Hour to hour syllabus for Math21b, Fall 2010Course Head: Oliver KnillAbstractHere is a brief outline of the lectures for the fall 2010 semester. The section numbers refer to the bookof Otto Bretscher, Linear algebra with applications.1. Week: Systems of Linear Equations and Gauss-Jordan1. Lecture: Introduction to linear systems, Section 1.1, September 8, 2010A central point of this week is Gauss-Jordan elimination. While the pr e cise procedure will be introduced inthe s econd lecture, we lear n in this first hour what a system of linear equations is by looking at examplesof systems of linear equations. The aim is to illustrate, where such systems can occur and how one could solvethem with ’ad hoc’ methods. This involves solving equations by combining e quations in a clever way or toeliminate variables until only one variable is left. We see examples with no s olution, several solutions or exactlyone solution.2. Lecture: Gauss-Jordan elimination, Section 1.2, September 10,2010We rewrite systems of linear equations using matrices and introduce Gauss- Jordan elimination steps:scaling of rows, swapping rows or subtract a multiple of one row to an other row. We also see an example,where one has not only one solution or no solution. Unlike in multi-variable calculus, we distinguish betweencolumn vectors and row vectors. Column vectors are n ×1 matrices, and row vectors are 1 ×m matrices. Ageneral n × m matrix has m columns and n rows. The output of Gauss-Jordan elimination is a matrix rref(A)which is in row reduced echelon form: the first nonzero entry in each row is 1, called leading 1, every columnwith a leading 1 has no other nonzero elements and every row above a row w ith a leading 1 has a leading 1 tothe left.2. Week: Linear Transformations and Geometry3. Lecture: On solutions of linear systems, Section 1.3, September 13,2010How many solutions does a s ystem of linear equations have? The goal of this lecture is to se e that there arethree possibilities: exac tly one solution, no solution or infinitely many solutions. This can be visualized andexplained geometrically in low dimensions. We also learn to determine which case we are in using Gauss-Jordanelimination by looking at the rank of the matrix A as well as the augmented matrix [A|b]. We also mentionthat one can see a system of linear equations Ax = b in two different ways: the column picture tells tha tb = x1v1+ ···+ xnvnis a sum of column vectors viof the matrix A, the row picture tells that the dot productof the row vectors wjwith x are the c omponents wj· x = bjof b.4. Lecture: Linear transformation, Section 2.1, September 15,20101This week provides a link between the geometric and algebraic description of linear transforma tions. Lineartransformations are introduced formally as transformations T (x) = Ax, where A is a matrix. We learn how todistinguish between linear and nonlinear, linear and affine transformations. The transformation T (x) = x + 5for example is not linear because 0 is not mapped to 0. We characterize linear transforma tions on Rnby threeprope rties: T (0) = 0, T (x + y) = T (x) + T (y) and T (sx) = sT (x), which means compatibility with the additivestructure on Rn.5. Lecture: Linear transformations in geometry, Section 2.2, September 17,2010We loo k at examples of rotations, dilations, projections, reflections, rotation-dilatio ns or shears. How are thesetransformations described algebraically? The main point is to see how to go forth and back between algebraicand geometric description. The key fact is that the column vectors vjof a matrix are the images vj= T ejofthe basis vectors ej. We derive for each of the mentioned geometric transformations the matrix form. Any ofthem will be important throughout the course.3. Week: Matrix Algebra and Linear Subspaces6. Lecture: Matrix product, Section 2.3, September 20, 2010The composition of linear transformations leads to the product of matrices. The inverse of a transformationis described by the inverse of the matrix. Square matrices can be treated in a similar way as numbers: wecan add them, multiply them with scalars and many matrices have inverses. There is two things to be ca refulabout: the product of two matrices is not commutative and many nonzer o matrices have no inverse. If we takethe product of a n × p matrix with a p × m matrix, we obtain a n × m matrix. The dot pr oduct as a specialcase of a matr ix product between a 1 × n matrix and a n × 1 matr ix. It produces a 1 ×1 matrix, a scalar.7. Lecture: The inverse, Section 2.4, September 22, 2010We first look at invertibility of maps f : X → X in general and then focus on the case of linear maps. If alinear map Rnto Rnis invertible, how do we find the inverse? We lo ok at examples when this is the case.Finding x such that Ax = y is equivalent to solving a system of linear equations. Doing this in paralle l gives usan eleg ant algorithm by row reducing the matrix [A|1n] to end up with [1n|A−1]. We also mig ht have time tosee how upper triangular block matricesAB0 Chave the inverseA−1−A−1BC−10 C−1.8. Lecture: Image and kernel, Section 3.1, September 24, 2010We define the notion of a linear subspace of n-dimensional space and the span of a set of vectors. Thisis a preparation for the more abstract definition of linear spaces which appear later in the course. The mainalgorithm is the computation of the kernel and the image of a linear transformation using row reduction. Theimage of a matrix A is spanned by the columns of A which have a leading 1 in rre f(A). The kernel of a matrixA is parametrized by ”free variables”, the va riables for which there is no leading 1 in rref(A). For a n × nmatrix, the kernel is trivial if a nd only if the matrix is invertible. The kernel is a lways nontrivial if the n × mmatrix satisfies m > n, that is if there are more variables than equations.4. Week: Basis and Dimension9. Lecture: Basis and linear independence, Section 3.2, September 27, 20102With the previously defined ”span” and the newly introduced linear independence, one can define what a basisfor a linear space is. It is a set of vectors which span the space and which ar e linear independent. The standardbasis in Rnis an example o f a basis. We show that if we have a basis, then every vector can be uniquelyrepresented as a linear combination of basis elements. A typical task is to find the basis o f the kernel and thebasis for the imag e of a linear transfor
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