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 Post subject: Exact sequencePosted: Wed Feb 17, 2016 7:34 pm
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Let $\displaystyle{\left(\mathbb{K},+,\cdot\right)}$ be a field. Consider the exact sequence

$\displaystyle{\left\{0\right\}\to V_1\to V_2\to V_3\to \left\{0\right\}}$ consisted of $\displaystyle{\mathbb{K}}$ -

vector spaces of finite dimension. Show that this sequence splits.

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 Post subject: Re: Exact sequencePosted: Mon Mar 07, 2016 2:33 pm
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 314
To show that the given short exact sequence of finite-dimensional vector spaces splits, we have to show that $Im(f)$ is a direct summand of $V_{2}$, where $f \ \colon V_{1} \longrightarrow V_{2}$ is the given $\mathbb{K}$-linear monomorphism.

Since $f$ is a $\mathbb{K}$-linear map, $Im(f)$ is a vector subspace of $V_{2}$. Since $V_{2}$ is a finite-dimensional vector space, say of dimension $n$, $Im(f)$ is also finite-dimensional, say of dimension $k$. Therefore, consider a $\mathbb{K}$-basis $\{ e_{i} \}_{i=1}^{k}$ of $Im(f)$ and extend it into a $\mathbb{K}$-basis of $V_{2}$. It follows that $Im(f)$ is a direct summand of $V_{2}$, as every element $\displaystyle v$ of $V_{2}$ is expressed uniquely as $\displaystyle v = \sum_{i=1}^{k} \lambda_{i}e_{i} + \sum_{i=k+1}^{n} \lambda_{i}e_{i}$and the first summand belongs to $Im(f)$.

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