Matrix diagonalization
Let A ∈ R n × n A \in \R^{n \times n} A ∈ R n × n n n n V j \mathbf V_j V j λ j \lambda_j λ j V = ( V 1 , V 2 , … , V n ) V = (\mathbf V_1, \mathbf V_2, \ldots, \mathbf V_n) V = ( V 1 , V 2 , … , V n )
Let Λ ∈ C n × n \Lambda \in \mathbb C^{n \times n} Λ ∈ C n × n A A A
Λ = ( λ 1 0 0 ⋯ 0 0 λ 2 0 ⋯ 0 0 0 λ 3 ⋮ ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ ⋯ λ n ) .
\Lambda = \begin{pmatrix}
\lambda_1 & 0 & 0 & \cdots & 0 \\
0 & \lambda_2 & 0 & \cdots & 0 \\
0 & 0 & \lambda_3 & & \vdots \\
\vdots & \vdots & & \ddots & \vdots \\
0 & 0 & \cdots & \cdots & \lambda_n
\end{pmatrix}.
Λ = λ 1 0 0 ⋮ 0 0 λ 2 0 ⋮ 0 0 0 λ 3 ⋯ ⋯ ⋯ ⋱ ⋯ 0 0 ⋮ ⋮ λ n . It follows that
A V = A ( V 1 , V 2 , … , V n ) = ( A V 1 , A V 2 , … , A V n ) = ( λ 1 V 1 , λ 1 V 2 , … , λ n V n ) = V Λ , A = V Λ V − 1 .
\begin{align*}
AV &= A(\mathbf V_1, \mathbf V_2, \ldots, \mathbf V_n) \\
&= (A\mathbf V_1, A\mathbf V_2, \ldots, A\mathbf V_n) \\
&= (\lambda_1 \mathbf V_1, \lambda_1 \mathbf V_2, \ldots, \lambda_n \mathbf V_n) \\
&= V \Lambda,\\
A &= V \Lambda V^{-1}.
\end{align*}
A V A = A ( V 1 , V 2 , … , V n ) = ( A V 1 , A V 2 , … , A V n ) = ( λ 1 V 1 , λ 1 V 2 , … , λ n V n ) = V Λ , = V Λ V − 1 . If we can write A = V Λ V − 1 A = V \Lambda V ^{-1} A = V Λ V − 1 A A A V V V Λ \Lambda Λ
Note
If A A A V \mathbf V V V − 1 = V ⊤ . \mathbf V^{-1} = \mathbf V^\top. V − 1 = V ⊤ .