Question

If F is a field, show that the vector space Fⁿ has dimension n over F.

Short answer

The vector space Fⁿ has dimension n over F.

Step-by-step solution

The Vector space:

If the vectors are linearly independent and every vector in vector space is a linear combination of the given set, then every set of elements in vector space V is called a basis.

A linearly independent spanning set is a basis.

To show that the vector space Fn has dimension n over F.

Assume that F is a field.

Now, to show that $F^n$has dimension n over F.

Proof:

Assume that the set of all row vectors $\left({\mathrm{x}}^1,{\mathrm{x}}^2,.....,{\mathrm{x}}^{\mathrm{n}}\right)$and column vectors$\left[\begin{array}{c}{\mathrm{x}}^1\\ {\mathrm{x}}^2\\ .\\ .\\ {\mathrm{x}}^{\mathrm{n}}\end{array}\right]$ and${\mathrm{x}}^1,{\mathrm{x}}^2,.....,{\mathrm{x}}^{\mathrm{n}}\in \mathrm{F}$is denoted by vector space $F^n$.

Now, consider the vectors,

$\left(\left[\begin{array}{c}1\\ 0\\ .\\ .\\ 0\end{array}\right]\left[\begin{array}{c}0\\ 1\\ .\\ .\\ 0\end{array}\right].....\left[\begin{array}{c}0\\ 0\\ .\\ .\\ 1\end{array}\right]\right)$

The above set has dimension $\mathrm{n}\times 1=\mathrm{n}$.

Now, the vector to be linearly independent, the following is considered:

${\mathrm{c}}_1\left[\begin{array}{c}1\\ 0\\ .\\ .\\ 0\end{array}\right]+{\mathrm{c}}_2\left[\begin{array}{c}0\\ 1\\ .\\ .\\ 0\end{array}\right]+....+{\mathrm{c}}_{\mathrm{n}}\left[\begin{array}{c}0\\ 0\\ .\\ .\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ .\\ .\\ 0\end{array}\right]$

The above condition implies that:

$\left[\begin{array}{c}{\mathrm{c}}_1\\ {\mathrm{c}}_2\\ .\\ .\\ {\mathrm{c}}_{\mathrm{n}}\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ .\\ .\\ 0\end{array}\right]$

Therefore, ${\mathrm{c}}_1=0,{\mathrm{c}}_2=0$,…., ${\mathrm{c}}_{\mathrm{n}}=0$

Hence, the above set of vectors is linearly independent.

Now, for span Fⁿ ,

Let $\left[\begin{array}{c}{\mathrm{a}}_1\\ {\mathrm{a}}_2\\ .\\ .\\ {\mathrm{a}}_{\mathrm{n}}\end{array}\right]\in {\mathrm{F}}^{\mathrm{n}}$

Then,

$\left[\begin{array}{c}{\mathrm{a}}_1\\ {\mathrm{a}}_2\\ .\\ .\\ {\mathrm{a}}_{\mathrm{n}}\end{array}\right]={\mathrm{a}}_1\left[\begin{array}{c}1\\ 0\\ .\\ .\\ 0\end{array}\right]+{\mathrm{a}}_2\left[\begin{array}{c}0\\ 1\\ .\\ .\\ 0\end{array}\right]+...+{\mathrm{a}}_3\left[\begin{array}{c}0\\ 0\\ .\\ .\\ 1\end{array}\right]$

Hence, the set spans Fⁿ over F.

Thus, the vector space Fⁿ has dimension n over F as the given set has n dimension.