Editing Characteristic polynomial (section)

==Characteristic polynomial of ''A''<sup>''k''</sup>==

If <math>\lambda</math> is an eigenvalue of a square matrix <math>A</math> with eigenvector <math>\mathbf{v},</math> then <math>\lambda^k</math> is an eigenvalue of <math>A^k</math> because
<math display=block>A^k \textbf{v} = A^{k-1} A \textbf{v} = \lambda A^{k-1} \textbf{v} = \dots = \lambda^k \textbf{v}.</math>

The multiplicities can be shown to agree as well, and this generalizes to any polynomial in place of <math>x^k</math>:<ref>{{Cite book | last1=Horn | first1=Roger A. | last2=Johnson | first2=Charles R. | title=Matrix Analysis | publisher=[[Cambridge University Press]] | isbn=978-0-521-54823-6 | year=2013 |edition=2nd|at=pp. 108–109, Section 2.4.2}}</ref>
{{math theorem 
  | name = Theorem
  | Let <math>A</math> be a square <math>n \times n</math> matrix and let <math>f(t)</math> be a polynomial. If the characteristic polynomial of <math>A</math> has a factorization
<math display=block>p_A(t) = (t - \lambda_1) (t - \lambda_2) \cdots (t-\lambda_n)</math>
then the characteristic polynomial of the matrix <math>f(A)</math> is given by
<math display=block>p_{f(A)}(t) = (t - f(\lambda_1)) (t - f(\lambda_2)) \cdots (t-f(\lambda_n)).</math>
}}
That is, the algebraic multiplicity of <math>\lambda</math> in <math>f(A)</math> equals the sum of algebraic multiplicities of <math>\lambda'</math> in <math>A</math> over <math>\lambda'</math> such that <math>f(\lambda') = \lambda.</math> 
In particular, <math>\operatorname{tr}(f(A)) = \textstyle\sum_{i=1}^n f(\lambda_i)</math> and <math>\operatorname{det}(f(A)) = \textstyle\prod_{i=1}^n f(\lambda_i).</math> 
Here a polynomial <math>f(t) = t^3+1,</math> for example, is evaluated on a matrix <math>A</math> simply as <math>f(A) = A^3+I.</math> 

The theorem applies to matrices and polynomials over any field or [[commutative ring]].<ref>{{Cite book |last=Lang |first=Serge |url=https://www.worldcat.org/oclc/852792828 |title=Algebra |publisher=Springer |year=1993 |isbn=978-1-4613-0041-0 |location=New York |oclc=852792828|at=p.567, Theorem 3.10}}</ref>
However, the assumption that <math>p_A(t)</math> has a factorization into linear factors is not always true, unless the matrix is over an [[algebraically closed field]] such as the complex numbers.

{{math proof|proof=
This proof only applies to matrices and polynomials over complex numbers (or any algebraically closed field).
In that case, the characteristic polynomial of any square matrix can be always factorized as 
<math display=block>p_A(t) = \left(t - \lambda_1\right) \left(t - \lambda_2\right) \cdots \left(t - \lambda_n\right)</math>
where <math>\lambda_1, \lambda_2, \ldots, \lambda_n</math> are the eigenvalues of <math>A,</math> possibly repeated. 
Moreover, the [[Jordan normal form|Jordan decomposition theorem]] guarantees that any square matrix <math>A</math> can be decomposed as <math>A = S^{-1} U S,</math> where <math>S</math> is an [[invertible matrix]] and <math>U</math> is [[upper triangular]]
with <math>\lambda_1, \ldots, \lambda_n</math> on the diagonal (with each eigenvalue repeated according to its algebraic multiplicity).
(The Jordan normal form has stronger properties, but these are sufficient; alternatively the [[Schur decomposition]] can be used, which is less popular but somewhat easier to prove).

Let <math display="inline">f(t) = \sum_i \alpha_i t^i.</math> 
Then 
<math display=block>f(A) = \textstyle\sum \alpha_i (S^{-1} U S)^i = \textstyle\sum \alpha_i S^{-1} U S S^{-1} U S \cdots S^{-1} U S = \textstyle\sum \alpha_i S^{-1} U^i S = S^{-1} (\textstyle\sum \alpha_i U^i) S = S^{-1} f(U) S.</math> 
For an upper triangular matrix <math>U</math> with diagonal <math>\lambda_1, \dots, \lambda_n,</math> the matrix <math>U^i</math> is upper triangular with diagonal <math>\lambda_1^i,\dots,\lambda_n^i</math> in <math>U^i,</math> 
and hence <math>f(U)</math> is upper triangular with diagonal <math>f\left(\lambda_1\right), \dots, f\left(\lambda_n\right).</math> 
Therefore, the eigenvalues of <math>f(U)</math> are <math>f(\lambda_1),\dots,f(\lambda_n).</math> 
Since <math>f(A) = S^{-1} f(U) S</math> is [[Similar matrix|similar]] to <math>f(U),</math> it has the same eigenvalues, with the same algebraic multiplicities.
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