# linearly independent?

• August 26th 2010, 07:47 PM
jht
linearly independent?
u=
(5)
(1)
(2)
v=
(1)
(2)
(1)
w=
(3)
(1)
(-2)

is it u+v+w = 0 if they are linearly independent?
And how do i find out if its form a basis of R^3
• August 26th 2010, 07:55 PM
tonio
Quote:

Originally Posted by jht
u=
(5)
(1)
(2)
v=
(1)
(2)
(1)
w=
(3)
(1)
(-2)

is it u+v+w = 0 if they are linearly independent?
And how do i find out if its form a basis of R^3

What you have to check is whether $au+bv+cw=0$ is possible with at least one of the scalars $a,b,c$ not zero. In our case you've to check whether:

$au+bv+cw=\begin{pmatrix}5a\\a\\2a\end{pmatrix}+\be gin{pmatrix}b\\2b\\b\end{pmatrix}+\begin{pmatrix}3 c\\c\\-2c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix }\iff a=b=c=0$.

If the above homogeneous linear system only has the trivial solution then the three vectors are lin. ind., otherwise they're lin. dep.

Tonio
• August 26th 2010, 08:31 PM
jht
Thanks, i also have a question other than part (b) above,
say if u,v,w is lin,indep
q=w-v+lu and l E R
show u,v,q is lin,indep.

I can see u,v,q is lin,indep if q=w-v+u
But what if the equaition at above with a l real number multiply by u in the equation?
• August 26th 2010, 08:47 PM
jht
so i found a=20/63 ; b = -5/9; c= 5/7
i am wondering is it right?
• August 27th 2010, 10:31 AM
HallsofIvy
Quote:

Originally Posted by jht
u=
(5)
(1)
(2)
v=
(1)
(2)
(1)
w=
(3)
(1)
(-2)

is it u+v+w = 0 if they are linearly independent?

No, if there exist any numbers, a, b, c, not all 0, so that au+ bv+ cw= 0 then they are dependent if u+ v+ w= 0 then a= b= c= 1 are not all 0 so they are dependent. Of course, with the u, v, w you give here, u+ v+ w is NOT 0.
Quote:

And how do i find out if its form a basis of R^3
If 3 vectors are independent, they they form a basis for a 3 dimensional space.
• August 27th 2010, 10:37 AM
HallsofIvy
Quote:

Originally Posted by jht
so i found a=20/63 ; b = -5/9; c= 5/7
i am wondering is it right?

Tonio told you that you needed
$au+bv+cw=\begin{pmatrix}5a\\a\\2a\end{pmatrix}+\be gin{pmatrix}b\\2b\\b\end{pmatrix}+\begin{pmatrix}3 c\\c\\-2c\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix }$
So you must have 5a+ b+ 3c= 0, a+ 2b+ c= 0, and 2a+ b- 2c= 0.

with a= 20/63, b= -5/9, c= 5/7, the second equation is
a+ 2b+ c= (20/63)+ 2(-5/9)+ 5/7= 20/63- 10/9+ 5/7= 20/63- 70/63+ 45/63= -5/63, not 0.
• August 27th 2010, 10:53 AM
HallsofIvy
Quote:

Originally Posted by jht
Thanks, i also have a question other than part (b) above,
say if u,v,w is lin,indep
q=w-v+lu and l E R
show u,v,q is lin,indep.

I can see u,v,q is lin,indep if q=w-v+u
But what if the equaition at above with a l real number multiply by u in the equation?

Suppose u, v, q were NOT linearly independent. The there exist a, b, c, not all 0, such that au+ bv+ cq= 0. But q= w- v+ lu so that is au+ bv+ c(w- v+ lu)= (a+ lc)u+ (b- c)v+ cw= 0. Since u, v, and w are independent, we must have a+ lc= 0, b- c= 0, and c= 0. Those reduce to a= 0, b= 0, and c= 0 contradicting the assertion that they are not all 0. Since c= 0, the "l" in lc does not matter.