Linear-2 6

Cayley-Hamilton theorem

\begin{matrix} A \in F^{n \times n} ⟹ P_{A} (A) = 0 \end{matrix}

\begin{matrix} Let A \in F^{n \times n} \\ Prove: \dim (s p {I, A, A^{2}, \dots}) \leq n \\ Proof: \\ P_{A} (A) = A^{n} + \sum_{i = 0}^{n - 1} α_{i} A^{i} = 0 \\ ⟹ A^{n} = \sum_{i = 0}^{n - 1} - α_{i} A^{i} \in s p {I, A, A^{2}, \dots, A^{n - 1}} \\ By Induction starting from n : \\ Let k \geq n \\ A^{k + 1} = A A^{k} = A (\sum_{i = 0}^{n - 1} β_{i} A^{i}) = \\ = \sum_{i = 0}^{n - 1} β_{i} A^{i + 1} \in s p {I, A, \dots, A^{k}} = s p {I, A, \dots, A^{n - 1}} \\ Note: by using m_{A} instead of P_{A} we can bound the dimension even further by \deg (m_{A}) \\ Note: dimension is exactly \deg (m_{A}) as otherwise m_{A} would not be minimal \end{matrix}

\begin{matrix} A^{2} = A ⟹ A^{2} - A = 0 ⟹ f (x) = x^{2} - x, f (A) = 0 \\ f (x) = x (x - 1) \\ ⟹ m_{A} \in {x, x - 1, x (x - 1)} \\ ⟹ Largest Jordan block in A_{J} is of size 1 \\ ⟹ A_{J} is diagonal ⟹ A is diagonalizable \end{matrix}

\begin{matrix} A \in R^{7 \times 7} is invertible \\ (A^{3} + A) (A - 2 I) = 0 \\ t r (A) = 2 \\ Find P_{A}, m_{A} \\ Solution: \\ (A^{3} + A) (A - 2 I) = A (A^{2} + I) (A - 2 I) = 0 \\ A is invertible ⟹ (A^{2} + I) (A - 2 I) = 0 \\ m_{A} (x) ∣ (x^{2} + 1) (x - 2) \\ Let m_{A} (x) = x^{2} + 1 \\ ⟹ P_{A} (x) = (x^{2} + 1)^{k} \\ ⟹ 2 k = 7 - Contradiction! \\ Let m_{A} (x) = x - 2 \\ ⟹ P_{A} (x) = (x - 2)^{7} ⟹ t r (A) = 14 - Contradiction! \\ ⟹ m_{A} (x) = (x^{2} + 1) (x - 2) \\ ⟹ P_{A} (x) = (x^{2} + 1)^{k} (x - 2)^{7 - 2 k} \\ P_{A} (x) = [\begin{matrix} (x^{2} + 1) (x - 2)^{5} \\ (x^{2} + 1)^{2} (x - 2)^{3} \\ (x^{2} + 1)^{3} (x - 2) \end{matrix} \\ Let us examine this over C \\ m_{A} (x) = (x - i) (x + i) (x - 2) \\ ⟹ A is diagonalizable over C \\ P_{A} (x) = (x - i)^{k} (x + i)^{k} (x - 2)^{7 - 2 k} \\ ⟹ t r (A) = k i - k i + 2 (7 - 2 k) = 2 (7 - 2 k) \\ ⟹ 7 - 2 k = 1 ⟹ k = 3 \\ ⟹ P_{A} (x) = (x^{2} + 1)^{3} (x - 2) \end{matrix}

\begin{matrix} Find Jordan form of A = (\begin{matrix} - 2 & 0 & 0 & 0 \\ - 1 & 1 & 0 & 0 \\ 1 & - 1 & 0 & - 1 \\ 1 & 1 & 1 & 2 \end{matrix}) over C \\ Solution: \\ P_{A} (x) = | \begin{matrix} x + 2 & 0 & 0 & 0 \\ 1 & x - 1 & 0 & 0 \\ - 1 & 1 & x & 1 \\ - 1 & - 1 & - 1 & x - 2 \end{matrix} | = (x + 2) (x - 1) | \begin{matrix} x & 1 \\ - 1 & x - 2 \end{matrix} | = \\ = (x + 2) (x - 1) (x^{2} - 2 x + 1) = (x + 2) (x - 1)^{3} \\ λ = 1 ⟹ (\begin{matrix} 3 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ - 1 & 1 & 1 & 1 \\ - 1 & - 1 & - 1 & - 1 \end{matrix}) \to (\begin{matrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}) ⟹ γ_{A} (1) = 2 \\ ⟹ A_{J} = J_{1} (- 2) \oplus J_{2} (1) \oplus J_{1} (1) \end{matrix}

\begin{matrix} Let A = (\begin{matrix} n & n - 1 & n - 2 & \dots & 1 \\ 0 & n & n - 1 & ⋱ & 2 \\ ⋮ & ⋱ & ⋱ & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & n - 1 \\ 0 & \dots & \dots & 0 & n \end{matrix}) \in R^{n \times n} \\ P_{A} (x) = (x - n)^{n} \\ r a n k (n I - A) = n - 1 ⟹ γ_{A} (n) = 1 \\ ⟹ A_{J} = J_{n} (n) \end{matrix}

\begin{matrix} Let A, B \in F^{n \times n} \\ Let P_{A} = P_{B} \\ Let m_{A} = m_{B} \\ Let \forall λ : γ_{A} (λ) = γ_{B} (λ) \\ Prove or disprove: A \sim B \\ Disproof: \\ A_{J} = J_{3} (λ) \oplus J_{2} (λ) \oplus J_{2} (λ) \\ B_{J} = J_{3} (λ) \oplus J_{3} (λ) \oplus J_{1} (λ) \end{matrix}

\begin{matrix} Let A \in C^{6 \times 6} \\ Let P_{A} (x) = (x - 1)^{4} (x - 2)^{2} \\ Let m_{A} (x) = (x - 1)^{2} (x - 2) \\ Let γ_{A} (1) = 2 \\ Find A_{J} \\ Solution: \\ A_{J} = J_{2} (1) \oplus J_{2} (1) \oplus J_{1} (2) \oplus J_{1} (2) \end{matrix}

\begin{matrix} Let A = (\begin{matrix} 3 & 1 & 0 \\ - 4 & - 1 & 0 \\ 4 & - 8 & - 2 \end{matrix}) \in C^{3 \times 3} \\ Find m_{A} \\ Solution: \\ P_{A} (x) = | \begin{matrix} x - 3 & - 1 & 0 \\ 4 & x + 1 & 0 \\ - 4 & 8 & x + 2 \end{matrix} | = (x + 2) (x^{2} - 2 x + 1) = (x + 2) (x - 1)^{2} \\ (A + 2 I) (A - I) = (\begin{matrix} 5 & 1 & 0 \\ - 4 & 1 & 0 \\ 4 & - 8 & 0 \end{matrix}) (\begin{matrix} 2 & 1 & 0 \\ - 4 & 0 & 0 \\ 4 & - 8 & - 3 \end{matrix}) = (\begin{matrix} 6 & * & * \\ * & * & * \\ * & * & * \end{matrix}) \neq 0 \\ ⟹ m_{A} (x) = (x + 2) (x - 1)^{2} \\ Alternative approach: \\ (\begin{array}{ccc} - 2 & - 1 & 0 \\ 4 & 2 & 0 \\ - 4 & 8 & 3 \end{array}) \to (\begin{array}{ccc} - 2 & - 1 & 0 \\ 0 & 0 & 0 \\ - 4 & 8 & 3 \end{array}) ⟹ γ_{A} (1) = 1 \\ ⟹ A_{J} = J_{1} (- 2) \oplus J_{2} (1) ⟹ m_{A} (x) = (x + 2) (x - 1)^{2} \end{matrix}