M341 Linear Algebra, Spring 2014, Travis Schedler Review Sheet
... . Conclude that the RHS of the above does not depend on the choice of v1 and v2 (as long as (v1 , v2 ) is an orthogonal basis of V ). Hint: recall the formula for projV : Fn → V from our Gram-Schmidt orthogonalization, which we also defined to be the unique linear map such that projV |V = I|V and pr ...
... . Conclude that the RHS of the above does not depend on the choice of v1 and v2 (as long as (v1 , v2 ) is an orthogonal basis of V ). Hint: recall the formula for projV : Fn → V from our Gram-Schmidt orthogonalization, which we also defined to be the unique linear map such that projV |V = I|V and pr ...
Comparative study of Geometric product and Mixed product
... Mixed product is directly consistent with Pauli matrix algebra and Dirac equation but Geometric product is not directly consistent with Pauli matrix algebra and Dirac equation. Mixed product can be used successfully in dealing with differential operators but Geometric product can not be used in deal ...
... Mixed product is directly consistent with Pauli matrix algebra and Dirac equation but Geometric product is not directly consistent with Pauli matrix algebra and Dirac equation. Mixed product can be used successfully in dealing with differential operators but Geometric product can not be used in deal ...
Math 310, Lesieutre Problem set #7 October 14, 2015 Problems for
... 4.1.5 Let P2 be the vector space of polynomials of degree at most 2. Is the set of polynomials of the form at2 a subset of P2 (where a is a scalar?) It is a subspace. We have (at2 ) + (bt2 ) = (a + b)t2 , which is also of the required form. So the set is closed under addition. Also, c(at2 ) = (ac)t2 ...
... 4.1.5 Let P2 be the vector space of polynomials of degree at most 2. Is the set of polynomials of the form at2 a subset of P2 (where a is a scalar?) It is a subspace. We have (at2 ) + (bt2 ) = (a + b)t2 , which is also of the required form. So the set is closed under addition. Also, c(at2 ) = (ac)t2 ...
Math 51H LINEAR SUBSPACES, BASES, AND DIMENSIONS
... Remark. If V is a subspace, then any linear combination of vectors in V must also be in V . For suppose A1 , . . . Ak are vectors in V and c1 , . . . , ck are scalars. P Then ci Ai is in V (closure under scalar multiplication) for each i. Therefore ci Ai is also in V (by closure under addition). A C ...
... Remark. If V is a subspace, then any linear combination of vectors in V must also be in V . For suppose A1 , . . . Ak are vectors in V and c1 , . . . , ck are scalars. P Then ci Ai is in V (closure under scalar multiplication) for each i. Therefore ci Ai is also in V (by closure under addition). A C ...
Math 261y: von Neumann Algebras (Lecture 13)
... In this lecture, we will begin the study of abelian von Neumann algebras. We first describe the prototypical example of an abelian von Neumann algebra. Let (X, µ) be a measure space: that is, X is a set equipped with a σ-algebra of subsets of X, called measurable set, and µ is a countably additive m ...
... In this lecture, we will begin the study of abelian von Neumann algebras. We first describe the prototypical example of an abelian von Neumann algebra. Let (X, µ) be a measure space: that is, X is a set equipped with a σ-algebra of subsets of X, called measurable set, and µ is a countably additive m ...
Math 315: Linear Algebra Solutions to Assignment 5
... Since span(u, v) is a linear subspace of R3 , it should contain the zero vector, which means that in the plane equation ax + by + cz + d = 0 the coefficient d is 0. Therefore, the plain is of the form ax + by + cz = 0. We know that each of the given vectors is in the plain, so it satisfies the plain ...
... Since span(u, v) is a linear subspace of R3 , it should contain the zero vector, which means that in the plane equation ax + by + cz + d = 0 the coefficient d is 0. Therefore, the plain is of the form ax + by + cz = 0. We know that each of the given vectors is in the plain, so it satisfies the plain ...
NOTES ON DUAL SPACES In these notes we introduce the notion
... Note in particular the following consequence of the above construction of the dual basis. If v ∈ V is an element such that f (v) = 0 for all f ∈ V ∗ , then v = 0. To see this, let (v1 , . . . , vn ) be a basis of V and let (f1 , . . . , fn ) be the dual basis. Write v as v = a1 v1 + · · · + an vn . ...
... Note in particular the following consequence of the above construction of the dual basis. If v ∈ V is an element such that f (v) = 0 for all f ∈ V ∗ , then v = 0. To see this, let (v1 , . . . , vn ) be a basis of V and let (f1 , . . . , fn ) be the dual basis. Write v as v = a1 v1 + · · · + an vn . ...
MTH6140 Linear Algebra II
... Or, symbolically, we are looking for solutions to f 0 (x) = 2ax + b = λax2 + λbx + λc = λf (x). Equating coefficients, we see that λa = 0, λb = 2a and λc = b. There are two cases. If λ = 0 then a = b = 0, leading to the eigenvector 1 identified above. If λ 6= 0, then a = b = c = 0 and there are no c ...
... Or, symbolically, we are looking for solutions to f 0 (x) = 2ax + b = λax2 + λbx + λc = λf (x). Equating coefficients, we see that λa = 0, λb = 2a and λc = b. There are two cases. If λ = 0 then a = b = 0, leading to the eigenvector 1 identified above. If λ 6= 0, then a = b = c = 0 and there are no c ...
CHARACTERS AS CENTRAL IDEMPOTENTS I have recently
... this document is to make sense of and explain this. Actually, I will show how characters can be defined in this way and the fact that they are scaled central idempotents immediately imply the usual and the skew orthogonality relations. 1. Endomorphisms Induced by Central Elements In this section, I ...
... this document is to make sense of and explain this. Actually, I will show how characters can be defined in this way and the fact that they are scaled central idempotents immediately imply the usual and the skew orthogonality relations. 1. Endomorphisms Induced by Central Elements In this section, I ...
General Vector Spaces
... Let (V , +, ·) be a vector space. Definition: A subset W of V is called a subspace if W is itself a vector space with the operations + and · defined on V . Theorem: Let W be a nonempty subset of V . Then W is a subspace of V if and only if it is closed under addition and scalar multiplication, in ot ...
... Let (V , +, ·) be a vector space. Definition: A subset W of V is called a subspace if W is itself a vector space with the operations + and · defined on V . Theorem: Let W be a nonempty subset of V . Then W is a subspace of V if and only if it is closed under addition and scalar multiplication, in ot ...
Chapter 3: Linear transformations
... – Remark: L(Fmx1,Fnx1) is same as Mmxn(F). • For each mxn matrix A we define a unique linear transformation Tgiven by T(X)=AX. • For each a linear transformation T has A such that T(X)=AX. We will discuss this in section 3.3. • Actually the two spaces are isomorphic as vector spaces. • If m=n, then ...
... – Remark: L(Fmx1,Fnx1) is same as Mmxn(F). • For each mxn matrix A we define a unique linear transformation Tgiven by T(X)=AX. • For each a linear transformation T has A such that T(X)=AX. We will discuss this in section 3.3. • Actually the two spaces are isomorphic as vector spaces. • If m=n, then ...
HILBERT SPACES Definition 1. A real inner product space is a real
... Proposition 3. (Gram-Schmidt) If {v n }n∈N is a countable set of linearly independent vectors in an inner product space then there exists a countable orthonormal set {u n }n∈N such that for every m ∈ N the span of {u n }n≥m is the same as the span of {v n }n≥m . Proposition 4. Every Hilbert space h ...
... Proposition 3. (Gram-Schmidt) If {v n }n∈N is a countable set of linearly independent vectors in an inner product space then there exists a countable orthonormal set {u n }n∈N such that for every m ∈ N the span of {u n }n≥m is the same as the span of {v n }n≥m . Proposition 4. Every Hilbert space h ...
Derivative Operators on Quantum Space(3)
... There has been a lot of interest in recent years in ‘noncommutative geometry’ or the principle of doing geometry on ‘coordinate rings’ which are noncommutative algebras (and hence could not be the ring of functions on any actual space). Central to most approaches, including that of Connes [8], is th ...
... There has been a lot of interest in recent years in ‘noncommutative geometry’ or the principle of doing geometry on ‘coordinate rings’ which are noncommutative algebras (and hence could not be the ring of functions on any actual space). Central to most approaches, including that of Connes [8], is th ...
O I A
... Suppose now that iρk and kρj . Let α ik ,α kj ∈ A with α ik ( i, k ) ≠ 0 , α kj ( k , j ) ≠ 0 . Then α = Dα α ik D k α kj D j belongs to A and α (i , j ) ≠ 0 , and thus iρj , proving the transitivity of ρ . Claim 2. The incidence algebra A( ρ ) coincides with A . The inclusion A ⊆ A( ρ ) is obvious. ...
... Suppose now that iρk and kρj . Let α ik ,α kj ∈ A with α ik ( i, k ) ≠ 0 , α kj ( k , j ) ≠ 0 . Then α = Dα α ik D k α kj D j belongs to A and α (i , j ) ≠ 0 , and thus iρj , proving the transitivity of ρ . Claim 2. The incidence algebra A( ρ ) coincides with A . The inclusion A ⊆ A( ρ ) is obvious. ...
Notes from Unit 1
... to make sense of that: In the bread-baking example earlier, what would the “product” (2·1.5, 1·1) = (3, 1) represent? On the other hand, there are good reasons to take that product vector, add its components and get a “dot product” of vectors in which the answer is not a vector but a scalar. (v1 , v ...
... to make sense of that: In the bread-baking example earlier, what would the “product” (2·1.5, 1·1) = (3, 1) represent? On the other hand, there are good reasons to take that product vector, add its components and get a “dot product” of vectors in which the answer is not a vector but a scalar. (v1 , v ...
Section 4.2 - Gordon State College
... If T: Rn → Rn is a linear operator, then a scalar λ is called an eigenvalue of T if there is a nonzero x in Rn such that T(x) = λx Those nonzero vectors x that satisfy this equation are called the eigenvectors of T corresponding to λ. ...
... If T: Rn → Rn is a linear operator, then a scalar λ is called an eigenvalue of T if there is a nonzero x in Rn such that T(x) = λx Those nonzero vectors x that satisfy this equation are called the eigenvectors of T corresponding to λ. ...
Exterior algebra
In mathematics, the exterior product or wedge product of vectors is an algebraic construction used in geometry to study areas, volumes, and their higher-dimensional analogs. The exterior product of two vectors u and v, denoted by u ∧ v, is called a bivector and lives in a space called the exterior square, a vector space that is distinct from the original space of vectors. The magnitude of u ∧ v can be interpreted as the area of the parallelogram with sides u and v, which in three dimensions can also be computed using the cross product of the two vectors. Like the cross product, the exterior product is anticommutative, meaning that u ∧ v = −(v ∧ u) for all vectors u and v. One way to visualize a bivector is as a family of parallelograms all lying in the same plane, having the same area, and with the same orientation of their boundaries—a choice of clockwise or counterclockwise.When regarded in this manner, the exterior product of two vectors is called a 2-blade. More generally, the exterior product of any number k of vectors can be defined and is sometimes called a k-blade. It lives in a space known as the kth exterior power. The magnitude of the resulting k-blade is the volume of the k-dimensional parallelotope whose edges are the given vectors, just as the magnitude of the scalar triple product of vectors in three dimensions gives the volume of the parallelepiped generated by those vectors.The exterior algebra, or Grassmann algebra after Hermann Grassmann, is the algebraic system whose product is the exterior product. The exterior algebra provides an algebraic setting in which to answer geometric questions. For instance, blades have a concrete geometric interpretation, and objects in the exterior algebra can be manipulated according to a set of unambiguous rules. The exterior algebra contains objects that are not just k-blades, but sums of k-blades; such a sum is called a k-vector. The k-blades, because they are simple products of vectors, are called the simple elements of the algebra. The rank of any k-vector is defined to be the smallest number of simple elements of which it is a sum. The exterior product extends to the full exterior algebra, so that it makes sense to multiply any two elements of the algebra. Equipped with this product, the exterior algebra is an associative algebra, which means that α ∧ (β ∧ γ) = (α ∧ β) ∧ γ for any elements α, β, γ. The k-vectors have degree k, meaning that they are sums of products of k vectors. When elements of different degrees are multiplied, the degrees add like multiplication of polynomials. This means that the exterior algebra is a graded algebra.The definition of the exterior algebra makes sense for spaces not just of geometric vectors, but of other vector-like objects such as vector fields or functions. In full generality, the exterior algebra can be defined for modules over a commutative ring, and for other structures of interest in abstract algebra. It is one of these more general constructions where the exterior algebra finds one of its most important applications, where it appears as the algebra of differential forms that is fundamental in areas that use differential geometry. Differential forms are mathematical objects that represent infinitesimal areas of infinitesimal parallelograms (and higher-dimensional bodies), and so can be integrated over surfaces and higher dimensional manifolds in a way that generalizes the line integrals from calculus. The exterior algebra also has many algebraic properties that make it a convenient tool in algebra itself. The association of the exterior algebra to a vector space is a type of functor on vector spaces, which means that it is compatible in a certain way with linear transformations of vector spaces. The exterior algebra is one example of a bialgebra, meaning that its dual space also possesses a product, and this dual product is compatible with the exterior product. This dual algebra is precisely the algebra of alternating multilinear forms, and the pairing between the exterior algebra and its dual is given by the interior product.