Prove or disprove:
- If is orthogonal to , then
- If
then is orthogonal to .
- Proof. Let and . Then we have
But, by hypothesis, so . Therefore,
since for all real
- Proof. Suppose otherwise, that . Then, we know that for all we have
for all . So, in particular is must hold for both
Both of these are nonzero since by assumption. But then,
and
But since we cannot have both and . Hence, this is a contradiction, so we must have .
b is false, take for example x=1, and A=kB. The inequality holds without A beeing orthogonal to B (for infinite values of k).
I don’t know what you mean, but here is a proof that b is true:
If B=0, the statement is clearly true, hence suppose B is not zero then norm(B)>0 now norm(A+xB)=sqrt(norm(A)^2 + 2x(A.B) + x^2norm(B)^2) >= norm(A) implies that 2x(A.B) + x^2norm(B)^2 >=0 and this is true for any real x, hence for any x>0 we may “solve” for x to get x >= -2(A.B)/norm(B)^2 this means that (A.B)>=0 otherwise, the term on the right will be a positive number less than or equal to all positive numbers which is absurd (There is a proof for this in the introduction). Similarly, if x <= 0 then by "solving" for x we get the similar inequality x <= -2(A.b)/norm(B)^2 this one implies that (A.B)=0 for otherwise -(A.B)/norm(B)^2 will be a negative number less than or equal to all negative numbers, this completes the proof.
Between the 4th and 5th formula, both side were divided by x. What if x is negative?
In (b)