show that if u (dot) v = 0 for all vectors v, then u = 0.
One of the Axioms an inner product has to satisfy is:
x dot x >=0 where equality only holds if x = 0
So, in your problem you take the special case v = u. Then:
u dot u = 0 ---->
u = 0
To prove that if u dot v = 0 for all vectors v, then u = 0, you can use the properties and axioms of the inner product.
First, let's consider the property of the inner product that states:
For any vector v, the inner product of v with itself (v dot v) is greater than or equal to 0, and it is equal to 0 only if v is the zero vector.
Now, if u dot v = 0 for all vectors v, we can choose v = u. Substituting this into the given condition, we get:
u dot u = 0
According to the property mentioned earlier, the inner product of a vector with itself is greater than or equal to 0, and it is equal to 0 only if the vector is the zero vector.
Since u dot u = 0 in this case, it implies that u is the zero vector. Therefore, if u dot v = 0 for all vectors v, then u must be the zero vector.
To prove that if u(dot)v = 0 for all vectors v, then u = 0, we can use a proof by contradiction.
Assume that u is not equal to 0. This means that there exists a nonzero vector u.
Let v = u. Since v = u, we have u(dot)v = u(dot)u.
Given that u(dot)v = 0 for all vectors v, we have u(dot)u = 0.
However, the property of the dot product states that for any vector x, x(dot)x >= 0, with equality holding only when x = 0.
Since u(dot)u = 0, this implies that u = 0.
But this contradicts our initial assumption that u is not equal to 0.
Therefore, our assumption that u is not equal to 0 leads to a contradiction, and we can conclude that if u(dot)v = 0 for all vectors v, then u = 0.