November 20, 2013 NORMED SPACES Contents 1. The Triangle
... Let Mm,n (F ) be the set of all m ⇥ n matrices with entries in F . Equipped with the operator norm, Mm,n (F ) is a normed space over F . The normed spaces of matrices are not inner product spaces (the norm does not come from an inner product). Normed algebras. The spaces of square matrices (and of l ...
... Let Mm,n (F ) be the set of all m ⇥ n matrices with entries in F . Equipped with the operator norm, Mm,n (F ) is a normed space over F . The normed spaces of matrices are not inner product spaces (the norm does not come from an inner product). Normed algebras. The spaces of square matrices (and of l ...
Sample pages 2 PDF
... (ii) For any b ∈ R n , Ax = b has a unique solution. (iii) There exists a unique matrix B such that AB = BA = I . Proof (i) ⇒ (ii). Since rank A = n we have C (A) = Rn and therefore Ax = b has a solution. If Ax = b and Ay = b then A(x − y) = 0. By 2.10, dim(N (A)) = 0 and therefore x = y. This prove ...
... (ii) For any b ∈ R n , Ax = b has a unique solution. (iii) There exists a unique matrix B such that AB = BA = I . Proof (i) ⇒ (ii). Since rank A = n we have C (A) = Rn and therefore Ax = b has a solution. If Ax = b and Ay = b then A(x − y) = 0. By 2.10, dim(N (A)) = 0 and therefore x = y. This prove ...
Full Text - J
... In [4] it is assumed that l10 ; . . . ; ln0 are not integers (assumption (i), page 693). In this paper, we allow any values of ðl10 ; . . . ; ln0 Þ A C n . Therefore, in order to generalize the method and the results of [4], we need to characterize the solutions of (3) for any ðl10 ; . . . ; ln0 Þ A ...
... In [4] it is assumed that l10 ; . . . ; ln0 are not integers (assumption (i), page 693). In this paper, we allow any values of ðl10 ; . . . ; ln0 Þ A C n . Therefore, in order to generalize the method and the results of [4], we need to characterize the solutions of (3) for any ðl10 ; . . . ; ln0 Þ A ...
Here
... echelon form. Comment: Later in the course we will also see that the rank of A is the dimension of the range of A; i.e., dim{Ax | x ∈ Rn }. The range of A is also equal to the linear span of the columns of A so the rank of A is equal to the dimension of the linear span of the columns of A. We will a ...
... echelon form. Comment: Later in the course we will also see that the rank of A is the dimension of the range of A; i.e., dim{Ax | x ∈ Rn }. The range of A is also equal to the linear span of the columns of A so the rank of A is equal to the dimension of the linear span of the columns of A. We will a ...
Solution for Linear Systems
... Elementary Row Operations on a Matrix There are three elementary row operations on a matrix. They are Interchange of any two Rows. Multiplication of the elements of any row with a non-zero scalar (or constant) Multiplication of elements of a row with a scalar and added to the corresponding elements ...
... Elementary Row Operations on a Matrix There are three elementary row operations on a matrix. They are Interchange of any two Rows. Multiplication of the elements of any row with a non-zero scalar (or constant) Multiplication of elements of a row with a scalar and added to the corresponding elements ...
A+B
... I.e., element (i, j ) of AB is given by the vector dot product of the i th row of A and the j th column of B (considered as vectors). Note: Matrix multiplication is not commutative! ...
... I.e., element (i, j ) of AB is given by the vector dot product of the i th row of A and the j th column of B (considered as vectors). Note: Matrix multiplication is not commutative! ...
GENERATING SETS 1. Introduction In R
... e1 , . . . , en . A notion weaker than a basis is a spanning set: a set of vectors in Rn is a spanning set if its linear combinations fill up the whole space. The difference between a spanning set and a basis is that a spanning set may contain more vectors than necessary to span the space. For insta ...
... e1 , . . . , en . A notion weaker than a basis is a spanning set: a set of vectors in Rn is a spanning set if its linear combinations fill up the whole space. The difference between a spanning set and a basis is that a spanning set may contain more vectors than necessary to span the space. For insta ...
17_ the assignment problem
... Solving Assignment Problems Recall that a permutation of a set N = {1, 2, . . . , n} is a function σ : N → N which is one-to-one and onto. For example, the function from {1, 2, 3, 4, 5} to itself where σ(1) = 5, σ(2) = 4, σ(3) = 2, σ(4) = 1, and σ(5) = 3, is a permutation which we denote 54213. Defi ...
... Solving Assignment Problems Recall that a permutation of a set N = {1, 2, . . . , n} is a function σ : N → N which is one-to-one and onto. For example, the function from {1, 2, 3, 4, 5} to itself where σ(1) = 5, σ(2) = 4, σ(3) = 2, σ(4) = 1, and σ(5) = 3, is a permutation which we denote 54213. Defi ...
