Try to explain the solution of this column vector equation in terms of a column image. Performing any of these row operations does not change the solution of the system of equations. At the end of the section, we will properly define 'pivot' when we see some more examples.
This is back substitution, and we will continue to do it with substitution operations to diagonalize the system. But we shall see that there is some value in using only the operations of elimination and substitution, if we can. Try to eliminate them using only the elimination operations and then support the replacement using only the reverse replacement operations.
We had to use a row switch or maybe some reverse swap operations in the elimination step. The third is the AB column view, where we look at the B a columns that provide the coefficients in the linear combinations of the A columns.
Inverses and Transposes
Now we can do almost exactly the same reduction on the augmented matrix. to solve the system of linear equations. or equivalently the matrix equation. What's really nice about this is that if we need to solve Ax=a for several different a's, we only have to do the elimination once. If the entries are large, the multiplication is more difficult, but we ignore that for now, we call each of 'add' or 'multiply' or 'add and multiply' a basic operation.).
We implicitly assume that we are working with n×n matrices and say that matrix addition and scalar multiplication both take n2 operations, or n2 time. The dot product of two vectors of lengths takes multiplication and additions, so we say it takes no operations. In elimination, to clear below the first pivot, we need to clear n−1 rows, and each row adds a scalar multiple of a vector to another vector, so takes n+ 1 operations.
Deleting across the second pivot takes only one operation per row, son-1 operations, deleting across the second pivot takes 2 operations, etc., and total replacement takes. Using Gauss-Jordan Elimination to find an inverse takes about twice as long, since you have to de-eliminate the augmented matrix I while eliminating A.
Triangular Factors and Row Exchanges
It is important that we only use elementary row operations here, not row permutation, if we use row swings then U may not be upper triangle. Show that if A is symmetric and has an LDU factorization, then it is of the form A=LDLT. We saw that Gaussian Elimination can fail for some matrices A, we also saw that we may need to use 'row swapping' and we said that there are no 'elementary row operations'.
A matrix that can be used to multiply on the left to get a permutation of rows is called a permutation matrix. An apermutation matrix is a matrix that can be obtained from an identity by permuting rows. When doing elimination, if we need to replace a row, we can pretend that we did it before we started the elimination.
The other row operations might change, but we'll still be able to do it.). So if we can eliminate A toI using row permutations, then we can eliminate P Ato I without using row permutations, for some permutation matrixP.
2 Vector Spaces
- Vector Spaces and Subspaces
- The subspace generated by a set
- Solving M x = 0 and M x = b
- Linear Independence
- The Four Fundamental Subspaces
- Graphs applications
- Linear Transformations
For a subset S of vectors in a vector space V, the subspace generated by S is the smallest subspace of V that contains S. Note that the column space of M has a useful interpretation as the set of vectors bso that the equation Mx =b has a solution. Show directly, without using the observation above, that the set S of vectors b such that Mx=b has a solution is a subspace of Rm.
The rank r ≤ n of anm×n matrixM is the number of pivots when it has been reduced to Echelon Form. A set of vectors is linearly independent if and only if the matrix M of which they are columns gives a unique solution to Mx=0; that is, if and only if M is nonsingular. Then we should be able to express b= (7,0,−8) as a linear combination of the other three vectors.
Note that if M has a left inverse L, then LT is the right inverse of AT: (ATLT) = (LA)T =IT =I. Here we have found a right inverseR of the wide matrixM, and all we needed was that there were m pivots. For us, however, when we say 'graph' we mean 'directed graph': the edges are ordered pairs rather than sets.
The component in a graph is a maximal subset C of vertices such that for every two vertices and cranes there is a path in Gbetweenuandv. The characteristic vector of a cycleC is the vector(x1, . . . , xm) such that xi = 1ifei is a forward arc inC, xi =−1ifei is a backward arc inC, and otherwise, xi= 0. The edges separating a face from the rest of the plane is a cycle, and it is not difficult to see that these 'face cycles' generate the cycle space of G (the left null space of MG). Generally in graph theory or topology we count the outer face as a face, but not for today!).
Thus the number of faces is exactly the dimension of the cycle space, and so we have f =m−n+c(G). We need the sum of the arc values around any cycle (by paying attention to orientation) to be 0. MGx=b has a solution if and only if the sum of the values inb corresponding to any cycle of G, sum to 0.
A good way to find the matrix that gives a transformation is to first look at what the transformation does to a basis of the vector space. For the homework you will need to know that the kernel of a transformationT is the set of vectors such that T(v) = 0.
3 Orthogonality
- Orthogonal Vectors and Subspaces
- Cosines and Projections
- Projections and least squares
- Orthogonal Basis and Gram Schmidt
- Fast Fourier Transforms
For a subspace V of Rn , the orthogonal complement V⊥ (read 'V perp') of V is the space of all vectors orthogonal to all vectors of V. Since the rows of a matrix contain a basis for its row space, we get that the row space and the null space of any matrix are autogonal. Indeed, since the null space is the set of all vectors xsuch that r·x= 0, the null space N(M) of a matrixM is exactly the orthogonal complement C(MT)⊥ of its row space.
What is the dimension of the orthogonal complement V⊥ of a space V spanned by the vectors (1,2,4) and (2,1,1). It is the vector of length kbkcosθ in the direction that we can write as. Finding an orthonormal basis of a higher-dimensional subspace is the same idea: from the third basis vector we need to remove the projections on the previous two.
The projection onto a vector is a linear transformation, so we can represent it as (multiplication on the left by) the matrix P. Error=b−Mx, it is perpendicular to the column space, so it is in the left null space of M. Since Mxˆ is the projection of xon onto the column space M, this matrix M(MTM)−1MT is called the projection matrix.
How about replacing w with the nearest vector wˆ in the column space, and finding the solution xˆ for it. Q(QTQ)QT =QIQT =QQT soQQT =:PQ is the matrix projecting onto the column space of Q. AsR is now triangular and invertible (as long as M has independent columns), so multiplying the left side by its inverse gives us.
There is no need to agree on other points, but since f is continuous as the intervals get smaller, even at side points in the interval, p approaches. At the value x=π, however, the values of cos(x) and cos(2x) are opposite, so that the values of f(2π) cancel out and we get −2a. Recall that in the complex plane C=R×iRa the point x+iy can be expressed as reiθ=r(cosθ+isinθ).
What this means is that for points on the unit circle in the complex plane, we can multiply two points by simply adding their angles from the x-axis. This sum is a sum of even vectors evenly distributed around the unit circle in the complex plane.
4 Determinants
- Introduction
- Properties of the Determinant
- Formulae for Determinants
- Applications of Determinants
We now know how the determinant works under row operations, so we can calculate the determinant using Gaussian elimination. We will prove them after developing some formulas for the determinant in the next section. Unless there was a nonzero entry in each column, the determinant of the summand was zero.
If there was one nonzero item per column, the determinant was the product of the items times a factor of −1 if the columns were out of order. In this decomposition, we get a non-zero summation by choosing a permuation of the columns and then multiplying the entries along the diagonal. This is the extension along the first row of a matrix with the same first row as M, but we replace the cofactor c1n with the cofactor c2n.
This is the (i, n) cofactor of the matrixM′ that we get from M by replacing the second row with the first. So the whole expression is the determinant of the matrix M′, whose first two rows are the first row of M. This is the expansion of the cofactor along the ith column of the matrix Bi, which we get from M by replacing the ith column ithit with the vector.
The most wonderful thing you've seen in days is that this continues to hold when the sides of the box are not orthogonal. We saw that det(M) does not change when we remove a multiple of one row from another, so by replacing r1 with the vector h, which is perpendicular to tor2, the determinant isb·needed. Where M is the matrix whose row vectors are the edges of a parallelepipedP,|det(M)| is the volume of P.
Since the determinant of M is preserved by taking the transposea, we can alternately see it as the volume of the figure spanned out by the columns of M. If we can successfully eliminate M into triangular T, let the ninth diagonal entry of the eliminated matrix be called theithpivot ten. Let Mm refer to the square matrix consisting of the intersection of the first m rows and m columns of M.
5 Eigenvectors and eigenvalues
Introduction
HOWEVER; when the matrix M has dimension greater than 3, the characteristic polynomial has degree greater than three, and it is usually difficult to find its roots. But before we get into this, we point out some larger matrices for which the above approach works, or for which we can use ad hoc reasoning.
