# linearly dependent

• Apr 19th 2009, 02:41 PM
amitmeh
linearly dependent
Hi guys, I need help with the following question:

Given that {X1, X2,..., Xk} is a linearly independent set, show that {X1, X2,..., Xk, y} is linearly dependent if and only if y is in the span {X1, X2,..., Xk}.
• Apr 19th 2009, 03:44 PM
Gamma
Linearly Independent
A set \$\displaystyle {x_1, x_2, ..., x_n}\$ is said to be linearly independent by definition if \$\displaystyle a_1 x_1 + a_2 x_2 + ... + a_n x_n=0\$ implies \$\displaystyle a_i=0\$ for all i.

Suppose \$\displaystyle {x_1, x_2,... x_n, y}\$ are linearly dependent, then there exist coefficients, not all of which are 0, such that \$\displaystyle a_1 x_1 + a_2 x_2 + ... + a_n x_n + a_{n+1} y=0 \$. Notice that if \$\displaystyle a_{n+1} = 0\$ then we would contradict the linear independence of \$\displaystyle {x_1, x_2, ..., x_n}\$. Thus we can say \$\displaystyle y=-(a_2 x_2 + ... + a_n x_n)/a_{n+1}\$ Showing it to be in the span as desired.

Conversely if y is in the span of \$\displaystyle {x_1, x_2, ..., x_n}\$ then there exist coefficients such that \$\displaystyle y=a_1 x_1 + a_2 x_2 + ... + a_n x_n\$ but then we would have\$\displaystyle 0=a_1 x_1 + a_2 x_2 + ... + a_n x_n + (-1)y\$ all of which are not zero, so \$\displaystyle {x_1, x_2,... x_n, y}\$ is linearly dependent.

QED
• Apr 19th 2009, 11:18 PM
scorpion007
Quote:

Originally Posted by Gamma
all of which are not zero...

You mean "some of which..."?
• Apr 20th 2009, 06:54 AM
Gamma
Yeah
yeah, i guess I can see the confusion there.
I mean all of them cannot be zero, so yeah some of them are nonzero, same thing.
• Apr 21st 2009, 05:35 AM
amitmeh
thanks
thank you so much!