Question

Show that, in n-dimensional space, any $\mathbf{(}\mathit{n}\mathbf{}\mathbf{+}\mathbf{}\mathbf{1}\mathbf{)}$ vectors are linearly dependent. Hint: See Section 8.

Short answer

In n-dimensional space, $(n+1)$vectors are linearly dependent.

Step-by-step solution

Given information from the question

$\text{n}$-dimensional space is given.

Linearly dependent vectors

If a nontrivial linear combination of vectors equals 0, a set of vectors is linearly dependent.

Assume that the n-dimensional space has a normalized set of bases

We assume that the n-dimensional space has a normalized set of bases $\left(e_1,e_2,\dots \dots \dots .,e_n\right)$, which is the standard basis for this space. Any vector in this

space is $V_k=\sum _i^n{C}_i{e}_i$; and in this case, k runs from $1\to n+1$i.e., we have $(\text{n}+1)$vectors. The first unit vector can be calculated as shown below.

$e_1=\frac{1}{C_1}(V_1-\sum _{i=2}^n)$

All the components of the other unit vectors were subtracted from $V_1$, except that of the first unit vector. Then they were divided by the corresponding coefficient.

From this we conclude that the $n$-dimensional space is spanned by the set $\left(V_1,e_2,\dots \dots \dots ,e_n\right)$.

If we continue this way for the remaining vectors up to vector $V_n$, we find that the $n$ dimensional space is spanned by the set $\left(V_1,V_2,\dots \dots \dots ,V_n\right)$ and hence the vectors $V_{n+1}$ is just a linear combination of the other $n$ vectors which make the set of $(n+1)$ vectors a linearly dependent set.