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let $W$ be a subspace of vector space $V$. Prove that there exist a linear mapping $$L\colon V\to V$$ such that $$\ker(L)=W.$$

I have totally no idea how to proceed this, so you are very welcomed to give any hint. We just learned the four fundamental subspaces and the nullity theorem if that helps. Thanks in advance!

edit:yes,this question is under finite dimensional assumption.

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Are your spaces finite dimensional? – uforoboa Sep 25 '12 at 14:34
yes.fry that i forget t put such important condition. =( – Will Sep 25 '12 at 15:51
up vote 4 down vote accepted

All of the answers given so far actually work without change when $W$ and $V$ are not necessarily finite-dimensional. To be completely explicit, notice that any basis $\mathcal{B}$ of $W$ is a linearly independent subset of $V$, and Zorn's lemma implies there is a maximal linearly independent subset $\mathcal{B}'$ of $V$ containing $\mathcal{B}$. It is not hard to show that $\mathcal{B}'$ is a basis of $V$, so we can construct a linear map $L : V \to V$ that sends every basis vector in $\mathcal{B}$ to $0$ while sending any basis vector in $\mathcal{B}' \setminus \mathcal{B}$ to itself. By construction, we have $\ker L = W$.

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Assuming $V$ is finite dimensional consider a basis $\{v_1,v_2\cdots v_k\cdots v_n\}$ for $V$ where $\{v_1,v_2\cdots v_k\}$ is a basis for $W$. Now let $L$ send each vector in $\{v_i:1\le i\le k\}$ to $0$ and the vectors $v_{k+1},\cdots v_n$ to themselves. This gives you the requisite transformation as any $v\in V$ may be expressed as $v=\sum_{i=1}^n \alpha_i v_i$ and you can define $L(v)=\sum_{i=1}^n \alpha_i L(v_i)$.

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I'm assuming that your vector spaces are finite-dimensional. So, you can take a base of W and then extend it to a base of V. Then remember that a linear map is completely defined by it's value on a base.

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Let $\{e_1,\ldots,e_k\}$ be a base of $W$. Complete it to a base $\{e_1,\ldots,e_k,e_{k+1},\ldots,e_n\}$ of $V$. Define $Le_i=0$ if $1 \leq i \leq k$, and $Le_i=e_i$ if $k+1 \leq i \leq n$. You can extend by linearity this map.

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