Linear combination proof

**Danneedshelp** · July 13th 2009, 01:27 PM

Q: provided that $S_{0}=\{\vec{u_{1}},...,\vec{u_{n}}\}$ is a linearly independent set of vectors and that the set $S_{1}=\{\vec{u_{1}},...,\vec{u_{n}},\vec{v}\}$ is linearly dependent, prove that $\vec{v}$ is a linear combination of the $\vec{u_{i}}$ 's.

A: Since $S_{0}$ is linearly independent the vector equation,

$c_{1}\vec{u_1}+c_{2}\vec{u_2}+. . .+c_{n}\vec{u_n}=\vec{0}$

has only the trival solution, $c_{1}=0, c_{2}=0,. . .,c_{k}=0$ . Conversely, the set $S_{1}$ is linearly dependent; thus, there exists a vector in $S_{1}$ that can be written as a linear combination. Since $S_{1}=\{\{S_{0}\},\vec{v}\}$ and $S_{0}$ is known to be linearly independent we can narrow our search to just one vector, the vector $\vec{v}$ . So, we have a new vector equation,

$c_{1}\vec{u_1}+c_{2}\vec{u_2}+. . .+c_{n}\vec{u_n}+k\vec{v}=\vec{0}$

where $k\in{\mathbb{R}}$

Solving for $\vec{v}$

$\vec{v}=-\frac{c_{1}}{k}\vec{u_1}-. . .-\frac{c_{n}}{k}\vec{u_n}$

$\therefore$ $\vec{v}$ can be written as linear comination of $\vec{u_{i}}$ 's.

I dunno, I feel a chunck is missing. Is that a sufficient proof?

Thanks

**Plato** · July 13th 2009, 01:46 PM

You need to explain why we know that $t\not=0$ .

**Danneedshelp** · July 13th 2009, 01:49 PM

Originally Posted by Plato

You need to explain why we know that $t\not=0$ .

are you talking about the constant $k$ ?

**Danneedshelp** · July 13th 2009, 02:38 PM

Well, $k\neq{0}$ , because the set $S_{1}$ is linear dependent and as I tried explaining, every vector in the set is linearly independent except for the vector $\vec{v}$ . This implies, when we write our new set as a linear combination of the zero vector, $k$ cannot equal zero, because if $k=0$ we would have only the trivail solution and in turn $S_{1}$ would be linearly independent. So, $S_{1}$ is dependent $\forall{k\neq{0}}\in{\mathbb{R}}$ .

Is that the right train of thought?

**Gamma** · July 14th 2009, 11:16 PM

I think things look good, you just gotta be more careful with how you say things and set up the proof. You are given that $\{u_1,...,u_n\}$ is L.I. and that when you add that vector v to it $\{u_1,...,u_n, v\}$ , it makes it L.D.

So you say since $\{u_1,...,u_n, v\}$ is LD, there exist a nontrivial linear combination such that

$c_1u_1+...+c_nu_n + kv=0$

Suppose for a moment that k=0. Then at least one of those $c_i\not = 0$ . But then you have just found a nontrivial linear combination of $\{u_1,...,u_n\}$ which sums to 0, in particular $c_1u_1+...+c_nu_n=0$ which contradicts the linear independence of $\{u_1,...,u_n\}$ . Thus, you know $k \not = 0$ and you can proceed using your argument.

Might be worth noting why $\frac{1}{k}$ exists. I don't know whether you are in algebra or linear algebra, but you can have modules which are basically vector spaces but the coefficients come from a ring where multiplicative inverses need not exist. You are in a vector space which is over a field, so it makes sense to divide by k since you have shown it to be a nonzero element of the field which means it is a unit (is invertible).

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