Conjugation of matrices is homomorphic

Dependencies:

  1. Matrix
  2. /complex-numbers/conjugation-is-homomorphic

Conjugation of matrices over complex numbers is homomorphic.

Proof

Let $A$ and $B$ be 2 $m$ by $n$ matrices.

\[ \overline{(A+B)}[i, j] = \overline{(A+B)[i, j]} = \overline{A[i, j] + B[i, j]} = \overline{A}[i, j] + \overline{B}[i, j] = (\overline{A} + \overline{B})[i, j] \]

Therefore, $\overline{A+B} = \overline{A} + \overline{B}$.

Let $c$ be a scalar and $A$ be a matrix.

\[ \overline{(cA)}[i, j] = \overline{(cA)[i, j]} = \overline{c(A[i, j])} = \overline{c}(\overline{A[i, j]}) = \overline{c}(\overline{A}[i, j]) = (\overline{c}\overline{A})[i, j] \]

Therefore, $\overline{cA} = \overline{c}\overline{A}$.

Let $A$ be an $m$ by $p$ matrix and $B$ be a $p$ by $n$ matrix.

\[ \overline{(AB)}[i, j] = \overline{(AB)[i, j]} = \overline{\sum_{k=1}^p A[i, k]B[k, j]} = \sum_{k=1}^p \overline{A[i, k]} \, \overline{B[k, j]} = \sum_{k=1}^p \overline{A}[i, k]\overline{B}[k, j] = (\overline{A} \, \overline{B})[i, j] \]

Therefore, $\overline{AB} = \overline{A}\,\overline{B}$.

Dependency for:

  1. Matrix of orthonormal basis change
  2. Real matrix with real eigenvalues has real eigenvectors

Info:

Transitive dependencies:

  1. /complex-numbers/conjugation-is-homomorphic
  2. Group
  3. Ring
  4. Semiring
  5. Matrix