Eigenvalues, Trace & Determinant

For a square matrix $A$, an eigenvalue $\lambda$ and eigenvector $v$ satisfy $Av = \lambda v$. The eigenvalues are the roots of the characteristic polynomial $\det(A - \lambda I) = 0$. The two golden identities for an $n \times n$ matrix with eigenvalues $\lambda_1, \ldots, \lambda_n$: $$\text{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n$$ $$\det(A) = \lambda_1 \cdot \lambda_2 \cdots \lambda_n$$ Concrete example: A $2 \times 2$ matrix has trace 5 and determinant 6. What are the eigenvalues? They satisfy $\lambda_1 + \lambda_2 = 5$ and $\lambda_1 \cdot \lambda_2 = 6$, so they are the roots of $\lambda^2 - 5\lambda + 6 = 0$, giving $\lambda = 2$ and $\lambda = 3$. Key properties: - $\text{tr}(AB) = \text{tr}(BA)$ (cyclic property) - Eigenvalues of $A^k$ are $\lambda_i^k$; eigenvalues of $A^{-1}$ are $1/\lambda_i$ - $A$ is invertible $\iff$ no eigenvalue is zero $\iff$ $\det(A) \ne 0$ - For symmetric matrices, all eigenvalues are real and eigenvectors are orthogonal When to use: Quant interviews heavily test the trace-determinant shortcuts for $2 \times 2$ matrices. For larger matrices, eigenvalue reasoning appears in Markov chains (stationary distribution is the eigenvector for $\lambda=1$), portfolio optimization (covariance matrix eigenvalues measure risk directions), and PCA (principal components are eigenvectors of the covariance matrix). Rayleigh quotient: For symmetric $A$, the maximum of $x^T A x / x^T x$ over nonzero $x$ equals the largest eigenvalue. This connects eigenvalues to optimization.