Question on n-dimensional space

[arenaninja]

Homework Statement

Show that in n-dimensional space, any n+1 vectors are linearly dependent.

Ok... I actually know this is true, but I'm lost as to how to show it.

Homework Equations

Vectors are linearly independent if the only solution to $$\alpha_{1}|v_{1}>+\alpha_{2}|v_{2}>+...+\alpha_{n}|v_{n}>+\alpha_{n+1}|v_{n+1}>=|0>$$ is the trivial solution (where $$\alpha_{1}=...=\alpha_{n}=\alpha_{n+1}=0$$).

The Attempt at a Solution

So far I'm only making a statement that each vector has n elements but there are n+1 unknowns, so if the vectors are in the vector space, then they must be linearly dependent.

Any suggestions to improve my answer would be greatly appreciated.

 

[hunt_mat]

Choose a basis for the n-dimensional space, what would the (n+1)th vector look like?

Mat

 

[micromass]

You need to use somewhere that your space is n-dimensional. What's your definition of an n-dimensional space

It's probably that there exists a basis of n elements, we need to use this somewhere. So let $$\{e_1,...,e_n\}$$ be a basis of your space. Since any basis spans the entire space, we can write every $$v_i$$ as

$$v_i=\sum_{j=1}^n{\beta_{i,j} e_i}$$

Now, we assume that

$$\alpha_1 v_1+...+\alpha_{n+1} v_{n+1}=0$$

This is equivalent to saying that

$$\alpha_1 \sum_{j=1}^n{\beta_{1,j} e_i}+...+\alpha_{n+1}\sum_{j=1}^n{\beta_{n+1,j} e_i}=0$$

A little bit of algebra shows that

$$\left(\sum_{i=1}^{n+1}{\alpha_i \beta_{i,1}}\right)e_1+...+ \left(\sum_{i=1}^{n+1}{\alpha_i \beta_{i,n}}\right)e_n=0$$

Now, we use that the basis is linearly independent. We get the following equations (with indeterminates the alpha_i

$$\left\{ \begin{array}{l}<br /> \alpha_1 \beta_{1,1}+...+\alpha_{n+1} \beta_{n+1,1}=0\\<br /> ...\\<br /> \alpha_1 \beta_{1,n}+...+\alpha_{n+1} \beta_{n+1,n}=0<br /> \end{array}\right.$$

This is a homogoneous system of n equations and n+1 indeterminates. Such a system only has the zero solution. So every alpha_i must be zero.