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example, real and complex case
The degree of this polynomial equals the length of the matrix's side. The number of roots therefore is not greater than this number.
The degree of this polynomial equals the length of the matrix's side. The number of roots therefore is not greater than this number.
== Example ==
Take <math>A = \begin{pmatrix}1 & 9 \\ 4 & 1\end{pmatrix}</math>
Then <math>f_A(\lambda) = \begin{vmatrix}1-\lambda & 9 \\ 4 & 1-\lambda\end{vmatrix}</math><math>=(1-\lambda)^2-36\,</math><math>=(-5-\lambda)(7-\lambda)\,</math>.
So, the roots of the characteristic polynomial are {-5, 7} - and these are the eigenvalues of the matrix. If you look at the slightly different matrix
<math>A' = \begin{pmatrix}1 & 9 \\ -4 & 1\end{pmatrix}</math>,
you find the characteristic polynomial
<math>f_{A'}(\lambda) = x^2 - 2x +37\,</math>.
This polynomial has no roots in <math>\mathbb{R}</math>, so if <math>A'\,</math> describes a linear map between to two dimensional real vector spaces, then this map has no eigenvalue. However, if <math>A'\,</math> is seen as a mapping of complex vector spaces, <math>f_{A'}(\lambda) = x^2 - 2x +37</math> can be factorized:
<math>f_{A'}(\lambda) = (1+ 6i -\lambda)(1-6i - \lambda)\,</math>.
For complex spaces, the sum of algebraic multiplicities of the eigenvalues equals the degree of the characteristic polynomial.
[[Category:Linear algebra]]
[[Category:Linear algebra]]