statement theorem
Eigenvectors with distinct eigenvalues are linearly independent
Statement
Let be a field, let be a vector space over , and let be a linear map. Let be eigenvectors of for eigenvalues that are pairwise distinct. Then are linearly independent.
Remarks
The proof is by induction on : apply to a linear relation, and the term of disappears. So the sum of the eigenspaces of for distinct eigenvalues is a direct sum. In particular, an endomorphism of an -dimensional space has at most 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.
What this node needs for W1
- an inventory row: the row of the inventory that the node accounts for (0 of 1)
The writing levels say how complete the text of a node is. They do not measure its truth.
What this needs
To read this node, read these 8 nodes in this order. The foundations come first, and the list ends with this node.
- no evidence level
- Vector space definition
- Linear combination definition
- Linear map definition
- Linear independence definition
- Endomorphism definition
- Eigenvalue definition
- Eigenvector definition
- Eigenvectors with distinct eigenvalues are linearly independent statement
What depends on this
1 node uses this node, directly or through other nodes.
- Distinct eigenvalues imply diagonalisable 0Y95Z7MT02
Sources
- Web https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors, Eigenspaces, geometric multiplicity, and the eigenbasis
What links here
Used by
- Distinct eigenvalues imply diagonalisable 0Y95Z7MT02