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So I had an exam two days ago and admittedly, I couldn't do this last question very well. I was wondering if someone could give me a guide on how to do this or how to think about it? I'll post what I did in the exam.

Question

Let $V$ be a vector space over a field $\mathbb{F}$ and $U$ a subspace of $V$. For any $\textbf{v}$ in $V$ we define the set
$$\textbf{v} + U = \{\textbf{v} + \textbf{u}|\textbf{u}\in U\}.$$
We write $V/U$ for the set of all of these sets, $$V/U = \{\textbf{v} + U|\textbf{v} \in V\},$$ and we define addition and scalar multiplication on $V/U$ by
$$(\textbf{v}_1 + U) + \textbf{v}_2 + U) = (\textbf{v}_1 + \textbf{v}_2) + U \quad \text{and} \quad \alpha(\textbf{v}+U) = (\alpha \textbf{v})+U.$$ You may assume these operations are well defined.
i) Find the identity element for addition in $V/U$. Prove that your answer is correct.
ii) Prove that the vector distributive law holds in $V/U$. (That is, the one involving a scalar and two vectors, not two scalars and a vector.)
iii) Let $\textbf{v}_1,\textbf{v}_2$ be in $V$. Show that $\textbf{v}_1 + U = \textbf{v}_2 + U$ if and only if $\textbf{v}_1 - \textbf{v}_2$ is in $U$.
iv) $\textbf{You may now assume}$ that $V/U$ is a vector space over $\mathbb{F}$, with operations defined as above. Prove that if $V$ is finite-dimensional, then $V/U$ is also finite-dimensional and
$$\dim V/U = \dim V - \dim U.$$

My attempts

Firstly, I'm not sure what the "standard" name for these type of subspaces are, as I've never encountered them before, so I was lost on what it even meant to "add" an element to another subspace (referring to $V/U$).
i) I claimed that the identity element is $U$ since for any $\textbf{w} = \textbf{v}_1 + U$ in $V/U$, the additive inverse is $-\textbf{w} = -(\textbf{v}_1 + U) = -\textbf{v}_1 + U$ (by definitions of the addition/scalar multiplication).
$\textbf{w} + (-\textbf{w}) = U +(-1)U = (1-1)U = 0U = U$.
ii) I used the same reasoning (and the definitions to show that $\lambda (\textbf{w}_1 + \textbf{w}_2) = \lambda \textbf{w}_1 + \lambda \textbf{w}_2$.
iii) & iv) I had no clue how to start or think about.

Thinking about it again, I'm confused how $U$ can be an identity element of a set... when $U$ is itself a set (with elements in it)..

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  • $\begingroup$ In an algebraic structure like a group, ring, field, vector space, etc., the members can be any objects at all. A vector sub-space U of the vector space V can also be the 0 vector of some other vector space. $\endgroup$ – DanielWainfleet Jun 26 '17 at 5:26
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The sets $\mathbf{v} +U$ are not subspaces technically (they are not closed under addition), but the standard name for them is "cosets" (of $U$). Your intuition for $(i)$ and $(ii)$ are correct.

For $(iii)$, suppose first that $\mathbf{v_0} = \mathbf{v}_1 - \mathbf{v}_2 \in U$. Then for any $\mathbf{u} \in U$, we have $\mathbf{v}_2+\mathbf{u} = \mathbf{v}_1 + (\mathbf{u}-\mathbf{v_0})$, which shows that $\mathbf{v}_2 + U \subseteq \mathbf{v}_1 + U$, and also $\mathbf{v}_1 + \mathbf{u} = \mathbf{v}_2 + (\mathbf{u}+\mathbf{v}_0)$, which shows that $\mathbf{v}_1+U \subseteq \mathbf{v}_2+U$. Conversely, suppose that $\mathbf{v}_1+U=\mathbf{v}_2+U$. Take any $\mathbf{u} \in U$. Then $\mathbf{v}_1+\mathbf{u} \in \mathbf{v}_1+U$, and so also by assumption, it is in $\mathbf{v}_2+U$. So there is some other $\mathbf{u}' \in U$ so that $\mathbf{v}_1+\mathbf{u} = \mathbf{v}_2+\mathbf{u}'$. Then $\mathbf{v}_1-\mathbf{v}_2=\mathbf{u}'-\mathbf{u} \in U$.

Part $(iv)$ is a well-known result that is known by the name "Rank-Nullity Theorem". A more abstract version is also known by the name "First Isomorphism Theorem". You should be able to find a proof by just searching on Google.

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