linear algebra proof help!

**gavin1989** · April 7th 2009, 12:21 AM

Let S = {V1,…vn} be a linearly independent set in a vector space V. Suppose w is in V, but w is not in span(S). Prove that the set T = {V1,…Vn,w} is a linearly independent set.

since it says w is in v, does it mean that w is a subspace of v? yet w is not in span(S). I am kinda confused with what it means?

I am thinking that suppose w is not = 0, and because w is in V.
a1v1+a2v2+...+anvn+a(n+1)w=0 which leads to a(n+1)=-a1v1-a2v2-...anvn
and then because S={V1,…vn} is a linearly independent set. a1v1, a2v2... are all zeros, so {V1,…Vn, W} would be linearly indep too. then w can be write in a linear combi. thus T is a linearly indep set.

I know it sounds bizzare, but just confused how to prove this one.

ANy ideas? thoughts?

**Twig** · April 7th 2009, 01:47 AM

hi

It means that your vectors $[v_{j}]_{j=1}^{p}$ that span
a subspace of V does not span all of V, since $\vec{w}$ is a vector that cannot "be reached" by v1,....,vn.

It says that $\vec{w}$ is in V, meaning that $\vec{w}$ is a vector in V.

The line through the origin and w is a subspace of V, that is, all scalar multiples of $\vec{w}$ is in the subspace spanned by $\vec{w}$ .

I am not sure either how to do this proof correct (I am myself currently studying linear algebra). But maybe something like this:
(If mr_fantastic or anybody wants to correct, plz do!)

Assume $c_{1}\vec{v_{1}} + ... + c_{n}\vec{v_{n}} = d\vec{w}$

$c_{1}\vec{v_{1}} + ... c_{n}\vec{v_{n}} -d\vec{v_{w}} = \vec{0}$

$c_{1}\vec{v_{1}} + ... c_{n}\vec{v_{n}} -(c_{1}\vec{v_{1}} + ... c_{n}\vec{v_{n}}= \vec{0}$

$(c_{1} - c_{1})\vec{v_{1}} + ... + (c_{n} - c_{n})\vec{v_{n}}) = \vec{0}$

Giving all coefficients equaling zero.

**ThePerfectHacker** · April 7th 2009, 09:04 AM

Say that $a_1v_1+a_2v_2+...+a_nv_n + bw = 0$ , there are two cases. Either $b=0$ or $b\not = 0$ . If $b=0$ then we end up with $a_1v_1+...+a_nv_n = 0$ and so $a_1=...=a_n=0$ , which shows that $\{v_1,...,v_n,w\}$ is linearly independent. Now say that $b\not = 0$ . In this case divide by $-b$ to get: $c_1v_1+....+c_nv_n = w$ where $c_j = -\tfrac{a_n}{b}$ . Thus, $w\in \text{spam}(S)$ which is a contradiction.

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