Linear Algebra: Eigenvalues & Eigenvectors

Av = \lambda v with v \neq 0 . The matrix only stretches (or flips) v by the factor \lambda ; its direction is unchanged.

\det(A - \lambda I) = 0 . Its roots are the eigenvalues.

(4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 10 = 0 \lambda = 2 and \lambda = 5

A - 5I = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix} , so v = \begin{pmatrix} 1 \\ 1 \end{pmatrix} (or any nonzero multiple)

The entries on the main diagonal.

Sum = \text{tr}(A) (sum of the diagonal) Product = \det A

The null space of A - \lambda I : all eigenvectors for \lambda together with the zero vector.

A is singular (not invertible), because \det A = product of eigenvalues = 0 .

A = PDP^{-1} , where the columns of P are eigenvectors and D is diagonal with the matching eigenvalues.

When it has n linearly independent eigenvectors. Having n distinct eigenvalues is enough (but not required).

A^k = P D^k P^{-1} , and D^k just raises each diagonal entry to the k th power.

\lambda^k , \dfrac{1}{\lambda} and \lambda + c (same eigenvectors)

A real symmetric matrix has real eigenvalues and can be orthogonally diagonalized: A = QDQ^T with orthonormal eigenvectors.

Algebraic: how many times λ is a root of the characteristic polynomial. Geometric: the dimension of its eigenspace. Geometric ≤ algebraic, and the matrix is diagonalizable exactly when they are equal for every λ.

No. \lambda^2 + 1 = 0 gives \lambda = \pm i . A 90° rotation leaves no real direction unchanged.

Principal component analysis (PCA), Markov chain steady states (eigenvalue 1), Google's PageRank, vibration modes in engineering, and stability of differential equations.