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.

  • 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

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.

  • $\begingroup$ 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

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