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Let $V$ be a vector space over a field $F$ and let $U$ be a subspace of $V$

a) Prove that the quotient $V/U$ has a natural structure of a vector space over $F$

b) Prove that $dim U + dim V/U = dim V$

My attempt: Every vector space has a basis so let {$V_1,V_2,..., V_n$} be a basis for $U$. Then this basis is a linearly independent set of vectors of $V$ so $V$ has a basis that contains {$V_1,V_2, ..., V_n$}. Now when you take the quotient $V/U$ you are left with the basis elements {$V_{n+1}, V_{n+2}, ...$}. I don't know where to go from here. A friend told me that I just have to show that $V/U$ inherits a natural basis, but I don't understand what he meant. I think what I wrote above is sufficient for showing (b)

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  • $\begingroup$ I think your first step is to make sure you know what a quotient is. What do elements of $V/U$ look like? $\endgroup$ – John Brevik Apr 7 '18 at 22:25
  • $\begingroup$ @JohnBrevik Equivalence classes of elements of V? $\endgroup$ – Vinny Chase Apr 7 '18 at 22:29
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You are in a good way for (b)

As you said take $\{v_1,..., v_n\}$ a basis of $U$ and expand this $\{v_1,...,v_n,v_{n+1},...v_{n+k}\}$ in a basis of $V$.

Now show that $\{v_{n+1}+U,..., v_{n+k}+U\}$ is a basis of the quotient $V/U$.

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  • $\begingroup$ @demonakos Thank you! Any insight into how to do a? $\endgroup$ – Vinny Chase Apr 7 '18 at 22:32
  • $\begingroup$ @demonakos or is this what I do for a and I just don't understand why its sufficient? $\endgroup$ – Vinny Chase Apr 7 '18 at 22:33
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    $\begingroup$ No in (a) you have to define how are you gonna add two elements of $V/U$ and how you will multiple elements from $F$ and $V/U$ . The natural definition is $(v_1 + U)+ (v_2+ U)= (v_1+v_2 + U)$ and $\lambda ( v +U)= \lambda v + U$ and then you have to prove the axioms of a vector space. But first you have to prove that these two actions are well defined. $\endgroup$ – dem0nakos Apr 7 '18 at 22:48
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    $\begingroup$ I still think OP needs to go back and make sure we know what $V/U$ means! Otherwise they won't get much out of the good stuff written above! $\endgroup$ – John Brevik Apr 8 '18 at 7:43
  • $\begingroup$ Maybe you are right , i assumed that he knew what $V/U$ means. $\endgroup$ – dem0nakos Apr 8 '18 at 8:12

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