Math 2.0

statement theorem

Eigenvectors with distinct eigenvalues are linearly independent

Statement

Let FF be a field, let VV be a vector space over FF, and let T ⁣:V→VT \colon V \to V be a linear map. Let v1,…,vkv_1, \dots, v_k be eigenvectors of TT for eigenvalues λ1,…,λk\lambda_1, \dots, \lambda_k that are pairwise distinct. Then v1,…,vkv_1, \dots, v_k are linearly independent.

Remarks

The proof is by induction on kk: apply T−λkid⁡VT - \lambda_k \id_V to a linear relation, and the term of vkv_k disappears. So the sum of the eigenspaces of TT for distinct eigenvalues is a direct sum. In particular, an endomorphism of an nn-dimensional space has at most nn distinct eigenvalues.

Explanations

L1 human

Eigenvectors for different eigenvalues point in genuinely different directions. None of them is a combination of the others. This is the key fact behind diagonalisation.

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Vector space, definition, no evidence levelVector spacedefinitionLinear combination, definition, no evidence levelLinear combinationdefinitionLinear map, definition, no evidence levelLinear mapdefinitionLinear independence, definition, no evidence levelLinear independencedefinitionEndomorphism, definition, no evidence levelEndomorphismdefinitionEigenvalue, definition, no evidence levelEigenvaluedefinitionEigenvector, definition, no evidence levelEigenvectordefinitionEigenvectors with distinct eigenvalues are linearly independent, statement, no evidence levelEigenvectors with dis…statement
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  1. Vector space definition
  2. Linear combination definition
  3. Linear map definition
  4. Linear independence definition
  5. Endomorphism definition
  6. Eigenvalue definition
  7. Eigenvector definition
  8. Eigenvectors with distinct eigenvalues are linearly independent statement

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Sources

  1. Web https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors, Eigenspaces, geometric multiplicity, and the eigenbasis

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