Question

Determine whether or not the vector functions are linearly dependent or independent.

u=(2-t, t, -2)^T
v=(t, -1, 2)^T
w=(2+t, t-2, 2)^T

Answer 1

What do you suspect? :)
u, v, and w are linearly independent if for any scalars a,b, and c
au + bv + cw = 0 implies a=b=c=0

Answer 2

but how do you test if there are any other solutions for a, b, c other than the trivial solution.

the book says to calculate the wronskian, but i dont know how to do that for 3x3. and you wouldn't use the determinent because these are functions right? so im confused on what to do to test for independence.

Answer 3

Maybe you could find a specific set of scalars?

Answer 4

Im sure you can guess and check and all that, but there has to be a specific way to test.

Can you take the determinent of vector functions? Or does that only work with vectors without variables?

Answer 5

I failed to note that this involved functions :)
I'm at a loss. But they're definitely linearly dependent :D
u + 2v - 2w = 0

Answer 6

Or, you could make the matrix :D
\[\huge \left[\begin{matrix}2-t & t & 2+t\\ t & -1 &t-2 \\ -2 & 2 & 2\end{matrix}\right]\]

Answer 7

you saying to row reduce to find a, b, c.... seems like that could get messy... or are you saying something different?

Answer 8

You could try to get its determinant :)

Answer 9

i did try the determinant, and I did get 0, but i didnt think you could do that for functions....

What is the difference between the wronskian and the determinant?

Answer 10

I don't know what a wronskian is :3

Answer 11

And determinant is absolute :)
If the determinant is zero, then no matter what the entries in the matrix, the columns were linearly dependent :D

Answer 12

sounds good to me. thank you for the help.

Answer 13

No problem :)

Answer 14

But do study the Wronskian, though, whatever that is. It might be essential :)

Answer 15

well i know what the wronskian is, and we used it for functions in ode, but we only used it when it was a 2x2. and in the book it says use the wronskian for a 3x3, the vector functions were much more complex though, and it calculating the determinant seems like alot of work if the functions are easy like this one, so I wondered if there is a different way.

Answer 16

are not easy i mean