Random exams

\begin{matrix} T : R^{3} \to R^{3} \\ T^{2} \neq 0, T^{3} = 0 \\ T^{2} \neq 0 ⟹ \exists v_{3} \in V : T^{2} (v_{3}) \neq 0 \\ T^{2} (v_{3}) \neq 0 ⟹ T (v_{3}) \neq 0 ⟹ v_{3} \neq 0 \\ Let v_{1} = T^{2} (v_{3}) \neq 0 \\ Let v_{2} = T (v_{3}) \neq 0 \\ Let α v_{1} + β v_{2} + γ v_{3} = 0 \\ ⟹ T (α v_{1} + β v_{2} + γ v_{3}) = α T (v_{1}) + β T (v_{2}) + γ T (v_{3}) = 0 \\ ⟹ α T^{3} (v_{3}) + β T^{2} (v_{3}) + γ T (v_{3}) = 0 \\ ⟹ β T^{2} (v_{3}) + γ T (v_{3}) = 0 ⟹ T (β T^{2} (v_{3}) + γ T (v_{3})) = 0 \\ ⟹ β T^{3} (v_{3}) + γ T^{2} (v_{3}) = 0 ⟹ γ T^{2} (v_{3}) = 0 ⟹ γ = 0 \\ ⟹ β T^{2} (v_{3}) = 0 ⟹ β = 0 \\ ⟹ α v_{1} = 0 ⟹ α = 0 \\ ⟹ B = {v_{1}, v_{2}, v_{3}} is a linear independence \\ ⟹ B is a basis of R^{3} \\ [T]_{B}^{B} = (\begin{matrix} \overset{|}{\underset{|}{[T (v_{1})]_{B}}} & \overset{|}{\underset{|}{[T (v_{2})]_{B}}} & \overset{|}{\underset{|}{[T (v_{3})]_{B}}} \end{matrix}) = (\begin{matrix} \overset{|}{\underset{|}{[T^{3} (v_{3})]_{B}}} & \overset{|}{\underset{|}{[T^{2} (v_{3})]_{B}}} & \overset{|}{\underset{|}{[T (v_{3})]_{B}}} \end{matrix}) = \\ = (\begin{matrix} \overset{|}{\underset{|}{[0]_{B}}} & \overset{|}{\underset{|}{[v_{1}]_{B}}} & \overset{|}{\underset{|}{[v_{2}]_{B}}} \end{matrix}) = (\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}) \end{matrix}

\begin{matrix} A \in C^{2 \times 2} \\ A A^{T} = 0 \overset{?}{⟹} A = 0 \\ A = (\begin{matrix} a + b i & c + d i \\ e + f i & g + h i \end{matrix}) \\ ⟹ A A^{T} = (\begin{matrix} (a + b i)^{2} + (c + d i)^{2} & (a + b i) (e + f i) + (c + d i) (g + h i) \\ (a + b i) (e + f i) + (c + d i) (g + h i) & (e + f i)^{2} + (g + h i)^{2} \end{matrix}) \\ (a + b i)^{2} = (a + b i) (a + b i) = a^{2} + 2 a b i - b^{2} \\ ⟹ {\begin{matrix} a^{2} + 2 a b i - b^{2} + c^{2} + 2 c d i - d^{2} = 0 \\ a e + (a f + b e) i - b f = 0 \\ e^{2} + 2 e f i - f^{2} + g^{2} + 2 g h i - h^{2} = 0 \end{matrix} ⟹ {\begin{matrix} a^{2} + c^{2} - b^{2} - d^{2} = 0 \\ a b + c d = 0 \\ a e - b f = 0 \\ a f + b e = 0 \\ e^{2} + g^{2} - f^{2} - h^{2} = 0 \\ e f + g h = 0 \end{matrix} \\ Let A = (\begin{matrix} 1 + i & - 1 + i \\ 0 & 0 \end{matrix}) \\ ⟹ A A^{T} = 0, A \neq 0 \end{matrix}

\begin{matrix} Let V be a finitely generated vector space \\ Let T : V \to V be a linear transformation such that \\ \forall linear transformations S : V \to V : S T = T S \\ Prove: \forall B = {v_{1}, v_{2}, \dots, v_{n}} bases of V : \forall i \in [1, n] : {T (v_{i}), v_{i}} is a linear dependence \\ Proof: \\ Let B = {v_{1}, v_{2}, \dots, v_{n}} be a basis of V \\ Let by contradiction \exists i \in [1, n] : {T (v_{i}), v_{i}} is a linear independence \\ ⟹ \exists \hat{B} = {v_{i}, T (v_{i}), u_{3}, u_{4}, \dots, u_{n}} basis of V \\ ⟹ By the defining theorem exists linear transformation S : V \to V such that \\ {\begin{matrix} S (v_{i}) = S (T (v_{i})) = v_{i} \\ \forall j \in [3, n] : S (u_{j}) = 0 \end{matrix} \\ S T = T S ⟹ v_{i} = S T (v_{i}) = T S (v_{i}) = T (v_{i}) \\ ⟹ {v_{i}, T (v_{i})} is a linear dependence - Contradiction! \\ ⟹ \forall i \in [1, n] : {v_{i}, T (v_{i})} is a linear dependence \end{matrix}

\begin{matrix} (\begin{matrix} a \\ a \\ 2 a \end{matrix}), (\begin{matrix} 0 \\ a + 1 \\ a + 1 \end{matrix}), (\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}), (\begin{matrix} a \\ a \\ a^{2} + 2 a + 1 \end{matrix}) \\ Find all values of a such that v_{4} \in s p ({v_{1}, v_{2}, v_{3}}) \\ (\begin{array}{cccc} a & 0 & 1 & a \\ a & a + 1 & 1 & a \\ 2 a & a + 1 & 2 & a^{2} + 2 a + 1 \end{array}) \to (\begin{array}{cccc} a & 0 & 1 & a \\ 0 & a + 1 & 0 & 0 \\ 0 & a + 1 & 0 & a^{2} + 1 \end{array}) \\ \to (\begin{array}{cccc} a & 0 & 1 & a \\ 0 & a + 1 & 0 & 0 \\ 0 & 0 & 0 & a^{2} + 1 \end{array}) \\ a^{2} + 1 > 0 ⟹ There are no solutions ⟹ \forall a \in R : v_{4} \notin s p ({v_{1}, v_{2}, v_{3}}) \\ For all values of a find dimension of s p ({v_{1}, v_{2}, v_{3}, v_{4}}) \\ v_{4} \notin s p ({v_{1}, v_{2}, v_{3}}) \\ Is {v_{1}, v_{2}, v_{3}} linear independence? \\ (\begin{matrix} a & 0 & 1 \\ a & a + 1 & 1 \\ 2 a & a + 1 & 2 \end{matrix}) \to \dots \to (\begin{matrix} a & 0 & 1 \\ 0 & a + 1 & 0 \\ 0 & 0 & 0 \end{matrix}) \\ Let a = 0 \\ (\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix}) ⟹ s p ({v_{1}, v_{2}, v_{3}}) = s p ({v_{2}, v_{3}}) ⟹ d i m (s p ({v_{1}, v_{2}, v_{3}, v_{4}})) = 3 \\ Let a = - 1 \\ (\begin{matrix} - 1 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) ⟹ s p ({v_{1}, v_{2}, v_{3}}) = s p ({v_{1}}) ⟹ d i m (s p ({v_{1}, v_{2}, v_{3}, v_{4}})) = 2 \\ Let a \neq 0, a \neq - 1 \\ {\begin{matrix} a \neq 0 \\ a + 1 \neq 0 \end{matrix} ⟹ (\begin{matrix} a & 0 & 1 \\ 0 & a + 1 & 0 \\ 0 & 0 & 0 \end{matrix}) ⟹ s p ({v_{1}, v_{2}, v_{3}}) = s p ({v_{1}, v_{2}}) \\ ⟹ d i m (s p ({v_{1}, v_{2}, v_{3}, v_{4}})) = 3 \end{matrix}

\begin{matrix} Let A \in F^{n \times n} \\ Let B \neq 0 \in F^{n \times n} symmetric such that A B = B A = 0 \\ Prove: d i m (N (A) \cap N (A^{T})) \neq 0 \\ Proof: \\ A B = 0 ⟹ C (B) \subseteq N (A) \\ B A = 0 ⟹ (B A)^{T} = A^{T} B^{T} = A^{T} B = 0 ⟹ C (B) \subseteq N (A^{T}) \\ ⟹ C (B) \subseteq N (A) \cap N (A^{T}) \\ B \neq 0 is symmetric ⟹ \exists i \in [1, n] : C_{i} (B) \neq 0 ⟹ C (B) \neq {0} \\ {0} \subset C (B) \subseteq N (A) \cap N (A^{T}) ⟹ N (A) \cap N (A^{T}) \neq {0} \\ Alternative proof: \\ B \neq 0 ⟹ \exists v \neq 0 \in F^{n} : B v \neq 0 \\ \underset{0}{\underset{⏟}{A B}} v = 0 ⟹ B v \in N (A) \\ B A = 0 ⟹ (B A)^{T} = A^{T} B^{T} = A^{T} B = 0 \\ ⟹ \underset{0}{\underset{⏟}{A^{T} B}} v = 0 ⟹ B v \in N (A^{T}) \\ ⟹ B v \in N (A) \cap N (A^{T}) ⟹ N (A) \cap N (A^{T}) \neq {0} \end{matrix}

\begin{matrix} B = {1, x + x^{2}, x^{3}, - x + x^{2}} basis of R_{3} [x] \\ C = {(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}), (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})} basis of R^{2 \times 2} \\ T : R_{3} [x] \to R^{2 \times 2} \\ [T]_{C}^{B} = (\begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 2 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & 0 \end{matrix}) \\ Find T \\ [T (1)]_{C} = 0 ⟹ T (1) = 0 \\ [T (x + x^{2})]_{C} = 0 ⟹ T (x + x^{2}) = 0 \\ [T (x^{3})]_{C} = 0 ⟹ T (x^{3}) = 0 \\ [T (- x + x^{2})]_{C} = (\begin{matrix} 0 \\ - 2 \\ 2 \\ 0 \end{matrix}) ⟹ T (- x + x^{2}) = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}) \\ T (x + x^{2}) + T (- x + x^{2}) = T (2 x^{2}) = 2 T (x^{2}) = (\begin{matrix} 0 & - 2 \\ 2 & 0 \end{matrix}) ⟹ T (x^{2}) = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}) \\ T (x + x^{2}) - T (- x + x^{2}) = T (2 x) = 2 T (x) = (\begin{matrix} 0 & 2 \\ - 2 & 0 \end{matrix}) ⟹ T (x) = (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) \\ ⟹ T (a + b x + c x^{2} + d x^{3}) = 0 + b T (x) + c T (x^{2}) + 0 = (\begin{matrix} 0 & b - c \\ c - b & 0 \end{matrix}) \end{matrix}

\begin{matrix} Let A, B \in R^{n \times n} \\ Prove or disprove: N (A) \cap C (B) = {0} ⟹ N (B) = N (A B) \\ Proof: \\ \forall v \in R^{n} : B v \in C (B) \\ N (A) \cap C (B) = {0} ⟹ \forall v \neq 0 \in R^{n} : A B v \neq 0 ⟹ N (A B) = {0} \\ B v = 0 ⟹ A B v = 0 ⟹ N (B) \subseteq N (A B) ⟹ N (B) = {0} \\ ⟹ N (B) = N (A B) \end{matrix}

\begin{matrix} Let A \in C^{n \times n} \\ Prove or disprove: A A^{T} = 0 ⟹ A = 0 \\ Disproof: \\ A = (\begin{matrix} 1 + i & - 1 + i \\ 0 & 0 \end{matrix}) \neq 0 \\ A A^{T} = (\begin{matrix} 1 + i & - 1 + i \\ 0 & 0 \end{matrix}) (\begin{matrix} 1 + i & 0 \\ - 1 + i & 0 \end{matrix}) = (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}) = 0 \end{matrix}

\begin{matrix} Let A \in R^{n \times n} \\ Prove or disprove: A A^{T} = 0 ⟹ A = 0 \\ Proof: \\ A A^{T} = 0 ⟹ \forall i \in [1, n] : (A A^{T})_{i i} = 0 \\ ⟹ \forall i \in [1, n] : (A A^{T})_{i i} = \sum_{k = 1}^{n} A_{i k} A_{k i}^{T} = \sum_{k = 1}^{n} A_{i k}^{2} = 0 \\ ⟹ \forall i \in [1, n] : \forall k \in [1, n] : A_{i k} = 0 ⟹ A = 0 \end{matrix}

\begin{matrix} Let A \in R^{n \times m}, B \in R^{n \times k} \\ Prove or disprove: \exists! C \in R^{k \times m} : A = B C ⟹ r a n k (B) = k \\ Proof: \\ Let \exists i \neq j \in [1, k] : {C_{i} (B), C_{j} (B)} is a linear dependence \\ CANT USE JUST TWO! But the idea is correct \\ Let {C_{1} (B), C_{2} (B)} be a linear dependence (WLOG) \\ ⟹ C_{2} (B) = α C_{1} (B) \\ C_{1} (A) = C_{1} (B C) = C_{1} (B) C_{11} + α C_{1} (B) C_{12} + x \\ Let C_{11} = β, C_{12} = γ \\ β C_{1} (B) + α γ C_{1} (B) + x = C_{1} (A) \\ Let α = 0 ⟹ γ can be anything ⟹ C is not unique - Contradiction! \\ ⟹ α \neq 0 \\ Let C_{11} = β + 1, C_{12} = γ - \frac{1}{α} \\ β C_{1} (B) + C_{1} (B) + α γ C_{1} (B) - C_{1} (B) + x = β C_{1} (B) + α γ C_{1} (B) + x = C_{1} (A) \\ ⟹ C is not unique - Contradiction! \\ ⟹ \forall i \neq j \in [1, k] : {C_{i} (B), C_{j} (B)} is a linear independence \\ ⟹ r a n k (B) = d i m (C (B)) = k \end{matrix}

\begin{matrix} Let T, S : V \to V be linear transformations \\ Prove or disprove: I m (T) \oplus I m (S) = V ⟹ T + S is injective \\ Disproof: \\ Let I m (T) \oplus I m (S) = V \\ I m (T + S) \subseteq I m (T) + I m (S) \\ T ((\begin{matrix} x \\ y \end{matrix})) = (\begin{matrix} x \\ 0 \end{matrix}) \\ S ((\begin{matrix} x \\ y \end{matrix})) = (\begin{matrix} 0 \\ x \end{matrix}) \\ (T + S) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x \\ x \end{matrix}) \\ (T + S) ((\begin{matrix} 1 \\ 2 \end{matrix})) = (T + S) ((\begin{matrix} 1 \\ 3 \end{matrix})) \\ ⟹ T + S is not injective \end{matrix}

\begin{matrix} Let T, S : V \to V be linear transformations \\ Prove or disprove: T + S is injective ⟹ I m (T) + I m (S) = V \\ Proof: \\ T + S is injective ⟹ T + S is surjective \\ ⟹ V = I m (T + S) \subseteq I m (T) + I m (S) \subseteq V ⟹ I m (T) + I m (S) = V \end{matrix}