Norm of the sum of complex vectors

Question

Say that we have $n$ complex-valued vectors $\mathbf{z}_{i}$ and we want to evaluate:

$$\left\|\sum_{i=1}^{n} \mathbf{z}_{i}\right\|_2^{2}$$

Now, I know that for real vectors $\mathbf{x}_{i}$ it holds:

$$\left\|\sum_{i=1}^{n} \mathbf{x}_{i}\right\|_2^{2}=\sum_{i=1}^{n}\left\|\mathbf{x}_{i}\right\|_2^{2}+\sum_{i \neq j} \mathbf{x}_{i} \cdot \mathbf{x}_{j}$$ But when dealing with complex vectors, the last term with the inner product will be a complex number in general so I am not sure that this formula generalizes for this case. What am I missing?

See https://math.stackexchange.com/questions/53148/interpretation-of-dot-product-of-complex-vectors — Chrystomath, Sep 14 '20 at 11:45

score 2 · Accepted Answer · answered Sep 14 '20 at 12:04

2

Generalizing @enzotib's answer to $n$ terms, since $\Vert z\Vert_2^2=z^\ast\cdot z$ we have$$\left\Vert\sum_iz_i\right\Vert_2^2=\sum_iz_i^\ast\cdot\sum_jz_j=\sum_{ij}z_i^\ast\cdot z_j=\sum_iz_i^\ast\cdot z_i+\sum_{i\ne j}z_i^\ast\cdot z_j.$$

answered Sep 14 '20 at 12:04

J.G.

115,835

Shouldn't the last term be the real part of that dot product? – Leonardo Massai Sep 14 '20 at 12:24
1

@LeonardoMassai it's real anyway because the term is $\sum_{i<j}(z_i^\ast\cdot z_j+z_j^\ast z_i)=2\Re\sum_{i<j}z_i^\ast\cdot z_j$. – J.G. Sep 14 '20 at 12:34
+1, more general – Vincenzo Tibullo Sep 14 '20 at 13:39

score 1 · Answer 2 · answered Sep 14 '20 at 11:42

1

For two complex numbers $$ \|z_1+z_2\|^2=(z_1+z_2)^*(z_1+z_2)=\|z_1\|^2+\|z_2\|^2+(z_1^*z_2+z_1z_2^*)=\\ \|z_1\|^2+\|z_2\|^2+2Re(z_1^*z_2) $$ easily extended to $n$.

answered Sep 14 '20 at 11:42

Vincenzo Tibullo

10,955

Norm of the sum of complex vectors

2 Answers2