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Eigenvalue Decomposition Worked Examples cheat sheet - grade college

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Math Grade college

Eigenvalue Decomposition Worked Examples Cheat Sheet

A printable reference covering eigenvalues, eigenvectors, diagonalization, spectral decomposition, and worked matrix examples for college linear algebra.

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Eigenvalue decomposition rewrites a square matrix in terms of its eigenvalues and eigenvectors. This cheat sheet helps college students follow worked examples without losing track of the algebra. It is especially useful for diagonalizing matrices, solving repeated matrix powers, and understanding symmetric matrices.

The focus is on clear steps, common patterns, and checks that confirm the decomposition is correct.

The main goal is to express a matrix as A=PDP−1A = PDP^{-1} when enough linearly independent eigenvectors exist. The columns of PP are eigenvectors, and DD contains the corresponding eigenvalues on its diagonal. For a real symmetric matrix, the stronger form A=QΛQTA = Q\Lambda Q^T uses orthonormal eigenvectors.

Worked examples usually begin with det⁡(A−λI)=0\det(A - \lambda I) = 0, then solve (A−λI)x=0(A - \lambda I)\mathbf{x} = \mathbf{0} for each eigenspace.

Key Facts

  • An eigenvalue λ\lambda and nonzero eigenvector v\mathbf{v} satisfy Av=λvA\mathbf{v} = \lambda \mathbf{v}.
  • Eigenvalues are found from the characteristic equation det⁡(A−λI)=0\det(A - \lambda I) = 0.
  • For each eigenvalue λ\lambda, eigenvectors come from the null space equation (A−λI)x=0(A - \lambda I)\mathbf{x} = \mathbf{0}.
  • A matrix AA is diagonalizable when it has nn linearly independent eigenvectors for an n×nn \times n matrix.
  • If P=[v1 v2 ⋯ vn]P = [\mathbf{v}_1\ \mathbf{v}_2\ \cdots\ \mathbf{v}_n] and D=diag⁡(λ1,λ2,…,λn)D = \operatorname{diag}(\lambda_1,\lambda_2,\ldots,\lambda_n), then A=PDP−1A = PDP^{-1}.
  • Matrix powers are easier after diagonalization because Ak=PDkP−1A^k = PD^kP^{-1}, where Dk=diag⁡(λ1k,λ2k,…,λnk)D^k = \operatorname{diag}(\lambda_1^k,\lambda_2^k,\ldots,\lambda_n^k).
  • A real symmetric matrix has an orthogonal eigenvalue decomposition A=QΛQTA = Q\Lambda Q^T, where QTQ=IQ^TQ = I.
  • The trace and determinant provide checks because tr⁡(A)=λ1+λ2+⋯+λn\operatorname{tr}(A) = \lambda_1 + \lambda_2 + \cdots + \lambda_n and det⁡(A)=λ1λ2⋯λn\det(A) = \lambda_1\lambda_2\cdots\lambda_n.

Vocabulary

Eigenvalue
An eigenvalue λ\lambda is a scalar that tells how much an eigenvector is stretched or reversed by a matrix transformation.
Eigenvector
An eigenvector v\mathbf{v} is a nonzero vector whose direction is unchanged by the transformation, so Av=λvA\mathbf{v} = \lambda\mathbf{v}.
Characteristic Polynomial
The characteristic polynomial is p(λ)=det⁡(A−λI)p(\lambda) = \det(A - \lambda I), whose roots are the eigenvalues of AA.
Diagonalization
Diagonalization is the process of writing a matrix as A=PDP−1A = PDP^{-1} using a basis of eigenvectors.
Eigenspace
The eigenspace for λ\lambda is the set of all vectors satisfying (A−λI)x=0(A - \lambda I)\mathbf{x} = \mathbf{0}, including 0\mathbf{0}.
Orthogonal Diagonalization
Orthogonal diagonalization writes a real symmetric matrix as A=QΛQTA = Q\Lambda Q^T using orthonormal eigenvectors.

Common Mistakes to Avoid

  • Using det⁡(λI−A)\det(\lambda I - A) in one step and det⁡(A−λI)\det(A - \lambda I) in another, then mixing signs, is wrong because inconsistent characteristic polynomials can change intermediate coefficients.
  • Forgetting that eigenvectors cannot be 0\mathbf{0} is wrong because A0=λ0A\mathbf{0} = \lambda\mathbf{0} holds for every λ\lambda and gives no direction information.
  • Putting eigenvalues in DD in a different order than the eigenvectors in PP is wrong because each column vi\mathbf{v}_i must match the diagonal entry λi\lambda_i.
  • Assuming repeated eigenvalues always give enough eigenvectors is wrong because algebraic multiplicity can be larger than geometric multiplicity.
  • Using Q−1Q^{-1} instead of QTQ^T without checking orthonormality is wrong because Q−1=QTQ^{-1} = Q^T only when QTQ=IQ^TQ = I.

Practice Questions

  1. 1 Find the eigenvalues of A=[2112]A = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix} using det⁡(A−λI)=0\det(A - \lambda I) = 0.
  2. 2 For A=[4013]A = \begin{bmatrix}4 & 0 \\ 1 & 3\end{bmatrix}, find eigenvectors for each eigenvalue and write A=PDP−1A = PDP^{-1} if possible.
  3. 3 Given A=PDP−1A = PDP^{-1} with D=[200−1]D = \begin{bmatrix}2 & 0 \\ 0 & -1\end{bmatrix}, compute D5D^5 and explain how it helps find A5A^5.
  4. 4 Explain why a real symmetric matrix can be diagonalized using orthonormal eigenvectors, and describe why this makes A=QΛQTA = Q\Lambda Q^T easier to use than A=PDP−1A = PDP^{-1}.