What are linearly dependent vectors in 4-space?

  • Thread starter Thread starter Dagger1
  • Start date Start date
  • Tags Tags
    Vectors
Dagger1
Messages
2
Reaction score
0
Homework Statement

Find three vectors in R^4 (none of which are the zero vector) such that they are linearly dependent and v3 is not in the Span of v1&v2.The attempt at a solution

My answer was v1=[1,0,0,0] v2=[0,1,0,0] & v3=[0,0,1,0]

Does this work? v3 is not in the span of v1&v2. If placed in a matrix set equal to zero, there is a free variable x4. So aren't they linearly dependent?
 
Physics news on Phys.org
No, they are linearly independent. c1*v1+c2*v2+c3*v3=0 has c1=c2=c3=0 as it's only solution. Nothing in the problem said v1 and v2 have to be linearly independent.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top