91影视

If a linear space V has a basis with n elements, then all other bases of V consist of n elements as well.

We say that n is the dimension of V that is dim(V)=n.

Consider two basis of V for a n-dimensional spaces as follows.

$$\begin{gathered} {\mathrm{B}}_1=\left({\mathrm{f}}_1,{\mathrm{f}}_2,...,{\mathrm{f}}_{\mathrm{n}}\right)\mathrm{where}\dim \left({\mathrm{B}}_1\right)=\mathrm{n} \\ {\mathrm{B}}_2=\left({\mathrm{g}}_1,{\mathrm{g}}_2,...,{\mathrm{g}}_{\mathrm{m}}\right)\mathrm{where}\dim \left({\mathrm{B}}_2\right)=\mathrm{m} \end{gathered}$$

To show that n is the largest possible dimension for an n-dimensional space that m=n.

Since, the m vectors of${\mathrm{B}}_2$ form a basis.

We know that as follows.

${\mathrm{c}}_1{\mathrm{g}}_1+...+{\mathrm{c}}_{\mathrm{m}}{\mathrm{g}}_{\mathrm{m}}=0$,By the definition of linear independence.

Also know that${\mathrm{B}}_1$forma a basis for the same space with n vectors and since the dimension of the space can鈥檛 change and as follows.

$\mathrm{m}=\mathrm{n}$

Since, m=n, that is there is no such basis m exists where $n鈮�m$for an n-dimensional space.

Thus, n is the largest possible dimension for an n-dimensional space that m=n