Alphabeta MathHow statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
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FALSE: A diagonalisable endomorphism must have a characteristic polynomial with distinct roots
Statement
False claim. A diagonalisable endomorphism must have a characteristic polynomial with distinct roots.
Facts & Assumptions
Given: The identity endomorphism of .
The standard unit vectors form an ordered basis of (The standard list with and for is an ordered basis of ; hence , and is the zero space with basis and dimension ).
An endomorphism is diagonalisable when it has a basis of eigenvectors (A diagonalisable endomorphism is one admitting a basis of eigenvectors, equivalently a diagonal matrix representation).
The characteristic polynomial of an operator is that of any matrix representation (The basis-independent characteristic polynomial of an endomorphism of a finite-dimensional space, including in dimension zero).
Refutation
Both standard basis vectors are eigenvectors of with eigenvalue , so [L1] and [L2] make diagonalisable.
Its matrix is diagonal with two entries , so [L3] gives , which has a repeated root.
Depends on
- A diagonalisable endomorphism is one admitting a basis of eigenvectors, equivalently a diagonal matrix representation
- The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$; hence $\dim_F F^{n} = n$, and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$
- The basis-independent characteristic polynomial $\chi_T$ of an endomorphism of a finite-dimensional space, including $\chi_T=1$ in dimension zero
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
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Sources
- Anthony W. Knapp, Basic Algebra, 2nd ed., Ch. V, §3, examples and Theorem 5.14 (standard reference, not scraped)