Linear-2 4

\begin{matrix} Let V be a finitely generated vector space over F \\ Let T, S : V \to V be linear transformations \\ Prove: Eigenvalues of S T are equal to eigenvalues of T S \\ Proof: \\ Let λ \neq 0 be an eigenvalue of T S \\ \exists v \neq 0 : T S v = λ v \\ ⟹ S T (S v) = S (T S v) = S (λ v) = λ S v \\ ⟹ λ is an eigenvalue of S T \\ Let λ be an eigenvalue of S T \\ Let λ = 0 be an eigenvalue of T S \\ ⟹ T S v = 0 v = 0 \\ ⟹ T S is not invertible ⟹ [\begin{matrix} T is not invertible \\ S is not invertible \end{matrix} ⟹ S T is not invertible \\ ⟹ k e r (S T) \neq {0} ⟹ λ is an eigenvalue of S T \end{matrix}

\begin{matrix} Let A \in R^{n \times n} of rank 1 \\ Prove: \forall x \neq y \in R ∖ {0} : [\begin{matrix} x I - A is invertible \\ y I - A is invertible \end{matrix} \\ Determine whether there is always x \neq 0 \in R : x I - A is not invertible \\ Proof: \\ Let x I - A be non-invertible and y I - A be non-invertible \\ ⟹ {\begin{matrix} E_{x} = N (x I - A) \neq {0} \\ E_{y} = N (y I - A) \neq {0} \end{matrix} \\ r a n k (A) = 1 ⟹ γ_{A} (0) = n - 1 ⟹ μ_{A} (0) \geq n - 1 \\ ⟹ {\begin{matrix} μ_{A} (x) + μ_{A} (y) = 1 \\ μ_{A} (x) \geq 1 \\ μ_{A} (y) \geq 1 \end{matrix} - Contradiction! \\ Solution: \\ No \\ A = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) ⟹ P_{A} (λ) = λ^{2} ⟹ \forall λ \neq 0 \in R : det (λ I - A) \neq 0 \end{matrix}

\begin{matrix} Let V be a finitely generated vector space over F \\ Let T : V \to V be an idempotentic linear transformation \\ Find eigenvalues of T \\ Determine whether T is diagonalizable \\ Solution: \\ E_{0} = k e r (T) \\ E_{1} = I m (T) ? \\ Let T v = v ⟹ v \in I m (T) \\ Let v \in I m (T) \\ \exists u \in V : T u = v ⟹ T (T u) = T v = v ⟹ v \in E_{1} \\ ⟹ E_{1} = I m (T) \\ \dim (E_{0}) + \dim (E_{1}) = n ⟹ There are no other eigenvalues and T is diagonalizable \\ {\begin{matrix} 1 is the only eigenvalue & T is invertible (T = I) \\ 0 is the only eigenvalue & T = 0 \\ 0, 1 are the only eigenvalues & otherwise \end{matrix} \end{matrix}

\begin{matrix} a_{n} = {\begin{matrix} 1 & n = 1, 2 \\ a_{n - 1} + 2 a_{n - 2} & n > 2 \end{matrix} \\ 1, 1, 3, 5, 11, 21, \dots \\ Let A = (\begin{matrix} 1 & 2 \\ 1 & 0 \end{matrix}) \\ \forall n > 2 : A \cdot (\begin{matrix} a_{n - 1} \\ a_{n - 2} \end{matrix}) = (\begin{matrix} a_{n} \\ a_{n - 1} \end{matrix}) \\ ⟹ A^{n - 2} \cdot (\begin{matrix} a_{2} \\ a_{1} \end{matrix}) = (\begin{matrix} a_{n} \\ a_{n - 1} \end{matrix}) \\ P_{A} (λ) = λ^{2} - λ - 2 = (λ - 2) (λ + 1) \\ E_{2} = s p {(\begin{matrix} 2 \\ 1 \end{matrix})} \\ E_{- 1} = s p {(\begin{matrix} 1 \\ - 1 \end{matrix})} \\ Let P = (\begin{matrix} 2 & 1 \\ 1 & - 1 \end{matrix}), D = (\begin{matrix} 2 & 0 \\ 0 & - 1 \end{matrix}) \\ A = P D P^{- 1} ⟹ A^{n} = P D^{n} P^{- 1} \\ ⟹ (\begin{matrix} a_{n} \\ a_{n - 1} \end{matrix}) = P (\begin{matrix} 2^{n - 2} & 0 \\ 0 & (- 1)^{n - 2} \end{matrix}) P^{- 1} (\begin{matrix} a_{2} \\ a_{1} \end{matrix}) \\ P^{- 1} = \frac{1}{3} (\begin{matrix} 1 & 1 \\ 1 & - 2 \end{matrix}) \\ A^{n - 2} = \frac{1}{3} (\begin{matrix} 2 & 1 \\ 1 & - 1 \end{matrix}) (\begin{matrix} 2^{n - 2} & 0 \\ 0 & (- 1)^{n - 2} \end{matrix}) (\begin{matrix} 1 & 1 \\ 1 & - 2 \end{matrix}) = \\ = \frac{1}{3} (\begin{matrix} 2 & 1 \\ 1 & - 1 \end{matrix}) (\begin{matrix} 2^{n - 2} & 2^{n - 2} \\ (- 1)^{n - 2} & - 2 \cdot (- 1)^{n - 2} \end{matrix}) = \frac{1}{3} (\begin{matrix} 2^{n - 1} + (- 1)^{n - 2} & 2^{n - 1} - 2 \cdot (- 1)^{n - 2} \\ * & * \end{matrix}) \\ ⟹ a_{n} = \frac{1}{3} (2^{n} - (- 1)^{n - 2}) \\ ⟹ a_{n} = \frac{4}{3} 2^{n - 2} - \frac{1}{3} (- 1)^{n - 2} \\ a_{n} = α λ_{1}^{n - 2} + β λ_{2}^{n - 2} \\ a_{n} = \sum_{i = 1}^{k} α_{i} λ_{i}^{n - k} \end{matrix}

\begin{matrix} Let A \in R^{n \times n} \\ P_{A} (λ) = λ^{3} (λ^{2} + 4) \\ Not diagonalizable over R \\ Over C ? \\ P_{A} (λ) = λ^{3} (λ - 2 i) (λ + 2 i) \\ A is diagonalizable ⟺ r a n k (A) = 2 \\ A = (\begin{matrix} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 i & 0 \\ 0 & 0 & 0 & 0 & - 2 i \end{matrix}) Incorrect :(, A must only have real values \\ A = (\begin{matrix} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & - 2 & 0 \end{matrix}) Correct! :) \\ A = (\begin{matrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & - 2 & 0 \end{matrix}) Correct! :) And a Jordan 3x3 block!! \end{matrix}