Dimension of the linear span of a vector

lolimcool
Messages
20
Reaction score
0

Homework Statement


find the dimension of the linear span of the given vectors
v1 = ( 2, -3, 1) v2 = ( 5, -8, 3) v3 = (-5, 9, -4)


Homework Equations





The Attempt at a Solution


so all i did was make it a matrix and put it in rref and i got
[1 0 0]
[0 1 0]
[0 0 1]

does this mean dimensions of the linear span = 3?
 
Physics news on Phys.org
Yes, it does. What you have shown is that those three vectors are independent and independent vectors form a basis for their span.
 
HallsofIvy said:
Yes, it does. What you have shown is that those three vectors are independent and independent vectors form a basis for their span.

It would mean that. Except I think you did the row reduction wrong. Those vectors aren't linearly independent.
 
Dick said:
It would mean that. Except I think you did the row reduction wrong. Those vectors aren't linearly independent.

oops i did i did it again and got

1 0 5
0 1 -3
0 0 0

so dimension = 2?
 
lolimcool said:
oops i did i did it again and got

1 0 5
0 1 -3
0 0 0

so dimension = 2?

Yes, dimension=2.
 
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