Polynomials can be applied to square matrices. For example,

`p`(`t`) = - 2 - `t` + 2`t`^{2} + `t`^{3} → `p`(` A`) = - 2

Moreover, factorizations are preserved,

- 2 - `t` + 2`t`^{2} + `t`^{3} = (`t` - 1)(`t` + 1)(`t` + 2) → - 2` I` -

The following summarizes what we know about eigenvalues and eigenvectors of `p`(` A`).

Let `p`(`t`) be a polynomial. Let ` A` be a square matrix.

is an eigenvector of**v**with eigenvalue**A**`λ`⇒is an eigenvector of**v**`p`() with eigenvalue**A**`p`(`λ`).is diagonalizable ⇔**A**`p`() is diagonalizable.**A**- Any eigenvalue of
`p`() is of the form**A**`p`(`λ`) for some eigenvalue`λ`of.**A**

Note that the third part is the "partial converse" of the first part. See the second example below and the subsequent discussion. Also note that the third part implicitly allows the use of complex numbers as eigenvalues.

Proof `p`(` A`) can be constructed from

To prove the third part, let `μ` be an eigenvalue of `p`(` A`). We may decompose the polynomial

`p`(`t`) - `μ` = `c`(`t` - `λ`_{1})(`t` - `λ`_{2})...(`t` - `λ _{k}`),

(this requires us to allow complex numbers) and get

`p`(` A`) -

Then by the property `det AB` =

`det`(`p`(` A`) -

Since `μ` is assumed to be an eigenvalue of `p`(` A`), we have

Example Let ` A` be a square matrix such that the eigenvalues of

1 + 2`λ` + `λ`^{2} = 0 (solution: `λ` = -1)

1 + 2`λ` + `λ`^{2} = 4 (solution: `λ` = 1, -3)

1 + 2`λ` + `λ`^{2} = 16 (solution: `λ` = 3, -5)

Thus `λ` must be one of the following five numbers: -5, -3, -1, 1, 3.

Moreover, the eigenvalues of ` A` must include at least one from each of the three groups above. For example, the eigenvalues of

= [A |
0 | 0 | ], = [B |
0 | 1 | ] |

0 | 0 | 0 | 0 |

satisfy **A**^{2} = **B**^{2} = ` O`. Note that although eigenvalues (= 0) of

In general, all we can say is that the inclusion of the eigenspace `nul`(` A` -

Example In an earlier example, we found the matrix

= [A |
13 | -4 | ] |

-4 | 7 |

has a basis {(1, 2), (-2, 1)} of eigenvectors, with eigenvalues 5, 15. For the characteristic polynomial `p`(`t`) = 75 - 20`t` + `t`^{2} = (5 - `t`)(15 - `t`), we know that {(1, 2), (-2, 1)} is a basis of eigenvectors for `p`(` A`) = 75

In another earlier example, we found the matrix

= [B |
1 | 3 | -3 | ] |

-3 | 7 | -3 | ||

-6 | 6 | -2 |

has a basis {(1, 1, 0), (1, 0, -1), (1, 1, 2)} of eigenvectors, with eigenvalues 4, 4, -2. For the polynomial `p`(`t`) = (4 - `t`)(- 2 - `t`) = - 8 - 2`t` + `t`^{2}, we know that {(1, 1, 0), (1, 0, -1), (1, 1, 2)} is a basis of eigenvectors for `p`(` B`) = - 8

Note that the equality - 8` I` - 2

B^{-1} = (1/8)(-2 + I) = (1/8)[B |
-1 | 3 | -3 | ]. |

-3 | 5 | -3 | ||

-6 | 6 | -4 |