Math Problem Statement

Define Hermitian matrix in C^n. Show that all the characteristic roots of a Hermitian matrix are real. OR Let F be a field with Char(F) = 0, and T ∈ A_F(V) be such that tr(T^i) = 0 for all i ≥ 1. Prove that T is nilpotent.

Solution

Let's solve both parts systematically:


Part 1: Definition of Hermitian Matrix and Proof of Real Eigenvalues

Definition of Hermitian Matrix:

A Hermitian matrix AA in Cn×n\mathbb{C}^{n \times n} satisfies: A=AA = A^* where AA^* denotes the conjugate transpose of AA. This means: aij=ajia_{ij} = \overline{a_{ji}} for all i,ji, j.

Proof: Eigenvalues of Hermitian Matrices are Real

  1. Let AA be a Hermitian matrix, and let λ\lambda be an eigenvalue of AA with eigenvector x0x \neq 0. Thus: Ax=λxA x = \lambda x

  2. Taking the inner product of AxA x with xx, we get: Ax,x=λx,x\langle A x, x \rangle = \langle \lambda x, x \rangle Since λ\lambda is a scalar: λx,x=λx,x\langle \lambda x, x \rangle = \lambda \langle x, x \rangle

  3. The Hermitian property implies: Ax,x=x,Ax\langle A x, x \rangle = \langle x, A x \rangle

  4. Substituting Ax=λxA x = \lambda x, this becomes: λx,x=λx,x\langle \lambda x, x \rangle = \lambda \langle x, x \rangle

  5. Since x,x>0\langle x, x \rangle > 0 (the norm of xx is nonzero), divide through by x,x\langle x, x \rangle: λ=λ\lambda = \overline{\lambda}

    Hence, λ\lambda is real.


Part 2: Proof of Nilpotency of TT

Given:

  1. Char(F)=0\text{Char}(F) = 0
  2. TAF(V)T \in A_F(V) satisfies tr(Ti)=0\text{tr}(T^i) = 0 for all i1i \geq 1.

Proof:

  1. Consider the Jordan canonical form of TT. Since TT is a linear transformation over a field of characteristic zero, it is similar to a block diagonal matrix consisting of Jordan blocks.

  2. The trace of TiT^i is the sum of λi\lambda^i over all eigenvalues λ\lambda of TT. Since tr(Ti)=0\text{tr}(T^i) = 0 for all i1i \geq 1, this implies: λi=0for all i1.\sum \lambda^i = 0 \quad \text{for all } i \geq 1.

  3. This system implies that all eigenvalues of TT are zero. To see this, note that if any eigenvalue λ0\lambda \neq 0, then λi0\sum \lambda^i \neq 0.

  4. Since all eigenvalues are zero, TT is nilpotent. Specifically, all Jordan blocks of TT correspond to the eigenvalue 00, meaning Tk=0T^k = 0 for some kk.


Do you have further questions or need detailed clarification on any part?


Related Questions:

  1. Can you explain why Hermitian matrices appear in quantum mechanics?
  2. How does the characteristic polynomial confirm the nilpotency of TT?
  3. Can a Hermitian matrix have complex eigenvalues in any case?
  4. What are examples of fields with Char(F)=0\text{Char}(F) = 0?
  5. How is the trace of a matrix invariant under similarity transformations?

Tip:

In linear algebra, analyzing eigenvalues often simplifies the problem significantly—always check their properties first!

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Matrix Theory
Hermitian Matrices
Nilpotent Matrices

Formulas

Hermitian condition: A = A* (A* is the conjugate transpose)
Trace property: tr(T^i) = sum of eigenvalues raised to the power i

Theorems

Eigenvalues of Hermitian matrices are real
Nilpotent matrices have eigenvalue 0 and satisfy T^k = 0 for some k

Suitable Grade Level

Undergraduate Mathematics