Polynomials can be applied to square matrices. For example,

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

Moreover, the product of polynomials can also be applied to matrix. For example,

- 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`).

is an eigenvector of**v**with eigenvalue**A**`λ`and`p`(`t`) is a polynomial ⇔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 The first part follows from this result, because `p`(` A`) is a combination of

The second part follows from this first part, and the use of basis of eigenvectors, similar to the proof for this result on similar matrices.

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

`p`(`t`) - `μ` = `λ`_{0}(`t` - `λ`_{1})(`t` - `λ`_{2})...(`t` - `λ _{k}`),

(this requires us to allow complex numbers) and get

`p`(` A`) -

Then

0 = `det`(`p`(` A`) -

where the first equality is due to the assumption that `μ` be an eigenvalue of `p`(` A`), and the second equality comes from the property

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 |

Then **A**^{2} = **B**^{2} = ` O`. The example shows that although we can determine eigenvalues (= 0) of

In general, all we can say is that 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 |

The polynomials 75 - 20`t` + `t`^{2} and - 8 - 2`t` + `t`^{2} of the example above are called minimal polynomials of the matrices.
For a *diagonalizable* matrix ` A` with

The minimial polynomials for non-diagonalizable matrices are more complicated. However, it still divides the characteristic polynomial. More detailed discussion on the minimal polynomial can be found here.