Is the Set Linearly Independant?

[bugatti79]

Homework Statement

Let V be a real vector space and $${b_1,b_2,b_3,b_4}$$ a linearly independent set of vectors in V
Is the set $$\left \{ b_1,b_2,b_3,b_1+b_4,b_2+b_4 \right \}$$

The Attempt at a Solution

$$\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4\left \{ b_1+b_4 \right \}+\alpha_5\left \{ b_2+b_4 \right \} =0$$

Rearranging

$$b_1(\alpha_1+\alpha_4)+b_2(\alpha_2+\alpha_5)+b_3(\alpha_3)+b_4(\alpha_4+\alpha_5)=0$$
Hence it is NOT linearly independent since the scalars are all not 0...?

 

[dacruick]

Hi there,

It's been a while since I've done this, and I'm sure that my methods are not sound but I think I have a pretty good idea of the answer here.

A set of vectors are linearly independent if you cannot write any on of the vectors in terms of any finite number of other vectors in the set. I can see right now that b2 + b4 = b2 + (b1 + b4) - b1

in other words if b1 = V1, b2 = V2...b2 +b4 = V5 then V5 = V2 + v4 - V1

Does that make sense to you?

 

[dacruick]

And I think that if a set of vectors are linearly independent they span the space correct? And it seems illogical that a set of 4 vectors can span 5 space :)

 

[bugatti79]

ok, that makes sense. Thanks :-)

 

[vela]

Homework Statement

Let V be a real vector space and $${b_1,b_2,b_3,b_4}$$ a linearly independent set of vectors in V
Is the set $$\left \{ b_1,b_2,b_3,b_1+b_4,b_2+b_4 \right \}$$

The Attempt at a Solution

$$\alpha_1b_1+\alpha_2b_2+\alpha_3b_3+\alpha_4\left \{ b_1+b_4 \right \}+\alpha_5\left \{ b_2+b_4 \right \} =0$$

Rearranging

$$b_1(\alpha_1+\alpha_4)+b_2(\alpha_2+\alpha_5)+b_3(\alpha_3)+b_4(\alpha_4+\alpha_5)=0$$
Hence it is NOT linearly independent since the scalars are all not 0...?

You're not quite finished. You need to show that all of the alpha's are necessarily equal to 0 if you want to show those vectors are linearly independent. Since the b's are independent, you know that equation can only be satisfied if
\begin{align*}
\alpha_1+\alpha_4 &= 0 \\
\alpha_2+\alpha_5&= 0 \\
\alpha_3&= 0 \\
\alpha_4+\alpha_5&= 0
\end{align*}If the only solution to this system is $\alpha_i = 0$ for i=1, 2, 3, 4, 5, the vectors are linearly independent. In this case, you'll find there are non-zero solutions to the system, so they are dependent.

And I think that if a set of vectors are linearly independent they span the space correct? And it seems illogical that a set of 4 vectors can span 5 space :)

You're thinking of a basis, which consists of linearly independent vectors that span the space. Any set of vectors, independent or not, will span some space or subspace.

You're close though. There's a theorem that says if you have more vectors than the dimension of the space they're in, they have to be dependent. If you've already covered this theorem, you can use it as well to solve this problem.

 

[bugatti79]

You're not quite finished. You need to show that all of the alpha's are necessarily equal to 0 if you want to show those vectors are linearly independent. Since the b's are independent, you know that equation can only be satisfied if
\begin{align*}
\alpha_1+\alpha_4 &= 0 \\
\alpha_2+\alpha_5&= 0 \\
\alpha_3&= 0 \\
\alpha_4+\alpha_5&= 0
\end{align*}If the only solution to this system is $\alpha_i = 0$ for i=1, 2, 3, 4, 5, the vectors are linearly independent. In this case, you'll find there are non-zero solutions to the system, so they are dependent.

Ok, do I use the Wronskian to find if the determinant is 0. if it is 0 then the vectors are linearly dependent? Thanks

 

[vela]

No, the Wronskian has nothing to do with this problem. I think you simply mean you want to find the determinant of the coefficient matrix. If it's 0, the system is dependent. It's probably less work just to go ahead and solve the system of equations though.

 

[bugatti79]

but can one not use the Wonrskian in this problem? Anyhow, will I solve this using gaussian elimination?

 

[vela]

No, the Wronskian is used to tell if a set of functions is linearly independent. You don't have a set of functions in this problem.

http://en.wikipedia.org/wiki/Wronskian