Linear functional problem

**skyking** · November 21st 2010, 05:40 AM

I have been strugling with this problem for a while now, would appreciate any direction.

If $l\neq 0$ is a linear functional on an n-dimensional vector space V.
Show that there exists a basis of V $B=\{v_{1},...v_{n}\}$ , such that for every $v\inV$ , if $v=\alpha_{1}v_{1}+...+\alpha_{n}v_{n}$ , then $l(v)=\alpha_{1}$ .

My only idea so far is that if I can show that there exists a basis of V $B=\{v_{1},...v_{n}\}$ such that $l(v_{1})=1$ and $l(v_{i})=0$ for $i=2,...,n$ then I will have what I need. I was thinking of chossing a $v$ such that $l(v)=1$ and then completing that to an orthonormal basis, but I can seem to get anywhere with this.

**tonio** · November 21st 2010, 07:05 AM

Originally Posted by skyking

I have been strugling with this problem for a while now, would appreciate any direction.

If $l\neq 0$ is a linear functional on an n-dimensional vector space V.
Show that there exists a basis of V $B=\{v_{1},...v_{n}\}$ , such that for every $v\inV$ , if $v=\alpha_{1}v_{1}+...+\alpha_{n}v_{n}$ , then $l(v)=\alpha_{1}$ .

My only idea so far is that if I can show that there exists a basis of V $B=\{v_{1},...v_{n}\}$ such that $l(v_{1})=1$ and $l(v_{i})=0$ for $i=2,...,n$ then I will have what I need. I was thinking of chossing a $v$ such that $l(v)=1$ and then completing that to an orthonormal basis, but I can seem to get anywhere with this.

Look at $\ker l$ . This is a subspace of dimension n -1, so choose from here a base $\{v_2,\ldots,v_n\}$ complete it to a base $\{v_1,\ldots,v_n\}$

of the whole space, and since then $l(v_1)=k\in\mathbb{F}\,,\,k\neq 0$ , instead $v_1$ take $\frac{1}{k}v_1$

Tonio

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