# Show that S in linearly Independent and how to extend S to a basis for V

• Oct 26th 2010, 07:05 PM
tn11631
Show that S in linearly Independent and how to extend S to a basis for V
Let
V={(x $_1$,x $_2$,x $_3$,x $_4$,x $_5$) $\in$R $^5$: x $_1$-2x $_2$+3x $_3$-x $_4$+2x $_5$=0}.

(a) Show that S={(0,1,1,1,0)} is a linearly independent subset of V.

(b) Extend S to a basis for V.
• Oct 27th 2010, 06:41 AM
HallsofIvy
Quote:

Originally Posted by tn11631
Let
V={(x $_1$,x $_2$,x $_3$,x $_4$,x $_5$) $\in$R $^5$: x $_1$-2x $_2$+3x $_3$-x $_4$+2x $_5$=0}.

(a) Show that S={(0,1,1,1,0)} is a linearly independent subset of V.

A set containing a single non-zero vector is always a linearly independent subset! If av= 0 and v is not 0 then you must have a= 0. Do you see how that verifies the definition of "linearly independent"? Here, all you need to do is show that (0, 1, 1, 1, 0) is in V. $x_1- 2x_2+ 3x_3- x_4+ 2x_5= 0$ becomes 0- 2(1)+ 3(1)- 1+ 2(0)= 0. Is that true?

Quote:

(b) Extend S to a basis for V.
" $x_1- 2x_2+ 3x_3- x_4+ 2x_5= 0$" is a single linear equation in 5 unknown values. You can solve for any one of them, say, $x_4= x_1- 2x_2+ 3x_3+ 2x_5$, leaving you "4 degrees of freedom"- that is your subspace has dimension 4 and a basis will require 4 vectors. Since you are already given one, you need to find 3 more that form an independent set.