This question discusses a special case of matrices where every non-zero vector is an eigenvector of the matrix $A$. Let's analyze each statement carefully:

1. All eigenvalues of $A$ are equal:

   • If every non-zero vector in $\mathbb{C}^n$ is an eigenvector of $A$, then $A$ must act as a scalar multiple of the identity matrix $I$. This implies $A = \lambda I$, where $\lambda$ is a scalar (eigenvalue of $A$). Consequently, all eigenvalues are equal to $\lambda$.
   • This statement is TRUE.

2. All eigenvalues of $A$ are distinct:

   • Since all eigenvalues are equal (from the reasoning above), this statement is false.
   • This statement is FALSE.

3. $A = \lambda I$ for $\lambda \in \mathbb{C}$, where $I$ is an $n \times n$ identity matrix:

   • The fact that every non-zero vector is an eigenvector necessarily implies $A = \lambda I$, as explained earlier. Thus, $A$ must be a scalar multiple of the identity matrix.
   • This statement is TRUE.

4. If $\chi_A$ and $m_A$ denote the characteristic polynomial and minimal polynomial of $A$, then $\chi_A = m_A$:

   • For $A = \lambda I$, the characteristic polynomial $\chi_A(t) = (t - \lambda)^n$ and the minimal polynomial $m_A(t) = t - \lambda$. Clearly, $\chi_A \neq m_A$ unless $n = 1$.
   • This statement is FALSE.

Correctness of Statements

• True: 1, 3
• False: 2, 4

Would you like a detailed breakdown of any specific point?

Related Questions:

1. What does it mean for every vector in $\mathbb{C}^n$ to be an eigenvector of a matrix?
2. How can we determine the minimal polynomial for matrices like $A$?
3. What are the implications if $\chi_A = m_A$ in general?
4. How do eigenvalues affect the diagonalization of a matrix?
5. What is the characteristic polynomial of a scalar matrix like $\lambda I$?

Tip: For matrices $A$ where $A = \lambda I$, eigenvectors are independent of the direction, making computations simpler!