Does the Equation U·U = ||U||^2 Hold for Complex Vectors?

[SoulofLoneWlf]

Homework Statement

well i am told basicly this
|u|^2 = U dot U except when in using imaginary numbers
i tried using the example below and a few others but it seems to always work for me :/
but i need the process or to show this is not true

Homework Equations

||u|| = U dot U in general not complex :/

The Attempt at a Solution

2i + 5

 

[CompuChip]

What are u and U?
Are they the same (U = u)?
Is u a vector, of which |u| is the norm, or just a number of which |u| is the absolute value (for real numbers) or the modulus (for complex ones). Or is it a matrix, for which |u| is some matrix norm?

If it is a number, it's rather easy to show.
The modulus (length) of 2i + 5 is $\sqrt{2^2 + 5^2}$ which is not equal to (2i + 5)^2.

In general, the correct expression is
|u|^2 = u . u*
where u* is the complex conjugate of u.

 

[nlightner]

j = 2i + 5j

In general, the formula ||u|| = U dot U holds true for any vector u, regardless of whether it contains imaginary numbers or not. This is because the magnitude of a vector, ||u||, is defined as the square root of the sum of the squares of its components. In other words, ||u|| = √(u1^2 + u2^2 + ... + un^2), where u1, u2, ..., un are the components of the vector u.

Similarly, the dot product of two vectors, U dot U, is defined as the product of their corresponding components summed together. In other words, U dot U = u1u1 + u2u2 + ... + unun.

Therefore, when we square both sides of the equation ||u|| = U dot U, we get ||u||^2 = (u1^2 + u2^2 + ... + un^2) = u1u1 + u2u2 + ... + unun = U dot U. This shows that ||u||^2 = U dot U and the original statement, U(dot)U = ||U||^2, is indeed true.

In conclusion, the formula ||u|| = U dot U holds true for all vectors, whether they contain imaginary numbers or not. This is because both ||u|| and U dot U represent the magnitude of a vector, just in different forms.