which of the following are linearly independent, given that each set is a subset of vectors from a standard vector space:
{ −2+x, x^2−2 x+3, x^2−3, −x }
{ [2,7,7], [0,6,−4], [0,−2,0] }
{ sin( 2 x ) , cos( x ) , e3 x, 1, 3 x, x2 }

Question

which of the following are linearly independent, given that each set is a subset of vectors from a standard vector space:
{ −2+x,    x^2−2 x+3,    x^2−3,    −x }
{ [2,7,7],    [0,6,−4],    [0,−2,0] }
{ sin( 2 x ) ,    cos( x ) ,    e3 x,     1,    3 x,    x2 }

anonymous · Answer

$x^{2} - 2x + 3 - (x^{2} - 3) + 3(-2 + x) - x = 0$, check my math

anonymous · Answer

Also: {$$\left[\begin{matrix}-2 & 1 \ 3 & 6\end{matrix}\right] , \left[\begin{matrix}0 & -1 \ 0 & 2\end{matrix}\right] , \left[\begin{matrix}0 & 0 \ -6 & 0\end{matrix}\right] , \left[\begin{matrix}-9 & 0 \ 0 & -5\end{matrix}\right] , \left[\begin{matrix}0 & -6 \ -4 & 0\end{matrix}\right]$$ }

zarkon · Answer

no need to check the first one..the basis for P_2 has 3 vectors...there are 4 vectors there...thus they have to form a dependent set

anonymous · Answer

Good point.

anonymous · Answer

Ee-gads I have no idea what that means! :S :(

anonymous · Answer

You can get every quadratic polynomial by taking multiples of $x^{2}, x, 1$

anonymous · Answer

Each of those terms being a "vector"

anonymous · Answer

Oooook - yep, sure - thanks :) (sorry for the delay - was writing this down :) )

anonymous · Answer

Do you understand the next one?

anonymous · Answer

Erm ... the second one IS linearly independent, because 2*0*0 + 7*6*-2 + 7*-4*0 /= 0 ??

anonymous · Answer

Is that correct? :/

anonymous · Answer

No.  What you want to try and do is express any of the vectors in terms of the vectors that come before it in the list (it is linearly independent)

anonymous · Answer

Can you write $[0, 6, -4]$ as a multiple of $[2, 7, 7]$ (not the 0 multiple)?

anonymous · Answer

No you cant? (sorry I'm not so good at this :( appreciating the help!)

anonymous · Answer

Sorry, I had to relocate.  No, you can't because there is no way you can get 2 times anything to be 0.

anonymous · Answer

Next we try to get $$\begin{bmatrix}0\-2\0\end{bmatrix} = a\begin{bmatrix}2\7\7\end{bmatrix} + b\begin{bmatrix}0\6\-4\end{bmatrix}$$.  Can you find an $a,b$?

anonymous · Answer

Does a solution exist for this? Other than zero? :'( *tears hair out*

anonymous · Answer

Nope, exactly.  If you pick anything other than 0 for $a$, then you will have a nonzero entry at the top (you can't cancel it with the second vector, because it will have a zero entry at the top).

anonymous · Answer

So $a = 0$.  But now, no matter what $b$ you pick, the bottom entry will be nonzero, so $b$ also has to be 0 and they are linearly independent.

anonymous · Answer

Notice that if we had found a nonzero a,b, we could have moved $[0, -2, 0]$ to the other side to match the definition of linear dependence.

anonymous · Answer

You have no idea how much of a hero you are to me at the moment!!!

anonymous · Answer

Thank you.  It's going to be the same kind of idea for the other parts.  It is always easier to prove linear dependence than independence (in my opinion).

anonymous · Answer

OK, so 1 = linearly dependent, 2 = linearly independent....
Is it too much to ask for help on 3 and 4? I didn't realize how long this would be, and you've already been such a great help!

anonymous · Answer

From the looks, I would guess independent and dependent, but I actually have to run to class now, sorry.

anonymous · Answer

I'm getting the hang of the theory, but now its the form of the others thats getting to me... How do you work with individual matrices etc...?

anonymous · Answer

OK no worries! thanks so much!!

anonymous · Answer

For the last one, can you think of 4 matrices that will give you all of your $2\times2$ matrices?  Then you know that 5 will be redundant because all bases have the same number of elements.