### Eigenspace and Multiplicity

##### 3. Multiplicity and diagonalization

For an `n` by `n` matrix **A**, we first find eigenvalues `λ`_{1}, `λ`_{2}, ..., `λ`_{k} and then a basis `B`_{i} = {**v**_{i1}, **v**_{i2}, ..., **v**_{imiG}} for each eigenspace **E**_{λ} = `nul`(**A** - `λ`**I**). Since **E**_{λ1} + **E**_{λ2} + ... + **E**_{λk} is a direct sum, by this result, the collection `B` = `B`_{1}∪`B`_{2}∪...∪`B`_{k} = {**v**_{ij}: all `i`, `j`} of all the eigenvectors we computed is linearly independent. The number of vectors in `B` is the sum `m`_{1}^{G} + `m`_{2}^{G} + ... + `m`_{k}^{G} of geometric multiplicities.

The diagonalizability of **A** is equivalent to whether `B` is a basis of **R**^{n}. Since `B` is already linearly independent, by this result, we have the following condition for diagonalizability.

An `n` by `n` matrix is diagonalizable ⇔ The sum of geometric multiplicities is `n`.

We already know the sum of algebraic multiplicities must be `n`. In view of the statement above, it is natural to ask how the algebraic and the geometric multiplicities are related.

The algebraic multiplicity is no less than the geometric multiplicity: `m`_{i}^{A} ≥ `m`_{i}^{G}.

Proof Let us consider the first eigenvalue `λ`_{1} of **A** and abbreviate its multiplicities: `m` = `m`_{1}^{G}, `m'` = `m`_{1}^{A}. Then we have `m` linearly independent eigenvectors `B`_{1} = {**v**_{1}, **v**_{2}, ..., **v**_{m}} with eigenvalue `λ`_{1}. Extend the vectors to a basis {**v**_{1}, **v**_{2}, ..., **v**_{m}, **w**_{m+1}, ..., **w**_{n}} of **R**^{n}. Then

**P** = [**v**_{1} **v**_{2} ... **v**_{m} **w**_{m+1} ... **w**_{n}] = [**P**_{1}** **`P`_{2}]

(**P**_{1} is the first `m` columns, **P**_{2} is the last `n` - `m` columns) is an invertible matrix. We partition **P**^{-1}`AP` accordingly (`m` and `n` - `m` columns, also `m` and `n` - `m` rows):

**P**^{-1}`AP` = [ |
**B**_{11} |
**B**_{12} |
]. |

**B**_{21} |
**B**_{22} |

Comparing

`AP` = [`λ`_{1}**v**_{1} `λ`_{1}**v**_{2} ... `λ`_{1}**v**_{m} **Aw**_{m+1} ... **Aw**_{n}] = [`λ`_{1}**P**_{1}** **`AP`_{2}]

with

**P**[ |
**B**_{11} |
**B**_{12} |
] = [**P**_{1}**B**_{11} + **P**_{2}**B**_{21} **P**_{1}**B**_{12} + **P**_{2}**B**_{22}], |

**B**_{21} |
**B**_{22} |

we get **B**_{11} = `λ`_{1}**I**, **B**_{21} = **O**. Thus

**P**^{-1}`AP` = [ |
`λ`_{1}**I** |
**B**_{12} |
] |

**O** |
**B**_{22} |

and (note that `λ`_{1}**I** is an `m` by `m` matrix)

`det`(**A** - `λ`**I**) = `det`(**P**^{-1}**AP** - `λ`**I**) = `det`[ |
(`λ`_{1} - `λ`)**I** |
**B**_{12} |
] = (`λ`_{1} - `λ`)^{m} det(**B**_{22} - `λ`**I**). |

**O** |
**B**_{22} - `λ`**I** |

In particular, `λ`_{1} appears as a root of the characteristic polynomial for at least `m` times. Since the algebraic multiplicity `m'` is the *total* number of times `λ`_{1} appears as a root, we conclude that `m'` ≥ `m`.

Based on `m`_{1}^{A} + `m`_{2}^{A} + ... + `m`_{k}^{A} = `n` and `m`_{i}^{A} ≥ `m`_{i}^{G}, we conclude that `m`_{1}^{G} + `m`_{2}^{G} + ... + `m`_{k}^{G} = `n` ⇔ `m`_{i}^{A} = `m`_{i}^{G}. Thus the earlier result on the diagonalizability has the following reformulation.

A square matrix is diagonalizable ⇔ Algebraic and geometric multiplicities are equal.