### Hermitian conjugate

The **Hermitian conjugate** of an $m\times n$ matrix $A=(a_{ij})$ over $\mathbb{C}^n$ (or any other vector space induced by a field with conjugates) is defined as the $n\times m$ matrix $A^\ast = (\overline{a_{ji}})$. It is sometimes also referred to as the **conjugate transpose**. A matrix which is equal to its Hermitian conjugate is called a **Hermitian matrix**. On the other hand, matrices equal to the additive inverse of their Hermitian conjugate, so that $A^\ast=-A$, are called **skew-Hermitian**.

**Proposition 1.** For arbitrary matrices $A,B$ with the same dimensions, $(A+B)^\ast = A^\ast + B^\ast$.

**Proposition 2.** For arbitrary matrices $A,B$ for which $AB$ is defined, we have $(AB)^\ast = B^\ast A^\ast$.

Here is an important theorem regarding the properties of Hermitian matrices:

**Theorem 1.** Hermitian matrices have purely real eigenvalues, and the eigenvectors corresponding to distinct eigenvalues are orthogonal.

*Proof.* Let $A$ be a Hermitian matrix, and $\lambda,v$ be an eigenvalue and a corresponding eigenvector of $A$ so that $Av=\lambda v$. Left-multiplying both sides of this equation by $v^\ast$ yields the equality $v^\ast Av = \lambda \lVert v\rVert^2$. However, if we take the Hermitian conjugate of both sides, obtaining $v^\ast A^\ast = \overline{\lambda} v^\ast$, or equivalently $v^\ast A = \overline{\lambda} v^\ast$, and then right-multiply both sides by $v$, we have that $v^\ast A v = \overline{\lambda} \lVert v\rVert ^2$. Thus, we have that $\lambda=\overline{\lambda}$, and so $\lambda$ must be purely real.

Suppose that $v_1,v_2$ are two eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda_1,\lambda_2$, so that $Av_1 = \lambda_1 v_1$ and $Av_2 = \lambda_2 v_2$. Left-multiplying the first equation by $v_2^\ast$ yields the equality and taking the Hermitian conjugate of the second equality and left-multiplying by $v_1$ yields the equality so we have that $\lambda_1 v_2^\ast v_1 = \overline{\lambda_2} v_2^\ast v_1$. Since the eigenvalues $\lambda_1,\lambda_2$ are real and distinct, we may conclude that $v_2^\ast v_1 = 0$ and the two eigenvectors are orthogonal. $\blacksquare$

We can use this to prove the following:

**Theorem 2.** Every Hermitian matrix is unitarily diagonalizable.

However, I haven't quite figured out how to prove this yet.

# linear-algebra

# matrices

back to home page