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Q53E

Expert-verifiedFound in: Page 177

Book edition
5th

Author(s)
Otto Bretscher

Pages
442 pages

ISBN
9780321796974

**Show that in an n-dimensional linear space we can find at most n linearly independent elements.**

The solution is m=n

**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.

${\mathrm{B}}_{1}=\left({\mathrm{f}}_{1},{\mathrm{f}}_{2},...,{\mathrm{f}}_{\mathrm{n}}\right)\mathrm{where}\mathrm{dim}\left({\mathrm{B}}_{1}\right)=\mathrm{n}\phantom{\rule{0ex}{0ex}}{\mathrm{B}}_{2}=\left({\mathrm{g}}_{1},{\mathrm{g}}_{2},...,{\mathrm{g}}_{\mathrm{m}}\right)\mathrm{where}\mathrm{dim}\left({\mathrm{B}}_{2}\right)=\mathrm{m}$

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’t change and as follows.

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

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

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

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