Prove infinitely many left inverses

  • Thread starter vintwc
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  • #1
vintwc
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Homework Statement



Let V be a vector space over K. Let L(V) be the set of all linear maps V->V. Prove that L(V) is a ring under the operations:
f+g:x -> f(x)+g(x) and fg:x -> f(g(x))

Now, let V=U+W be the direct sum of two vector spaces over K such that the dimension of both U and W are countable. Then V has countable dimension. Choosing a linear bijection between U and V gives us an element f:V->U of L(V). Prove that there are infinitely many [tex]x \in R = L(V)[/tex] such that xf=1_R. Prove that there is no [tex]y \in R[/tex] such that fy=1_R.

Homework Equations


Direct sum of two vector spaces U and W is the set U+W of pairs of vectors (u,w) in U and W with operations:
a(u,w)+b(u'+w')=(au+bu',aw+bw')

The Attempt at a Solution


For the first bit, I managed to show that L(V) is indeed a ring. In the second part, I'm not sure how to approach this problem. Should I define a bijective function f such that xf=1_R? Also, is linear bijection essentially means an isomorphism?
 

Answers and Replies

  • #2
vintwc
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ok so linear bijection is an isomorphism. i define f(v_1,v_2)=(u(v_1,v_2),0) but i'm still not sure how to proceed from there.
 
  • #3
Dick
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Pick bases for U, {u_i} and W, {w_i}. A basis for V, {v_i} is then the union of the two. And the two are disjoint. That's really what direct sum means in this case. You don't have to worry about the ordered pair definition business. Lets also pick the bases so that f(v_i)=u_i is your bijection. Do you see it now? In the case xf=1_R, f maps everything 1-1 onto U. To undo that, you just have to make sure everything in U goes back to the corresponding vector in V. What you define x to be for elements of W doesn't matter (hence the infinite number). Can you see why fy can't be 1_R?
 

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