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Let V be a vector space, V X V the product vector space and Z = {(x,x): xeV}. called the diagonal of VXV . Prove that Z is a linear subspace of V X V and that (VXV) / Z is isomorphic to V

your help is appreciated

2007-10-19 08:07:29 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

its a 2nd year uni course, brutal stuff, for a hint it says see this excercise

if T:VXV--->V is the linear mapping T(x,y) = x-y, determine the kernen and range of T.

i dont see how that helps as a hint.

2007-10-19 09:19:47 · update #1

kernel** .......

2007-10-19 09:20:17 · update #2

4 answers

First, we verify that Z is a subspace. We must verify three things: the zero vector is in Z, Z is closed under addition, and Z is closed under scalar multiplication.

1: The zero vector of V×V is (0, 0), and that is indeed an element of Z.

2: Let a and b be arbitrary vectors in Z. Then a=(x, x) and b=(x, y) for some x and y in V. Then a+b = (x, x) + (y, y) = (x+y, x+y) ∈ Z, so Z is closed under addition.

3: Let a be an arbitrary vector in Z and k be an arbitrary scalar. Then a=(x, x) for some x∈V, so ka = k(x, x) = (kx, kx) ∈ Z, so Z is closed under scalar multiplication.

Therefore, Z is a subspace of V×V, so the quotient set (V×V)/Z is well-defined. Now, to show that it is isomorphic to V, let f:(V×V)/Z → V be the function that sends [(a, b)] ↦ a-b (where [(a, b)] is the equivalence class of (a, b)). We need to show that this function is well-defined, which means we must check that this assignment does not depend on our choice of representatives. Suppose (a, b) ~ (c, d), then (a, b) - (c, d) ∈ Z, so (a, b) - (c, d) = (x, x) for some x∈V. Then (a, b) = (x, x) + (c, d) = (x+c, x+d), so a-b = (x+c) - (x+d) = x + c - x - d = c-d. Thus, the result does not depend on our choice of representatives, so this function is well-defined.

Now, I claim that f is an isomorphism. First, we check that it is injective: Suppose f([(a, b)]) = f([(c, d)). Then a-b = c-d, then adding b-c to both sides yields a-c = b-d, so (a, b) - (c, d) = (a-c, b-d) ∈ Z, so (a, b) ~ (c, d), thus [(a, b)] = [(c, d)]. Therefore, f is an injection. To see that it is a surjection, note that for any vector x∈V, f([(x, 0)]) = x-0 = x. So f is also surjective and thus a bijection. To show that f is a homomorphism, we note that f([(a, b)] + [(c, d)]) = f([(a+c, b+d)]) = (a+c) - (b+d) = (a-b) + (c-d) = f([(a, b)]) + f([(c, d)]), and f(k[(a, b)]) = f([(ka, kb)]) = ka-kb = k(a-b) = kf([(a, b)]). So f is a bijective homomorphism and thus an isomorphism.

The hint you were given was obvious -- the range of T was all of V and the kernel of T was precisely Z. In general, if T:V → W is a surjective homomorphism, then S: V/ker(T) → W given by S([x]) = T(x) will be an isomorphism (you may want to prove this if you haven't done so already, it's a useful theorem).

2007-10-19 13:13:01 · answer #1 · answered by Pascal 7 · 0 0

The matrix version of the subject is AX = b the place A is a 2x2 matrix [4, ok; ok, a million] X is a vector variable [x; y] and b is a vector consistent [7; 0] there will be a diverse answer as long as A is non-singular. there heavily isn't a diverse answer if A *is* singular. For A to be singular det(A) = 0 the place det() is the "determinant" of a sq. matrix for a 2x2 matrix [a, b; c, d] det([a, b; c, d]) = advert - bc subsequently on your case 4*a million - ok*ok = 0 ok^2 = 4 ok = +/- 2

2016-12-18 11:58:39 · answer #2 · answered by eisenhauer 4 · 0 0

Is this highschool or college math? cause my college math is nowhere near as hard as this!

2007-10-19 08:09:34 · answer #3 · answered by Ashley 3 · 0 0

Can you give a little bit more information. if i had a little bit more i would solve it.

2007-10-19 08:10:31 · answer #4 · answered by ceej 2 · 0 0

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