1
$\begingroup$

Let $V$ be a finite dimensional vector space over $k$ and $V_1, V_2 \subseteq V$ be two subspace of $V$. Suppose that $W \subseteq V$ is a subspace such that $V_1, V_2 \subseteq W$. We have two inclusions $i_1:V_1 \cap V_2 \hookrightarrow V_1$ and $i_2:V_1 \cap V_2 \hookrightarrow V_2$. We obtain an exact sequence of subspaces $$ 0 \to V_1 \cap V_2 \overset{f}{\longrightarrow} V_1 \oplus V_2 \overset{g}{\longrightarrow} W $$ with $f:V_1 \cap V_2 \to V_1 \oplus V_2$ and $g:V_1 \oplus V_2 \to W$ given by $f(x)=(-i_2(x), i_1(x))$ and $g(x, y)=x+y$, respectively.

My question is:

(1) Is possible to find the image of $g$ and its dimension? Is it $Span(V_1, V_2)$?

(2) How can I adapt this sequence to an internal direct sum? In this case, I would need $V_1 \cap V_2=\{0\}$ and what else could we say?

Thank you!

$\endgroup$
3
  • 1
    $\begingroup$ (1) The image of $g$ is indeed $\textrm{Span}(V_1,V2)$, and its dimension is $\textrm{dim}((V_1\bigoplus V_2)/V_1\cap V_2) = \textrm{dim}(V_1) + \textrm{dim}(V_2) - \textrm{dim}(V_1 \cap V_2)$. It is so because $g$ induces an injective map $V_1 \bigoplus V_2/V_1\cap V_2 \rightarrow W$ (if you do not want to speak about quotient, you can take a supplementary $S$ of $V_1 \cap V_2$ in $V_1\bigoplus V_2$ and show that the map $g\restriction_S : S \rightarrow W$ is injective). $\endgroup$
    – Enguerrand Moulinier
    Commented Nov 6 at 16:16
  • 1
    $\begingroup$ But we cannot say more on $W$, as it can be anything that contains $V_1$ and $V_2$ and is in $V$. $V$ always satisfies the hypothesis, but smaller sbspace also does in general,for eample if $V_1=V_2$, then $W=V_1$ also satisfies the hypothesis but is smaller than $V$. $\endgroup$
    – Enguerrand Moulinier
    Commented Nov 6 at 16:17
  • $\begingroup$ For (2), I must admit that I do not fully understand the question. Can you be a bit more precise ? Thanks ! $\endgroup$
    – Enguerrand Moulinier
    Commented Nov 6 at 16:17

0

Reset to default

You must log in to answer this question.