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(A T ) -1
... 24. If V is a subspace of R and X is a vector n in R , then vector proj V X must be orthogonal to vector X- Proj V X True. The projection is perpendicular to the space and proj V X is in the space, so proj V X is perpendicular to X-proj V X ...
... 24. If V is a subspace of R and X is a vector n in R , then vector proj V X must be orthogonal to vector X- Proj V X True. The projection is perpendicular to the space and proj V X is in the space, so proj V X is perpendicular to X-proj V X ...
Chapter A.1. Basic Algebra
... (70 ) and (700 ) depends upon the particular value of the modulus n. For instance, all are true when n = 2. For n = 6 we have 2 · 3 = 0 (mod 6) (whence 2 · 3 = 2 · 0 (mod 6)); yet neither 2 nor 3 equals 0 in the integers mod6. Therefore each of (7), (70 ), and (700 ) is false in Z6 . Although the ar ...
... (70 ) and (700 ) depends upon the particular value of the modulus n. For instance, all are true when n = 2. For n = 6 we have 2 · 3 = 0 (mod 6) (whence 2 · 3 = 2 · 0 (mod 6)); yet neither 2 nor 3 equals 0 in the integers mod6. Therefore each of (7), (70 ), and (700 ) is false in Z6 . Although the ar ...
11 Linear dependence and independence
... 4. If a set of vectors is linearly dependent, then one of them can be written as a linear combination of the others: (We just do this for 3 vectors, but it is true for any number). Suppose c1 x1 + c2 x2 + c3 x3 = 0, where at least one of the ci 6= 0 (that is, supppose the vectors are linearly depend ...
... 4. If a set of vectors is linearly dependent, then one of them can be written as a linear combination of the others: (We just do this for 3 vectors, but it is true for any number). Suppose c1 x1 + c2 x2 + c3 x3 = 0, where at least one of the ci 6= 0 (that is, supppose the vectors are linearly depend ...
Linear Transformations
... S:U 6 V and T:V 6 W are linear transformations with the matrix representation of S w.r.t. B and BN being C and the matrix representation of T w.r.t. BN and BO the matrix A. Then the composition SBT has as its matrix representation w.r.t. B and BO, the matrix product CA. Now consider a linear transfo ...
... S:U 6 V and T:V 6 W are linear transformations with the matrix representation of S w.r.t. B and BN being C and the matrix representation of T w.r.t. BN and BO the matrix A. Then the composition SBT has as its matrix representation w.r.t. B and BO, the matrix product CA. Now consider a linear transfo ...
L1-2. Special Matrix Operations: Permutations, Transpose, Inverse
... 3. Suppose A and B are both nxn symmetric matrices. Is matrix multiplication of A and B commutative? Why or why not? Note: A proof is a complete chain of logical statements relating a claim or assertion to a verifiable conclusion. It is therefore never enough to just supply a single example when ask ...
... 3. Suppose A and B are both nxn symmetric matrices. Is matrix multiplication of A and B commutative? Why or why not? Note: A proof is a complete chain of logical statements relating a claim or assertion to a verifiable conclusion. It is therefore never enough to just supply a single example when ask ...
Matrix Summary Matrices and Matrix Math It will be useful to
... It will be useful to remember (or perhaps learn) a few things about matrices so we can use them to solve problems. We will start by defining a matrix we will call a. é a11 é é é a = é a21 é éa é é 31 é Matrix a is called a column matrix, because its three elements are arrayed in a vertical column. M ...
... It will be useful to remember (or perhaps learn) a few things about matrices so we can use them to solve problems. We will start by defining a matrix we will call a. é a11 é é é a = é a21 é éa é é 31 é Matrix a is called a column matrix, because its three elements are arrayed in a vertical column. M ...
Two new direct linear solvers in the QR family
... equals 1m , AG becomes the identity matrix 1m , and the condition is manifestly satisfied. However, if A is not of full row rank, then this condition is not in general satisfied. Finally, the fourth Penrose condition is seen to be true: ...
... equals 1m , AG becomes the identity matrix 1m , and the condition is manifestly satisfied. However, if A is not of full row rank, then this condition is not in general satisfied. Finally, the fourth Penrose condition is seen to be true: ...
Linear Algebra, Section 1.9 First, some vocabulary: A function is a
... is never mentioned anymore). Normally, the question is whether the function is onto its codomain. For example, y = x2 is not onto the real line, but is onto its range, which is the interval [0, ∞). If we don’t want to specify that a function is onto its codomain, we will say that f maps x into the c ...
... is never mentioned anymore). Normally, the question is whether the function is onto its codomain. For example, y = x2 is not onto the real line, but is onto its range, which is the interval [0, ∞). If we don’t want to specify that a function is onto its codomain, we will say that f maps x into the c ...
Math 215 HW #4 Solutions
... meaning that the rank of A must be both 2 and 1 simultaneously. This is clearly impossible, so we see that the row space and the nullspace can’t both be two-dimensional, so no two vectors can be a basis for both the row space and the nullspace of any 3 × 3 matrix. 7. Problem 2.4.32. Describe the fou ...
... meaning that the rank of A must be both 2 and 1 simultaneously. This is clearly impossible, so we see that the row space and the nullspace can’t both be two-dimensional, so no two vectors can be a basis for both the row space and the nullspace of any 3 × 3 matrix. 7. Problem 2.4.32. Describe the fou ...
Jordan normal form
In linear algebra, a Jordan normal form (often called Jordan canonical form)of a linear operator on a finite-dimensional vector space is an upper triangular matrix of a particular form called a Jordan matrix, representing the operator with respect to some basis. Such matrix has each non-zero off-diagonal entry equal to 1, immediately above the main diagonal (on the superdiagonal), and with identical diagonal entries to the left and below them. If the vector space is over a field K, then a basis with respect to which the matrix has the required form exists if and only if all eigenvalues of the matrix lie in K, or equivalently if the characteristic polynomial of the operator splits into linear factors over K. This condition is always satisfied if K is the field of complex numbers. The diagonal entries of the normal form are the eigenvalues of the operator, with the number of times each one occurs being given by its algebraic multiplicity.If the operator is originally given by a square matrix M, then its Jordan normal form is also called the Jordan normal form of M. Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. In spite of its name, the normal form for a given M is not entirely unique, as it is a block diagonal matrix formed of Jordan blocks, the order of which is not fixed; it is conventional to group blocks for the same eigenvalue together, but no ordering is imposed among the eigenvalues, nor among the blocks for a given eigenvalue, although the latter could for instance be ordered by weakly decreasing size.The Jordan–Chevalley decomposition is particularly simple with respect to a basis for which the operator takes its Jordan normal form. The diagonal form for diagonalizable matrices, for instance normal matrices, is a special case of the Jordan normal form.The Jordan normal form is named after Camille Jordan.