#prove it's linearly independent
41 messages · Page 1 of 1 (latest)
- Wait patiently for a helper to come along.
- Once someone helps you, say thank you and close the thread with:
+close
- Feel free to nominate the person for helper of the week in #helper-nominations
- Do not ping the mods, unless someone is breaking the rules.
- If you're happy with the help you got here, and the server overall, you can contribute financially as well:
If u and v are linearly independent, what do a+b and a-b have to equal here?
is it a=b=0 so a-b=0 and a+b=0?
Yep
What happens when you add the two equations together
Add $a-b = 0$ to $a + b = 0$
MicMac
its zero
On the right it's zero and on the left its (a-b) + (a+b), i.e., $(a - b) + (a + b) = 0$
MicMac
What does that simplify to?
2a?
MicMac
so i can just get rid of the u and v?
Not exactly. Your goal should be to show that if $a(u + v) + b(u - v) = 0$, then $a = 0$ and $b = 0$.
MicMac
At this point, you're halfway there (more than that really) because we just saw that a must equal 0
Now you just have to show that b = 0, which will be way easier than everything else so far.
But if you mean that we don't have to do anything else with u and v in this problem, you're totally right
All you gotta do is show that both a and b are zero
is it impossible for u or v to be 0
Yeah, they can't be zero because the 0-vector is linearly dependent with everything
Since u and v are linearly independent they both gotta be nonzero
Like if $u$ were the zero vector, then $cu$ would equal zero for any real number c. But linear independence requires that to happen only when c is zero.
MicMac
ohh i see
but i dont understand how we take 0 = u(a+b) + v(a-b) and turn it into 0 = (a+b) + (a-b)
So, you said back at the beginning that since u and v are linearly independent, then the numbers multiplying them here, a+b and a-b, have to be 0
So you made two equations $a+b = 0$ and $a-b = 0$
MicMac
I suggested that you add the two equations together to eliminate b
ohhh sorry i get it now
It's all good
thank u so much micmac
Sure thing! Take care