Diagonalizable Matrix

Definition

$\mathbf{A}=\mathbf{P\Lambda }{\mathbf{P}}^{-1}$ For square matrix $\mathbf{A}$, There exist invertible matrix $\mathbf{P}$ that satisfies ${\mathbf{P}}^{-1}\mathbf{AP}=\mathbf{D}$ where $\mathbf{D}$ is Diagonal Matrix

Orthogonally Diagonalizable Matrix

$\mathbf{A}=\mathbf{P\Lambda }{\mathbf{P}}^{⊺}$ For square matrix $\mathbf{A}$, There exist orthogonal matrix $\mathbf{P}$ that satisfies ${\mathbf{P}}^{⊺}\mathbf{AP}=\mathbf{D}$ where $\mathbf{D}$ is Diagonal Matrix

Unitary Diagonalizable Matrix

$\mathbf{A}=\mathbf{U\Lambda }{\mathbf{U}}^{†}$ For square matrix $\mathbf{A}$, There exist Unitary Matrix $\mathbf{U}$ that satisfies ${\mathbf{U}}^{†}\mathbf{AU}=\mathbf{D}$ where $\mathbf{D}$ is Diagonal Matrix

Diagonalization

$\mathbf{A}=\mathbf{P\Lambda }{\mathbf{P}}^{-1}$

Let $\mathbf{P}$ be a matrix of eigenvectors of $\mathbf{A}$ and $\mathbf{\Lambda }=\text{diag}({\lambda }_1,{\lambda }_2,\dots )$ be a Diagonal Matrix which has eigenvalues of $\mathbf{A}$. Then, by formula of Eigendecomposition $\mathbf{AP}=\mathbf{P\Lambda }$ If $\mathbf{P}$ is full rank matrix, then $\mathbf{P}$ is invertible. So, ${\mathbf{P}}^{-1}\mathbf{AP}=\mathbf{\Lambda }$ holds.

Facts

$\mathbf{A}$ is diagonalizable $\iff$ $\mathbf{A}$ has $n$ linearly independent eigenvectors

The matrix $\mathbf{P}$ is not unique.

The order of $p_i$ and ${\lambda }_i$ in $\mathbf{S}$ and $\mathbf{\Lambda }$ should be the same.

Not all matrices are diagonalizable^[Since the property of eigen decomposition]

${\mathbf{A}}^k\mathbf{v}={\lambda }^k\mathbf{v}\Rightarrow {\mathbf{A}}^k=\mathbf{P}{\mathbf{\Lambda }}^k{\mathbf{P}}^{-1}$

the matrix is diagonalizable $\iff$ If the Geometric Multiplicity is equal to the Algebraic Multiplicity for all eigenvalues

Unitary Diagonalizable Matrix $\iff$ Normal Matrix

Spectral Theorem

Diagonalized matrix is similar to the original matrix.