# direct sum

• Nov 16th 2010, 07:03 PM
mathbeginner
direct sum
Quote:

Suppose x1,...., xn (where n>= 2) is a basis of a vector space V. Choose r in {1,....,n-1} and define

M= Span{x1,...,xr},
N= Span{xr+1,....xn}
show that V = M (direct sum) N
since need to show direct sum that I have to show v= M+N
and the intersection of M, N is {0}.
But I don't know how to show that the intersection of M, N is {0}.

Plx help
• Nov 16th 2010, 07:22 PM
Drexel28
Quote:

Originally Posted by mathbeginner
since need to show direct sum that I have to show v= M+N
and the intersection of M, N is {0}.
But I don't know how to show that the intersection of M, N is {0}.

Plx help

It's patently true that $\displaystyle M+N=V$, right? Now, the reason why the intersection is trivial is because if $\displaystyle v\in M\cap N$ then $\displaystyle v=\alpha_1x_1+\cdots+\alpha_rx_r$ and $\displaystyle v=\alpha_{r+1}x_{r+1}+\cdots+\alpha_nx_n$ and thus $\displaystyle \alpha_1x_1+\cdots+\alpha_rx_r=\alpha_{r+1}x_{r+1} +\cdots+\alpha_nx_n$ or equivalently $\displaystyle \alpha_1x_1+\cdots+\alpha_rx_r+(-\alpha_{r+1})x_{r+1}+\cdots+(-\alpha_n)x_n=\bold{0}$. But, what does that tell us?
• Nov 16th 2010, 07:37 PM
mathbeginner
Quote:

Originally Posted by Drexel28
It's patently true that $\displaystyle M+N=V$, right? Now, the reason why the intersection is trivial is because if $\displaystyle v\in M\cap N$ then $\displaystyle v=\alpha_1x_1+\cdots+\alpha_rx_r$ and $\displaystyle v=\alpha_{r+1}x_{r+1}+\cdots+\alpha_nx_n$ and thus $\displaystyle \alpha_1x_1+\cdots+\alpha_rx_r=\alpha_{r+1}x_{r+1} +\cdots+\alpha_nx_n$ or equivalently $\displaystyle \alpha_1x_1+\cdots+\alpha_rx_r+(-\alpha_{r+1})x_{r+1}+\cdots+(-\alpha_n)x_n=\bold{0}$. But, what does that tell us?

[tex]\alpha_1=...=\alpha_n=0[tex] they are linearly independent, am I right?
• Nov 16th 2010, 07:39 PM
Drexel28
Quote:

Originally Posted by mathbeginner
[tex]\alpha_1=...=\alpha_n=0[tex] they are linearly independent, am I right?

Oui oui!