L2 norm for complex valued vector

[ericm1234]

Let's say I have a vector (4+2i, 1-i), how do I take an L2 norm?
Dont tell me I simply do sqrt(16+4+1+1)..?

 

[jbunniii]

Let's say I have a vector (4+2i, 1-i), how do I take an L2 norm?
Dont tell me I simply do sqrt(16+4+1+1)..?

The short answer is yes, you simply do sqrt(16+4+1+1). Here is why:

The L2 norm of a vector $v$ is $\sqrt{v^* v}$, where $v^*$ is the conjugate transpose. In your case,
$$v = \left[\begin{matrix} 4 + 2i \\ 1 - i \end{matrix}\right]$$
so
$$v^* = \left[\begin{matrix} 4 - 2i & 1 + i \end{matrix}\right]$$
and
$$v^* v = \left[\begin{matrix} 4 - 2i & 1 + i \end{matrix}\right] \left[\begin{matrix} 4 + 2i \\ 1 - i \end{matrix}\right] = (4-2i)(4+2i) + (1+i)(1-i) = 16 + 4 + 1 + 1$$
and therefore
$$||v||_2 = \sqrt{v^H v} = \sqrt{16 + 4 + 1 + 1}$$