Hey widenerl194.
Hint: Remember that {x,y,z,w} is independent iff for real numbers a,b,c,d that ax + by + cz + dw = 0 holds iff a,b,c,d are all 0. You can assume the statement for {w,x,y,z} now try adjusting this to get the next statement.
I'm not sure where to start with this: (these are all matrices -- are supposed to have bars over them, but couldn't figure out how to insert those)
Prove that if {w, x, y, z} is a linearly dependent subset of V, then so is {w+z, x+y, y+z, w-z}.
Any help or guidance would be much appreciated!!
Hey widenerl194.
Hint: Remember that {x,y,z,w} is independent iff for real numbers a,b,c,d that ax + by + cz + dw = 0 holds iff a,b,c,d are all 0. You can assume the statement for {w,x,y,z} now try adjusting this to get the next statement.
Okay so this is what I did:
Let {w, x, y, z} be a linearly independent subset of V. For a,b,c,d, in the reals, aw+bx+cy+dz=0 because a=b=c=d=0.
We want to show {w+x, x+y, y+z, w-z} is linearly independent.
So, a(w+x) + b(x+y) + c(y+z) + d(w-z) = 0.
Then, (a+d)w + (a+b)x + (b+c)y + (c-d)z = 0.
So then we have: a + d=0, a+b=0, b+c=0, c-d=0. I put that into an augmented matrix and row reduced getting
[1 0 0 0]
[0 1 0 0]
[0 0 0 1]
[0 0 1 0]
So, a=b=c=d=0. Therefore {w+x, x+y, y+z, w-z} is linearly independent.