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- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/04%3A_R/4.05%3A_Geometric_Meaning_of_Scalar_MultiplicationThen, by using Definition 4.4.1, the length of this vector is given by \[\sqrt{\left( \left( k u_{1}\right) ^{2}+\left( k u_{2}\right) ^{2}+\left( k u_{3}\right) ^{2}\right) }=\left\vert k \right\vert...Then, by using Definition 4.4.1, the length of this vector is given by √((ku1)2+(ku2)2+(ku3)2)=|k|√u21+u22+u23 Thus the following holds. ‖k→u‖=|k|‖→u‖ In other words, multiplication by a scalar magnifies or shrinks the length of the vector by a factor of \left\vert k \right\vert.
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/09%3A_Appendices/9.04%3A_Well_Ordering_and_InductionLet T consist of all integers larger than or equal to a which are not in S. The theorem will be proved if T=\emptyset . If T\neq \emptyset then by the well ordering principle, ther...Let T consist of all integers larger than or equal to a which are not in S. The theorem will be proved if T=\emptyset . If T\neq \emptyset then by the well ordering principle, there would have to exist a smallest element of T, denoted as b. It must be the case that b>a since by definition, a\notin T. Thus b\geq a+1, and so b-1\geq a and b-1\notin S because if b-1\in S, then b-1+1=b\in S by the assumed property of S. Therefore, \(b-…
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/01%3A_Systems_of_Equations/1.04%3A_Uniqueness_of_the_Reduced_Row-Echelon_FormAs we have seen in earlier sections, we know that every matrix can be brought into reduced row-echelon form by a sequence of elementary row operations. Here we will prove that the resulting matrix is ...As we have seen in earlier sections, we know that every matrix can be brought into reduced row-echelon form by a sequence of elementary row operations. Here we will prove that the resulting matrix is unique; in other words, the resulting matrix in reduced row-echelon does not depend upon the particular sequence of elementary row operations or the order in which they were performed.
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/02%3A_Matrices/2.10%3A_LU_FactorizationAn LU factorization of a matrix involves writing the given matrix as the product of a lower triangular matrix (L) which has the main diagonal consisting entirely of ones, and an upper triangular matri...An LU factorization of a matrix involves writing the given matrix as the product of a lower triangular matrix (L) which has the main diagonal consisting entirely of ones, and an upper triangular matrix (U) in the indicated order.
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/02%3A_Matrices/2.08%3A_Elementary_MatricesWe now turn our attention to a special type of matrix called an elementary matrix.
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/05%3A_Vector_Spaces/5.08%3A_The_Kernel_and_Image_of_a_Linear_MapHere we consider the case where the linear map is not necessarily an isomorphism. First here is a definition of what is meant by the image and kernel of a linear transformation.
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/04%3A_R/4.09%3A_The_Cross_ProductRecall that the dot product is one of two important products for vectors. The second type of product for vectors is called the cross product.
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/05%3A_Vector_Spaces/5.09%3A_The_Matrix_of_a_Linear_TransformationYou may recall from Rn that the matrix of a linear transformation depends on the bases chosen. This concept is explored in this section, where the linear transformation now maps from one arbitrary v...You may recall from Rn that the matrix of a linear transformation depends on the bases chosen. This concept is explored in this section, where the linear transformation now maps from one arbitrary vector space to another.
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/01%3A_Systems_of_Equations/1.03%3A_Gaussian_EliminationThe work we did in the previous section will always find the solution to the system. In this section, we will explore a less cumbersome way to find the solutions. First, we will represent a linear sys...The work we did in the previous section will always find the solution to the system. In this section, we will explore a less cumbersome way to find the solutions. First, we will represent a linear system with an augmented matrix. A matrix is simply a rectangular array of numbers. The size or dimension of a matrix is defined as m×n where m is the number of rows and n is the number of columns.
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/06%3A_Linear_Transformations/6.03%3A_Properties_of_Linear_TransformationsLet T: \mathbb{R}^n \mapsto \mathbb{R}^m be a linear transformation. Then there are some important properties of T which will be examined in this section.
- https://math.libretexts.org/Courses/SUNY_Schenectady_County_Community_College/A_First_Journey_Through_Linear_Algebra/02%3A_Matrices/2.E%3A_ExercisesAn LU factorization of the coefficient matrix is \[\left [ \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 5 & 4 \end{array} \right ] = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 1 ...An LU factorization of the coefficient matrix is \left [ \begin{array}{rrr} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 5 & 4 \end{array} \right ] = \left [ \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & 1 \end{array} \right ] \left [ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & -1 & -5 \\ 0 & 0 & 0 \end{array} \right ]\nonumber First solve \[\left [ \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & 1 \end{array} \right ] \left [ \begin{array}{c} u \\ v \\ w \end{array} \right ] =\left [ \begin{arra…