Can Linear Independence Be Determined Without Row Reduction?

[charlies1902]

Suppose that S = {v1, v2, v3} is a basis for a
vector space V.
a. Determine whether the set T = {v1, v1 +
v2, v1 + v2 + v3} is a basis for V.
b. Determine whether the set
W = {−v2 + v3, 3v1 + 2v2 + v3, v1 −
v2 + 2v3} is a basis for V.


I must check if they're linearly independent.

For a:
c1v1+c2v1+c2v2+c3v1+c3v2+c3v3=0 c's are constants
v1(c1+c2)+v2(c2+c3)+v3(c3)
Forming the matrix gives
1 1 0
0 1 1
0 0 1
rref of this matrix is the identity matrix, thus it's linearly independent.


For b:
the same thing was done except the rref of the matrix was not the identity matrix, thus it's not a basis.


My question is is there an easier way to do this problem? It seems i made it harde/longer.

 

[LCKurtz]

Suppose that S = {v1, v2, v3} is a basis for a
vector space V.
a. Determine whether the set T = {v1, v1 +
v2, v1 + v2 + v3} is a basis for V.
b. Determine whether the set
W = {−v2 + v3, 3v1 + 2v2 + v3, v1 −
v2 + 2v3} is a basis for V.


I must check if they're linearly independent.

For a:
c1v1+c2v1+c2v2+c3v1+c3v2+c3v3=0 c's are constants

It would be clearer if you start$$c_1v_1+c_2(v_1+v_2)+c_3(v_1+v_2+v_3)=0$$

v1(c1+c2)+v2(c2+c3)+v3(c3)

and maybe you wouldn't have made that mistake, should be
$$(c_1+c_2+c_3)v_1+(c_2+c_3)v_2 +c_3v_3=0$$

Forming the matrix gives
1 1 0
0 1 1
0 0 1
rref of this matrix is the identity matrix, thus it's linearly independent.

My question is is there an easier way to do this problem? It seems i made it harde/longer.

Yes, you can do it in your head. Looking at my last equation you see $c_3=0$. Then since $(c_2+c_3)=0$ you know $c_2=0$ so...

Sometimes the equations are so simple it isn't worth the time to row reduce.