0
$\begingroup$

I have a problem that asks to verify $U = \{ [u]_w : u \in V\}$ (small $u$ is the vector) is a vector space with respect to addition and multiplication operations. ($+$ and $\ast$ are already given.) $[u]_w + [v]_w = [u + v]_w$ and $r\ast [u]_w = [ru]_w$ for all $u,v \in V$ and $r \in R$.

So all I need to do is verify associative, commutative, zero-element, inverse, multiplication, and distributive with given addition and multiplication operations?

Thank you.

$\endgroup$
  • 3
    $\begingroup$ what do you mean by $[u]_w$? Is this a coordinate vector? I'm not clear on the structure which you ascribe to $U$... $\endgroup$ – James S. Cook Sep 6 '13 at 4:46
  • $\begingroup$ Seconding James' question. It may denote a component in the direction of a fixed vector $w$ (if so, how is that defined without an inner product?), or it may denote something else. We don't know, you have to tell us :-) $\endgroup$ – Jyrki Lahtonen Sep 6 '13 at 5:01
  • $\begingroup$ Also, often when students (or readers of a textbook) are asked to prove that something is a vector space, the simplest way of doing that is to show that the set in question is a subspace of an already known space. The axioms are then given, if we can show that the subset is closed under the vector space operations. $\endgroup$ – Jyrki Lahtonen Sep 6 '13 at 5:03
  • $\begingroup$ @JyrkiLahtonen I assumed U is a subspace of a vector space V over R. Then I verified the properties of a vector space. All vectors u and v are in V, and so on. Did I in a correct way? $\endgroup$ – therexists Sep 6 '13 at 5:49
  • $\begingroup$ No. The vector space axioms hold in $V$, so they hold in any subset of $V$ irrespective of whether that subset is a subspace or not. The problem is with the "zeroth" axiom: namely checking that addition and scalar multiplication won't take you outside of $U$. In other words, you have to prove that $U$ is a subspace of $V$. Because $V$ is known to be a vector space, the axiom's come free of charge (other than the zeroth axiom). But all the above was pure speculation based on a guess that $U$ is a subset of $V$. You didn't answer the key question: what does $[u]_w$ mean? $\endgroup$ – Jyrki Lahtonen Sep 6 '13 at 5:59
0
$\begingroup$

It seems that one is given a real vector space $(V,+,\cdot)$, a set $U$ and a function $f:V\to U$ such that $f(V)=U$, and that the question is to show that $U$ is a vector space when endowed with the addition $\oplus$ and the scalar multiplication $\odot$ defined by $$ f(u)\oplus f(v)=f(u+v),\qquad r\odot f(u)=f(r\cdot u), $$ for every $r$ in $\mathbb R$ and $u$ and $v$ in $V$.

In the question, $f$ is denoted $[\ \ ]_w$ (and the meaning of this notation is not given, despite several proddings to this effect in the comments) but this is irrelevant.

On the other hand, a crucial hypothesis is missing, to make sure that these definitions even make sense, for example it is necessary that for every $u$, $u'$, $v$ and $v'$ in $V$ such that $f(u)=f(u')$ and $f(v)=f(v')$, one has $f(u+v)=f(u'+v')$, otherwise the sum $x\oplus y$ of two elements $x$ and $y$ of $U$ might depend on the choice of their preimages $u$ and $v$ in $V$, in which case $\oplus$ would be undefined. Similarly for the scalar multiplication $\odot$.

A way to ensure the soundness of the definition of $\oplus$ and $\odot$ is to assume that $f$ is bijective but this is not the only one. Here again, to know what is $[\ \ ]_w$ would be helpful.

All this being taken care of... it is highly probable that the lecture notes include a general result showing why $(U,\oplus,\odot)$ is a vector space. Otherwise, a direct verification of the axioms is possible, and not difficult.

$\endgroup$
  • $\begingroup$ I read your answer just now. Very helpful. Thank you. : )* $\endgroup$ – therexists Sep 7 '13 at 22:07
  • $\begingroup$ Welcome. $ $ $ $ $\endgroup$ – Did Sep 8 '13 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.