Finding the basis.

mauzel · Postby **mauzel** » Fri Oct 30, 2009 9:03 pm

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 · Postby **stapel_eliz** » Fri Oct 30, 2009 10:08 pm

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:

. . . . . $\left<x_1,\, x_2,\, x_3,\, x_4\right>\, =\, \left<-\frac{5}{3}r\, -\, \frac{8}{3}s\, -\, \frac{4}{3}t,\, r,\, s,\, t\right>$

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

mauzel · Postby **mauzel** » Fri Oct 30, 2009 11:01 pm

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

Thanks!

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