Diagonalising A Matrix

There are many methods to obtain a diagonalisable matrix.

Using Eigenvectors and Eigenvalues §

We can use eigenvectors to make things easier:

Diagonalisable Matrices using Eigenvectors ^theorem

A matrix, $\mathbf{A}\in {\mathbf{M}}_{m,n}(\mathbb{F})$ is diagonalisable over the field $\mathbb{F}$ ($\mathbb{F}$ must be $\mathbb{R}$ or $\mathbb{C}$) if and only if it has $n$ linearly independent eigenvectors. Then:

$$\mathbf{A}={\mathbf{PDP}}^{-1}$$

$$\mathbf{P}=\left[\begin{array}{cccc}{{\vec{v}}_{\lambda }}_1& {{\vec{v}}_{\lambda }}_2& \dots & {{\vec{v}}_{\lambda }}_n\end{array}\right]$$

\boldsymbol{D} = \begin{bmatrix} \lambda_{1} & 0 & 0 & 0 \\

0 & \lambda_{2} & 0 & 0 \ 0 & 0 & \ddots & 0 \ 0 & 0 & 0 & \lambda_{i}\end{bmatrix} = \text{diag}(\lambda_{1}, \lambda_{2}, \dots, \lambda_{i})

Where each eigenvector corresponds to a unique eigenvalue. Alternatively, for every eigenvalue, $\lambda$ the algebraic multiplicity must equal the geometric multiplicity

Note that this requires obtaining the eigenvectors from a matrix.

Using Orthogonal Matrices §

If $\mathbf{A}$ is both real and a symmetric matrix, then the diagonalising matrix, $\mathbf{P}$ is an orthogonal matrix

Orthogonal Diagonalisation ^theorem2

If $\mathbf{A}$ is a real symmetric matrix of size $n\times n$, it can be diagonalised in ${\mathbb{R}}^n$, with the inner product being the dot product:

1. All eigenvalues of $\mathbf{A}$ are real: ${\lambda }_{1\dots n}\in \mathbb{R}$
2. Each eigenvector that corresponds to the eigenvalue is orthonormal. The set of these eigenvectors is an orthonormal basis.
3. $\mathbf{A}$ is diagonalisable with:

$$\mathbf{A}={\mathbf{QDQ}}^{-1}$$

where:

$$\mathbf{D}=\text{diag}({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)$$

\boldsymbol{Q} = \begin{bmatrix} {\vec{v}_{\lambda}}_{1} & {\vec{v}_{\lambda}}_{2} & \dots & {\vec{v}_{\lambda}}_{n}

\end{bmatrix}