Finding the basis.

Linear spaces and subspaces, linear transformations, bases, etc.
mauzel
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Finding the basis.

Find a basis of the subspace of R4 that satisfies the following equation:
-3x1-5x2-8x3-4x4=0

The basis should be three vectors with 4 components each

I tried V1 as (-3,0,0,0), V2 as (0,-5,0,0), V3 as (0,0,-8,-4). But that was incorrect.

Wouldn't the subspace for this be the Vectors (-3,0,0,0), (0,-5,0,0), (0,0,-8,0), (0,0,0,-4) or any multiple? I feel like the basis would have to be in R4 because it is linearly independent with 4 vectors.

stapel_eliz
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Find a basis (three vectors with 4 components each) of the subspace of R4 that satisfies the following equation: -3x1-5x2-8x3-4x4=0
Since you have only one equation, you'll be expressing one of the variables in terms of the other three. This is what will cause you to have three vectors. (And you're in 4-space, which is why those vectors will have four components each.)

You have:

. . . . .$-3x_1\, -\, 5x_2\, -\, 8x_3\, -\, 4x_4\, =\, 0$

Then:

. . . . .$-\frac{5}{3}x_2\, -\, \frac{8}{3}x_3\, -\, \frac{4}{3}x_4\, =\, x_1$

$\mbox{Let }\, x_2\, =\, r,\, x_3\, =\, s,\, \mbox{ and }\, x_4\, =\, t.\, \mbox{ Then:}$

. . . . .$x_1\, =\, -\frac{5}{3}r\, -\, \frac{8}{3}s\, -\, \frac{4}{3}t$

This means that your basic vector will be of the form:

. . . . .$\left\, =\, \left<-\frac{5}{3}r\, -\, \frac{8}{3}s\, -\, \frac{4}{3}t,\, r,\, s,\, t\right>$

Now decompose this into three vectors multiplied by r, s, and t, and I believe you'll have your basis vectors.

mauzel
Posts: 2
Joined: Fri Oct 30, 2009 8:49 pm
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Re: Finding the basis.

Oh wow. I tried something like this but apparently I failed
Thanks!
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