Linear-1 5

\begin{matrix} A, B \in R^{n \times n} \\ A - A^{2} B - 2 I = 0 \\ 3 B A^{2} + A^{2} - 3 A = 0 \\ Prove: \exists A^{- 1}, B^{- 1} \\ Proof: \\ A - A^{2} B = 2 I \\ A (I - A B) = 2 I ∣: 2 \\ A \underset{A^{- 1}}{\underset{⏟}{(\frac{1}{2} I - \frac{1}{2} A B)}} = I \\ 3 B A^{2} + A^{2} - 3 A = 0 ∣ \cdot [\dots] A^{- 1} \\ 3 B A + A - 3 I = 0 \\ (3 B + I) A = 3 I ∣: 3 \\ \underset{A^{- 1}}{\underset{⏟}{(B + \frac{1}{3} I)}} A = I \\ B + \frac{1}{3} I = \frac{1}{2} I - \frac{1}{2} A B \\ B + \frac{1}{2} A B = \frac{1}{6} I \\ (I + \frac{1}{2} A) B = \frac{1}{6} I ∣ \cdot 6 \\ \underset{B^{- 1}}{\underset{⏟}{(6 I + 3 A)}} B = I \end{matrix}

PLU decomposition

\begin{matrix} A \in R^{n \times n} \\ \exists P - permutation matrix (row-switching) \\ \exists L - lower-triangle matrix \\ \exists U - upper-triangle matrix \\ Such that P A = L U \end{matrix}

(\begin{array}{ccc} A & I & I \end{array}) \to (\begin{array}{ccc} U & L & P \end{array})

\begin{matrix} 1. Biggest element in the column must be on the main diagonal \\ 2. Elementary transformations \\ 2.1 α R_{i} - forbidden \\ 2.2 R_{i} \leftrightarrow R_{j}, U and P are affected as usual, L is only affected at the main diagonal \\ 2.3 α R_{i} = R_{i} + α R_{j} - U is affected as usual, P is not affected, L_{i j} = - α \end{matrix}

\begin{matrix} (\begin{array}{ccccccccc} 6 & - 2 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\ 9 & - 1 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\ 3 & 7 & 5 & 0 & 0 & 1 & 0 & 0 & 1 \end{array}) \overset{R_{1} \leftrightarrow R_{2}}{\to} (\begin{array}{ccccccccc} 9 & - 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 \\ 6 & - 2 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 3 & 7 & 5 & 0 & 0 & 1 & 0 & 0 & 1 \end{array}) \\ \overset{R_{2} - \frac{2}{3} R_{1}}{\underset{R_{3} - \frac{1}{3} R_{1}}{\to}} (\begin{array}{ccccccccc} 9 & - 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & - \frac{4}{3} & - \frac{2}{3} & \frac{2}{3} & 1 & 0 & 1 & 0 & 0 \\ 0 & \frac{22}{3} & \frac{14}{3} & \frac{1}{3} & 0 & 1 & 0 & 0 & 1 \end{array}) \\ \overset{R_{2} \leftrightarrow R_{3}}{\to} (\begin{array}{ccccccccc} 9 & - 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & \frac{22}{3} & \frac{14}{3} & \frac{1}{3} & 1 & 0 & 0 & 0 & 1 \\ 0 & - \frac{4}{3} & - \frac{2}{3} & \frac{2}{3} & 0 & 1 & 1 & 0 & 0 \end{array}) \\ \overset{R_{3} + \frac{2}{11} R_{2}}{\to} (\begin{array}{ccccccccc} 9 & - 1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & \frac{22}{3} & \frac{14}{3} & \frac{1}{3} & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & \frac{2}{11} & \frac{2}{3} & - \frac{2}{11} & 1 & 1 & 0 & 0 \end{array}) \end{matrix}

\begin{matrix} P = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}), L = (\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ - 1 & 3 & 1 \end{matrix}), U = (\begin{matrix} 3 & 1 & - 1 \\ 0 & 2 & \frac{1}{2} \\ 0 & 0 & 1 \end{matrix}) \\ Find solution(s) to the system: A x = (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) \\ A x = (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) \\ P A x = P \cdot (\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}) = (\begin{matrix} 2 \\ 3 \\ 1 \end{matrix}) \\ L U x = (\begin{matrix} 2 \\ 3 \\ 1 \end{matrix}) U x = U \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} \hat{x} \\ \hat{y} \\ \hat{z} \end{matrix}) \\ L \cdot (\begin{matrix} \hat{x} \\ \hat{y} \\ \hat{z} \end{matrix}) = (\begin{matrix} 2 \\ 3 \\ 1 \end{matrix}) \\ \hat{x} = 2, \hat{y} = - 1, \hat{z} = 6 \\ U \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 2 \\ - 1 \\ 6 \end{matrix}) \\ z = 6, y = - 2, x = \frac{10}{3} \end{matrix}

\begin{matrix} A \in R^{n \times n} \\ Calculate P L U decomposition of A \\ If P = I, then A = L U \end{matrix}

\begin{matrix} V is called vector space over field F if ...(was in lecture) \end{matrix}

\begin{matrix} 1. V = F^{n} = {{(\begin{matrix} a_{1} & a_{2} & \dots & a_{n} \end{matrix})}^{T} | \forall i \in [1, n] : a_{i} \in F} \\ 2. V = F^{m \times n} = {A | \forall i \in [1, m], j \in [1, n] : A_{i j} \in F} \\ 3. V = F_{n} [x] = {\sum_{i = 0}^{n} a_{i} \cdot x^{i} | \forall i \in [0, n] : a_{i} \in F} \\ 4. V = F [x] = {\sum_{i = 0}^{\infty} a_{i} \cdot x^{i} | \forall i \in {0} \cup N : a_{i} \in F} \\ 5. V = {f | f - function} \\ \dots \end{matrix}

\begin{matrix} Given: V - vector space over F \\ W is called vector subspace of V if W is a vector space and W \subseteq V \end{matrix}

\begin{matrix} V = R^{2} \\ W = {(\begin{matrix} x \\ y \end{matrix}) | x \geq 0} \\ Not a subspace of V \\ \dots I’m lazy bro \end{matrix}

\begin{matrix} V = F^{n \times n} \\ W_{1} = {A | A = A^{T}} - subspace of V \\ W_{2} = {A | A = - A^{T}} - subspace of V \\ W_{3} = {A | A = \pm A^{T}} - not a subspace of V \\ \dots I’m lazy bro \end{matrix}

\begin{matrix} V = R_{3} [x] \\ W = {P (x) | P (1) = 0 \land P (2) = P (0)} - subspace of V \\ \dots I’m lazy bro \end{matrix}