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The conjugate of vector-scalar multiplication in C^n is the product of the conjugates.

Created over 8 years ago, updated 3 days ago

Theorem: For any scalar (a \in \mathbb{C}) and any vector (\mathbf{z} \in \mathbb{C}^n), the conjugate of their product equals the product of their conjugates:

[\overline{a \cdot \mathbf{z}} = \overline{a} \cdot \overline{\mathbf{z}}]

Proof: For each component (j):

[\overline{(a \cdot \mathbf{z})}_j = \overline{a \cdot z_j} = \overline{a} \cdot \overline{z_j} = (\overline{a} \cdot \overline{\mathbf{z}})_j]

The second equality uses the fact that conjugation distributes over multiplication for complex numbers: (\overline{(a+bi)(c+di)} = \overline{(ac-bd) + (ad+bc)i} = (ac-bd) - (ad+bc)i = (a-bi)(c-di) = \overline{a+bi} \cdot \overline{c+di}).

Example:

  • (a = 1+i), (\mathbf{z} = (2, i))
  • (a \cdot \mathbf{z} = (2+2i, -1+i)) → (\overline{a \cdot \mathbf{z}} = (2-2i, -1-i))
  • (\overline{a} \cdot \overline{\mathbf{z}} = (1-i) \cdot (2, -i) = (2-2i, -1-i)) ✓

Special case: When (a \in \mathbb{R}), we have (\overline{a} = a), so (\overline{a \cdot \mathbf{z}} = a \cdot \overline{\mathbf{z}}).