Linear-1 12

\begin{matrix} Let A, B \in R^{m \times n} \\ r a n k (A + B) = r a n k (A) + r a n k (B) ⟹ C (A) \cap C (B) = {0} \\ Proof: \\ d i m (C (A + B)) = d i m (C (A)) + d i m (C (B)) = d i m (C (A) + C (B)) + d i m (C (A) \cap C (B)) \\ C (A + B) \subseteq C (A) + C (B) \\ ⟹ d i m (C (A + B)) \leq d i m (C (A) + C (B)) \\ ⟹ d i m (C (A) + C (B)) + d i m (C (A) \cap C (B)) \leq d i m (C (A) + C (B)) \\ ⟹ d i m (C (A) \cap C (B)) \leq 0 ⟹ d i m (C (A) \cap C (B)) = 0 \\ ⟹ C (A) \cap C (B) = {0} \end{matrix}

\begin{matrix} Let V be a vector space of dimension n \\ Let A \in R^{n \times n} \\ Prove or disprove: \exists B, C bases of V : [I]_{C}^{B} = A \\ Disproof: \\ Let A be non-invertible \\ \forall B, C : [I]_{C}^{B} is invertible ⟹ [I]_{C}^{B} \neq A \\ Let A be invertible \\ ⟹ C (A) is a linear independence \\ ⟹ B = {C_{1} (A), \dots, C_{n} (A)} is a basis of V \\ Let C = S_{V} \\ ⟹ [I]_{C}^{B} = A \end{matrix}

\begin{matrix} Let A \neq 0 \in R^{n \times n} \\ r a n k (A) = r a n k (A^{2}) \\ Prove or disprove: A is invertible \\ Disproof: \\ Let A = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}) \\ A^{2} = A ⟹ r a n k (A^{2}) = r a n k (A) \\ A is not invertible \end{matrix}

\begin{matrix} Let V be a vector space over F \\ Let T : V \to V be a linear transformation, T = T^{2} \\ Prove: I m (T) \oplus k e r (T) = V \\ Proof: \\ Let v \in I m (T) \cap k e r (T) \\ ⟹ T (v) = 0, \exists w \in V : T (w) = v \\ T (w) = T (T (w)) = T (v) = 0 ⟹ v = 0 \\ ⟹ I m (T) \cap k e r (T) = {0} \\ ⟹ d i m (I m (T) + k e r (T)) = d i m (I m (T)) + d i m (k e r (T))) = d i m (V) \\ I m (T) + k e r (T) \subseteq V \land d i m (I m (T) + k e r (T)) = d i m (V) \\ ⟹ I m (T) + k e r (T) = V ⟹ I m (T) \oplus k e r (T) = V \end{matrix}

\begin{matrix} Let V be a vector space over F \\ Let T : V \to V be a linear transformation, T = T^{2} \\ Prove: k e r (I - T) \oplus k e r (T) = V \\ Proof: \\ Let v \in k e r (I - T) \\ (I - T) (v) = I (v) - T (v) = v - T (v) = 0 ⟹ T (v) = v ⟹ v \in I m (T) \\ ⟹ k e r (I - T) \subseteq I m (T) \\ Let v \in I m (T) \\ \exists w \in V : T (w) = v \\ (I - T) (v) = v - T (v) = T (w) - T (T (w)) = T (w) - T (w) = 0 \\ ⟹ I m (T) \subseteq k e r (I - T) \\ ⟹ I m (T) = k e r (I - T) ⟹ k e r (I - T) \oplus k e r (T) = V \end{matrix}

\begin{matrix} Let A, B \in F^{n \times n} \\ r a n k (A) = k \\ Prove: r a n k (B) < n - k ⟹ A + B is non-invertible \\ Proof: \\ r a n k (A + B) \leq r a n k (A) + r a n k (B) < n - k + k = n \\ ⟹ A + B is non-invertible \end{matrix}

\begin{matrix} Let A, B \in F^{n \times n} \\ r a n k (A) = k \\ Prove or disprove: \\ A \neq 0 and non-invertible ⟹ \exists B : r a n k (B) = n - k : A + B is non-invertible \\ Proof: \\ A \neq 0 ⟹ \exists i \in [1, n] : C_{i} (A) \neq 0 (W L O G) \\ Let C_{i} (B) = - C_{i} (A) \\ Let {v_{2}, \dots, C_{i} (B), \dots, v_{n - k}} be a linear independence \\ Let n - k columns of B be {v_{2}, \dots, C_{i} (B), \dots, v_{n - k}} \\ And k columns of B be 0 \\ ⟹ r a n k (B) = n - k \\ And C_{i} (A + B) = 0 ⟹ A + B is non-invertible \end{matrix}

\begin{matrix} Let A, B \in F^{n \times n} \\ r a n k (A) = k \\ Prove: \exists B : r a n k (B) = n - k : A + B is invertible \\ Proof: \\ Let {C_{i_{1}} (A), C_{i_{2}} (A), \dots, C_{i_{k}} (A)} be a linear independence \\ Let C_{i_{1}} (B) = C_{i_{2}} (B) = \dots = C_{i_{k}} (B) = 0 \\ For n - k columns left: \\ Let {v_{1}, \dots, v_{n - k}} be a linear independence \\ Let \forall j \in [1, n - k] : C_{i_{j}} (B) = v_{j} - C_{i_{j}} (A) \end{matrix}