Math Problem Statement

Define Hermitian matrix in Cn. Show that all the characteristic roots of a Hermitian matrix are real.

Solution

Definition of Hermitian Matrix in Cn\mathbb{C}^n

A Hermitian matrix AA is a square matrix ACn×nA \in \mathbb{C}^{n \times n} (complex nn-dimensional space) that satisfies the following condition:

A=AA = A^*

where AA^* is the conjugate transpose (also known as Hermitian transpose) of AA. In other words, if A=[aij]A = [a_{ij}], then:

A=AT=[aji],A^* = \overline{A}^T = [\overline{a_{ji}}],

where aji\overline{a_{ji}} denotes the complex conjugate of ajia_{ji}. Hence, AA is Hermitian if:

aij=ajifor all i,j.a_{ij} = \overline{a_{ji}} \quad \text{for all } i, j.

Proof: Characteristic Roots of a Hermitian Matrix Are Real

Let AA be a Hermitian matrix, and let λ\lambda be a characteristic root (eigenvalue) of AA. This means there exists a non-zero vector vCn\mathbf{v} \in \mathbb{C}^n such that:

Av=λv.A \mathbf{v} = \lambda \mathbf{v}.

To show that λ\lambda is real, consider the following steps:

  1. Compute the inner product of both sides: Take the inner product of AvA \mathbf{v} with v\mathbf{v}:

    Av,v=λv,v.\langle A \mathbf{v}, \mathbf{v} \rangle = \langle \lambda \mathbf{v}, \mathbf{v} \rangle.

    Using the properties of inner products, λv,v=λv,v\langle \lambda \mathbf{v}, \mathbf{v} \rangle = \lambda \langle \mathbf{v}, \mathbf{v} \rangle.

    Hence: Av,v=λv,v.\langle A \mathbf{v}, \mathbf{v} \rangle = \lambda \langle \mathbf{v}, \mathbf{v} \rangle.

  2. Express the inner product Av,v\langle A \mathbf{v}, \mathbf{v} \rangle: Since AA is Hermitian, A=AA^* = A, and by the definition of the inner product:

    Av,v=v,Av.\langle A \mathbf{v}, \mathbf{v} \rangle = \langle \mathbf{v}, A \mathbf{v} \rangle.

    Substituting Av=λvA \mathbf{v} = \lambda \mathbf{v}, this becomes:

    v,Av=v,λv=λv,v,\langle \mathbf{v}, A \mathbf{v} \rangle = \langle \mathbf{v}, \lambda \mathbf{v} \rangle = \overline{\lambda} \langle \mathbf{v}, \mathbf{v} \rangle,

    where λ\overline{\lambda} is the complex conjugate of λ\lambda.

  3. Equating the two expressions for Av,v\langle A \mathbf{v}, \mathbf{v} \rangle: From step 1, Av,v=λv,v\langle A \mathbf{v}, \mathbf{v} \rangle = \lambda \langle \mathbf{v}, \mathbf{v} \rangle, and from step 2, Av,v=λv,v\langle A \mathbf{v}, \mathbf{v} \rangle = \overline{\lambda} \langle \mathbf{v}, \mathbf{v} \rangle.

    Therefore: λv,v=λv,v.\lambda \langle \mathbf{v}, \mathbf{v} \rangle = \overline{\lambda} \langle \mathbf{v}, \mathbf{v} \rangle.

  4. Simplify to conclude λ\lambda is real: Since v,v\langle \mathbf{v}, \mathbf{v} \rangle is the norm of v\mathbf{v}, it is a positive real number. Dividing through by v,v\langle \mathbf{v}, \mathbf{v} \rangle:

    λ=λ.\lambda = \overline{\lambda}.

    This implies that λ\lambda is real.

Conclusion

Every characteristic root (eigenvalue) of a Hermitian matrix AA is real.


Let me know if you would like further clarifications or examples! Here are some related questions you might explore:

  1. What are some properties of eigenvalues and eigenvectors of Hermitian matrices?
  2. Can a Hermitian matrix have complex eigenvalues?
  3. How do Hermitian matrices relate to symmetric matrices over Rn\mathbb{R}^n?
  4. What is the spectral theorem for Hermitian matrices?
  5. How are unitary matrices connected to Hermitian matrices?

Tip: When working with Hermitian matrices, always leverage the fact that their eigenvectors corresponding to distinct eigenvalues are orthogonal.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Hermitian Matrices
Eigenvalues and Eigenvectors
Inner Product Spaces

Formulas

A = A* (Hermitian condition)
λ⟨v, v⟩ = ⟨Av, v⟩ (eigenvalue property)

Theorems

Hermitian Matrix Property: Eigenvalues of Hermitian matrices are real

Suitable Grade Level

Undergraduate