NLA MT25, Eigenvalue decomposition

Created: October 14, 2025 | Updated: October 14, 2025 | About these notes | View in context | Study these flashcards

Flashcards

@State a theorem that gives any symmetric matrix an eigenvalue decomposition.

Suppose $A \in \mathbb R^{n \times n}$ is a symmetric matrix. Then there is the decomposition

\[A = V \Lambda V^\top\]

where:

  • $V$ is orthogonal
  • $V^\top V = I _ n = V V^\top$
  • $\lambda = \text{diag}(\lambda _ 1, \ldots, \lambda _ n)$ is a diagonal matrix of eigenvalues

@Prove that every symmetric matrix has an eigenvalue decomposition, i.e. that if $A \in \mathbb R^{n \times n}$, then there is a decomposition

\[A = V \Lambda V^\top\]

where:

  • $V$ is orthogonal
  • $V^\top V = I _ n = V V^\top$
  • $\lambda = \text{diag}(\lambda _ 1, \ldots, \lambda _ n)$

@todo.



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