Then \(\ker \left( T\right) \subseteq V\) and \(\mathrm{im}\left( T\right) \subseteq W\). Hence \(\mathbb{F}^n\) is finite-dimensional. \end{aligned}\end{align} \nonumber \]. The second important characterization is called onto. 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 cannot possibly pick values for \(x\) and \(y\) so that \(2x+2y\) equals both 0 and 4. as a standard basis, and therefore = More generally, =, and even more generally, = for any field. \[\left[\begin{array}{ccc}{1}&{1}&{1}\\{2}&{2}&{2}\end{array}\right]\qquad\overrightarrow{\text{rref}}\qquad\left[\begin{array}{ccc}{1}&{1}&{1}\\{0}&{0}&{0}\end{array}\right] \nonumber \], 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. Theorem 5.1.1: Matrix Transformations are Linear Transformations. [3] What kind of situation would lead to a column of all zeros? For this reason we may write both \(P=\left( p_{1},\cdots ,p_{n}\right) \in \mathbb{R}^{n}\) and \(\overrightarrow{0P} = \left [ p_{1} \cdots p_{n} \right ]^T \in \mathbb{R}^{n}\). 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. GATE-CS-2014- (Set-2) Linear Algebra. As before, let \(V\) denote a vector space over \(\mathbb{F}\). AboutTranscript. To find two particular solutions, we pick values for our free variables. 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. ), Now let us confirm this using the prescribed technique from above. Performing the same elementary row operation gives, \[\left[\begin{array}{ccc}{1}&{2}&{3}\\{3}&{k}&{10}\end{array}\right]\qquad\overrightarrow{-3R_{1}+R_{2}\to R_{2}}\qquad\left[\begin{array}{ccc}{1}&{2}&{3}\\{0}&{k-6}&{1}\end{array}\right] \nonumber \]. { "1.4.01:_Exercises_1.4" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "1.01:_Introduction_to_Linear_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.02:_Using_Matrices_to_Solve_Systems_of_Linear_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.03:_Elementary_Row_Operations_and_Gaussian_Elimination" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.04:_Existence_and_Uniqueness_of_Solutions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "1.05:_Applications_of_Linear_Systems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, { "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "01:_Systems_of_Linear_Equations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "02:_Matrix_Arithmetic" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "03:_Operations_on_Matrices" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "04:_Eigenvalues_and_Eigenvectors" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "05:_Graphical_Explorations_of_Vectors" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass228_0.b__1]()" }, 1.4: Existence and Uniqueness of Solutions, [ "article:topic", "authorname:apex", "license:ccbync", "licenseversion:30", "source@https://github.com/APEXCalculus/Fundamentals-of-Matrix-Algebra", "source@http://www.apexcalculus.com/" ], https://math.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fmath.libretexts.org%2FBookshelves%2FLinear_Algebra%2FFundamentals_of_Matrix_Algebra_(Hartman)%2F01%253A_Systems_of_Linear_Equations%2F1.04%253A_Existence_and_Uniqueness_of_Solutions, \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}}}\) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\), Definition: Consistent and Inconsistent Linear Systems, Definition: Dependent and Independent Variables, Key Idea \(\PageIndex{1}\): Consistent Solution Types, Key Idea \(\PageIndex{2}\): Inconsistent Systems of Linear Equations, source@https://github.com/APEXCalculus/Fundamentals-of-Matrix-Algebra. After moving it around, it is regarded as the same vector. Legal. The statement \(\ker \left( T \right) =\left\{ \vec{0}\right\}\) is equivalent to saying if \(T \left( \vec{v} \right)=\vec{0},\) it follows that \(\vec{v}=\vec{0}\). 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 \(v_1,v_2,\ldots,v_m\in V\).