Linear Algebra

This problem is easy - first we want = 0, which means

3 u1 v1 + 2 u2 v2 = 0

But we also want them to be not orthogonal in the Euclidean sense, which means

u1 v1 + u2 v2 =/= 0

Simply scaling the second equation by 2. We know u1 v1 cannot be 0. To make things easy let make them integers first, and it has to be even because we want to other side to be integers, too. So let's pick (2, 2) and try.

3 (2) (2) + 2 u2 v2 = 0

Now it is obvious! Let just pick (-3, 2) for (u2, v2), so we get orthogonal vectors. <(2, -3), (2, 2)> = 0, call these vectors x', y'.

To make them othronormal. All we need to do is to normalize it.

The norm of x' is sqrt(<(2, -3), (2, -3)>) = sqrt(30) - so to normalize x', we have
x = 1/sqrt(30) (2, -3)

The norm of y' is sqrt(<(2, 2), (2, 2)>) = sqrt(20) - so to normalize y', we have
y = 1/sqrt(20) (2, 2)

These are the answer
x = 1/sqrt(30) (2, -3).
y = 1/sqrt(20) (2, 2).

A simple check can verify these answers are correct. The vectors are in fact orthonormal in the new inner product and not even orthogonal in Euclidean sense.

i don't know sorry

你要求 ||x|| = 1, ||y|| = 1, and = 0

Let x = (a, b), y = (c, d)

a² + b² = 1
c² + d² = 1
3ac + 2bd = 0
ac + bd ≠ 0

是這樣嗎？

2015-01-27 15:01:26 補充：
以上是指 < x , y > = 0.