Midterm solutions.
... False. (IMPORTANT) This is true in Rn but not in general. For a counterexample consider an in nite discrete metric space. ...
... False. (IMPORTANT) This is true in Rn but not in general. For a counterexample consider an in nite discrete metric space. ...
Lecture 14: Orthogonal vectors and subspaces
... onal to its nullspace, and its column space is orthogonal to its left nullspace. row space dimension r ...
... onal to its nullspace, and its column space is orthogonal to its left nullspace. row space dimension r ...
16. Homomorphisms 16.1. Basic properties and some examples
... check in this case is ϕ(x + y) = ϕ(x) + ϕ(y). Verification is given below: ϕ(x) + ϕ(y) = [x] + [y] = [x + y] = ϕ(x + y) (where equality [x] + [y] = [x + y] holds by definition of addition in Zn ). Example 2. Let F be a field, n > 1 and integer, G = GLn (F ) and H = (F \ {0}, ·). Define the map ϕ(A) ...
... check in this case is ϕ(x + y) = ϕ(x) + ϕ(y). Verification is given below: ϕ(x) + ϕ(y) = [x] + [y] = [x + y] = ϕ(x + y) (where equality [x] + [y] = [x + y] holds by definition of addition in Zn ). Example 2. Let F be a field, n > 1 and integer, G = GLn (F ) and H = (F \ {0}, ·). Define the map ϕ(A) ...
Pushouts and Adjunction Spaces
... R. Now we put F = g −1 (G) and work in X. The two maps vB ◦ f : A → R and u ◦ (g|F ): F → R agree on A ∩ F and so define a map A ∪ F → R. Because X is normal, this extends to a map vX : X → R. Since vX |A = vB ◦ f , we find a map v: Y → R that satisfies v ◦ g = vX and v|B = vB . By construction, v e ...
... R. Now we put F = g −1 (G) and work in X. The two maps vB ◦ f : A → R and u ◦ (g|F ): F → R agree on A ∩ F and so define a map A ∪ F → R. Because X is normal, this extends to a map vX : X → R. Since vX |A = vB ◦ f , we find a map v: Y → R that satisfies v ◦ g = vX and v|B = vB . By construction, v e ...
EXAM 2 Prof. Alexandru Suciu MTH 1230 LINEAR ALGEBRA
... (c) Does the equation A · x = 0 only have the solution x = 0, or does it have other solutions? Explain your answer. (d) Does the equation A · x = b have a solution for every choice of b in R4 ? Explain your answer. ...
... (c) Does the equation A · x = 0 only have the solution x = 0, or does it have other solutions? Explain your answer. (d) Does the equation A · x = b have a solution for every choice of b in R4 ? Explain your answer. ...
Introduction to topological vector spaces
... Conversely, suppose kvkC = r < 1. If r = 0, then cv lies in C for all c in R. Otherwise, v/r is on the boundary of C —v/η ∈ C for η > r but v/η ∈ / C for η < r. Since r < 1, there then exists some r+ < 1 such that v/r+ ∈ C , and since C is convex and v lies between 0 and v/r+ the vector v also lies ...
... Conversely, suppose kvkC = r < 1. If r = 0, then cv lies in C for all c in R. Otherwise, v/r is on the boundary of C —v/η ∈ C for η > r but v/η ∈ / C for η < r. Since r < 1, there then exists some r+ < 1 such that v/r+ ∈ C , and since C is convex and v lies between 0 and v/r+ the vector v also lies ...
Some results on the syzygies of finite sets and algebraic
... map a2(ç) = (a2 being This is in turn verified by an explicit calculation. Specifically, choose a basis s, , Sr+11 of V so that Sl(xJ) 03B4ij, and denote by e, the evident element of H° (X, , OX1(2)) supported at x,. Then ker(b1) is spanned by elements of the form ...
... map a2(ç) = (a2 being This is in turn verified by an explicit calculation. Specifically, choose a basis s, , Sr+11 of V so that Sl(xJ) 03B4ij, and denote by e, the evident element of H° (X, , OX1(2)) supported at x,. Then ker(b1) is spanned by elements of the form ...