Answer:
(-5 1 0 0) and (-3 0 1 0)
Step-by-step explanation:
Based on my memory, you can use the dot product to find orthogonality meaning
u1 . u2 = 0
In this case, we let u be the arbitrary vector represented by (x y z w)
u . (1 5 3 0) can be expressed as an equation given by:
{(x y z w) | x + 5y + 3z + 0w= 0}
We solve this equation letting y = s and z = t and x = -5s -3t
Reconstruct this according to their variable will give
(-5 1 0 0)s + (-3 0 1 0)t = 0
So the basis vectors are (-5 1 0 0) and (-3 0 1 0).
Let c1 and c2 be arbitrary constants:
if you solve for c1(-5 1 0 0) + c2(-3 0 1 0) = 0
c1 and c2 should have only trivial solution, which proves its
linear independence and so it forms a basis vector space.
