Suppose that \(\lambda\) and \(\rho\) are two different eigenvalues of the square matrix \(A\). We want to prove that the intersection of the eigenspaces for these two eigenvalues is trivial, i.e., \(E_A(\lambda) \cap E_A(\rho) = \{0\}\).
Assume the opposite, that there exists a non-zero vector \(v\) in both \(E_A(\lambda)\) and \(E_A(\rho)\). This implies:
\[ Av = \lambda v \quad \text{and} \quad Av = \rho v \]
Combining these equations, we get:
\[ \lambda v = \rho v \]
Rearranging terms:
\[ (\lambda - \rho) v = 0 \]
If \(v\) is non-zero, this implies \(\lambda = \rho\). However, we started with the assumption that \(\lambda\) and \(\rho\) are different eigenvalues, leading to a contradiction.
Therefore, the intersection of \(E_A(\lambda)\) and \(E_A(\rho)\) must be trivial, and we conclude that \(E_A(\lambda) \cap E_A(\rho) = \{0\}\).