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Let $V = \{ (x_1, x_2, x_3,x_4)\in R^4: x_1-2x_2+x_4 = 0, 2x_1-3x_2+x_3 = 0 \}$

I am trying to find a basis for V. Subtracting the constraints from each other yields $x_4= x_2/2+x_3/2$ this means that we have 3 degrees of freedom so the number of elements in our basis should be 3.

$(1,0,0,0), (0,1,0,1/2), (0,0,1,1/2)$ are three linear independent vectors with span of V.

Then is this a basis for V? Am I allowed to subtract the constraint equations like so?

Thanks.

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  • $\begingroup$ When you reduced two equations to one, you went from 2 degrees of freedom to 3. You will find that NONE of your three vectors satisfy both original equations. $\endgroup$ – vadim123 Mar 11 '15 at 15:38
  • $\begingroup$ It is not. $V$ is defined by $2$ linearly independent equations, hence it has codimension 2. As it's a subspace of $\mathbf R^4$, this means it has dimension $4-2$. $\endgroup$ – Bernard Mar 11 '15 at 15:42
  • $\begingroup$ What if th equation were not linearly independant? $\endgroup$ – Dan Mar 11 '15 at 15:43
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Take

$$u=(1,0,-2,-1)$$ $$v=(0, 1, 3, 2)$$

and you are done.

Every vector in V has a representation with these two vectors, as you can check with ease. And from the first two components of u and v, you see, u and v are linear independet.

You have two equations in four unknowns, so rank is two. You can't find more then two linear independent vectors.

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