Systematic way of extending a set to a basis

[negation]

Homework Statement

I want to extend the below U set of vectors to R4.
u1 = (0, 0, 0, -4), u2 = (0, 0, -4, 3), u3 = (3, 2, 3, -2).

The Attempt at a Solution

For a set of vectors to form a basis for Rn, the vectors must be LI and spans Rn(has n vectors)

u1, u2 and u3 are linearly independent and spans R3 but not R4.
To extend u1, u2 and u3 to R4. I require another vector, w. [itex]w\notin span{u1,u2,u3}[/itex]

One of the solution is trial-and-error which I am not keen.


I started with this:

0λ1 + 0λ2 +3λ3 +λ4w1= 0
0λ1 + 0λ2 +2λ3 +λ4w2= 0
0λ1 -4λ2 +3λ3 + λ4w3= 0
-4λ1 +3λ2 -2λ3 + λ4w4 = 0

which is a brick wall.

The other approach I had was UW = 0
I want to find the W set of vectors that maps the coefficient matrix U into the zero-vector.

 

[vela]

In the second approach, how are you defining U and what's stopping you from doing it?

 

[negation]

In the second approach, how are you defining U and what's stopping you from doing it?

There's nothing stopping me from doing so. The calculation can easily be done but I'd appreciate if anyone could shed some light on the idea behind using the definition of UW=0 to solve for the W set of vectors.

 

[vela]

There's nothing stopping me from doing so. The calculation can easily be done but I'd appreciate if anyone could shed some light on the idea behind using the definition of UW=0 to solve for the W set of vectors.

Well, again, you need to define what U and W are first. We can't read your mind.

 

[Ray Vickson]

Homework Statement

I want to extend the below U set of vectors to R4.
u1 = (0, 0, 0, -4), u2 = (0, 0, -4, 3), u3 = (3, 2, 3, -2).

The Attempt at a Solution

For a set of vectors to form a basis for Rn, the vectors must be LI and spans Rn(has n vectors)

u1, u2 and u3 are linearly independent and spans R3 but not R4.
To extend u1, u2 and u3 to R4. I require another vector, w. [itex]w\notin span{u1,u2,u3}[/itex]

One of the solution is trial-and-error which I am not keen.


I started with this:

0λ1 + 0λ2 +3λ3 +λ4w1= 0
0λ1 + 0λ2 +2λ3 +λ4w2= 0
0λ1 -4λ2 +3λ3 + λ4w3= 0
-4λ1 +3λ2 -2λ3 + λ4w4 = 0

which is a brick wall.

The other approach I had was UW = 0
I want to find the W set of vectors that maps the coefficient matrix U into the zero-vector.

There is a way that is easy in this problem, but is not really systematic and applicable to all problems without modification. Say the fourth vector you want is ##v = (v_1,v_2,v_3,v_4)##. If you want v to be orthogonal to all the u1, u2 and u3, then ##v \perp u1 \rightarrow v_4 = 0##.
So, now ##v = (v_1,v_2,v_3,0)##. The condition ##v \perp u2 \rightarrow v_3=0##. So, now ##v = (v_1,v_2,0,0)##, and ##v \perp u3 \rightarrow 3v_1+2v_2 = 0##. Take any v_1 and v_2 that satisfy this last equation; then v is perpendicular to all the ui, so is linearly independent of them.

 

[Mugged]

A basis need not be orthogonal. i.e. (1,1), (1,0) is a basis for R2

 

[micromass]

A basis need not be orthogonal. i.e. (1,1), (1,0) is a basis for R2

Nobody said it was. But adding vectors orthogonal to the original set is an easy solution.