Exam 2023 (B)

1

\begin{matrix} Dimension theorem: \\ Let V be a finitey generated vector space over F \\ Let U, W \subseteq V be vector subspaces of V \\ d i m (U + W) = d i m (U) + d i m (W) - d i m (U \cap W) \\ Proof: \\ Let B = {v_{1}, v_{2}, \dots, v_{n}} be a basis of U \cap W \\ Let B_{U} = B \cup {u_{1}, u_{2}, \dots, u_{k}} be a basis of U \\ Let B_{W} = B \cup {w_{1}, w_{2}, \dots, w_{t}} be a basis of W \\ Let \hat{B} = B_{U} \cup B_{W} \\ U + W = s p (B_{U} \cup B_{W}) = s p (\hat{B}) \\ Let \sum_{i = 1}^{n} α_{i} v_{i} + \sum_{i = 1}^{k} β_{i} u_{i} + \sum_{i = 1}^{t} γ_{i} w_{i} = 0 \\ ⟹ \sum_{i = 1}^{n} α_{i} v_{i} + \sum_{i = 1}^{k} β_{i} u_{i} = - \sum_{i = 1}^{t} γ_{i} w_{i} \\ ⟹ - \sum_{i = 1}^{t} γ_{i} w_{i} \in s p (B_{U}) = U \\ - \sum_{i = 1}^{t} γ_{i} w_{i} \in s p (B_{W}) = W ⟹ - \sum_{i = 1}^{t} γ_{i} w_{i} \in U \cap W \\ ⟹ - \sum_{i = 1}^{t} γ_{i} w_{i} = \sum_{i = 1}^{n} δ_{i} v_{i} ⟹ \underset{\in s p (B_{W})}{\underset{⏟}{\sum_{i = 1}^{n} δ_{i} v_{i} + \sum_{i = 1}^{t} γ_{i} w_{i}}} = 0 \\ ⟹ δ_{1} = δ_{2} = \dots = δ_{n} = γ_{1} = γ_{2} = \dots = γ_{t} = 0 \\ ⟹ \underset{\in s p (B_{U})}{\underset{⏟}{\sum_{i = 1}^{n} α_{i} v_{i} + \sum_{i = 1}^{k} β_{i} u_{i}}} = 0 \\ ⟹ α_{1} = α_{2} = \dots = α_{n} = β_{1} = β_{2} = \dots = β_{k} = 0 \\ ⟹ \hat{B} is a linear independence and a basis of U + W \\ ⟹ d i m (U + W) = | \hat{B} | = | B_{U} \cup B_{W} | = | B_{U} | + | B_{W} | - \underset{B}{\underset{⏟}{| B_{U} \cap B_{W} |}} \\ ⟹ d i m (U + W) = d i m (U) + d i m (W) - d i m (U \cap W) \end{matrix}

2a

\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}

2b

\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}

3a

\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}

3b

\begin{matrix} Let V be a finitely generated vector space, d i m (V) = n \\ Let T : V \to V, T \neq 0 \\ Prove: \exists B = {v_{1}, v_{2}, \dots, v_{n}} basis of B : \forall i \in [1, n] : T (v_{i}) \neq 0 \\ Proof: \\ T \neq 0 ⟹ \exists v_{1} \in V : T (v_{1}) \neq 0 \\ Let B = {v_{1}, v_{2}, \dots, v_{n}} be a basis of V \\ Let \hat{B} = {u_{1}, u_{2}, \dots, u_{n}} : \forall i \in [1, n] : u_{i} = {\begin{cases} v_{i} & T (v_{i}) \neq 0 \\ v_{i} + v_{1} & T (v_{i}) = 0 \end{cases} \\ \forall i \in [1, n] : T (v_{i}) = 0 ⟹ T (u_{i}) = T (v_{i} + v_{1}) = T (v_{i}) + T (v_{1}) = T (v_{1}) \neq 0 \\ Let k \in [1, n - 1] : \forall i \in [1, k] : T (v_{i}) \neq 0 (WLOG) \\ Let \sum_{i = 1}^{n} α_{i} u_{i} = 0 \\ \sum_{i = 1}^{n} α_{i} u_{i} = \sum_{i = 1}^{k} α_{i} v_{1} + \sum_{i = 1}^{n} α_{i} v_{i} = \underset{\in s p (B)}{\underset{⏟}{(2 α_{1} + \sum_{i = 2}^{k} α_{i}) v_{1} + \sum_{i = 2}^{n} α_{i} v_{i}}} = 0 \\ ⟹ α_{2} = α_{3} = \dots = α_{n} = 0 ⟹ α_{1} = 0 \\ ⟹ \hat{B} is a linear independence and a basis of V \\ \forall u \in \hat{B} : T (u) \neq 0 \end{matrix}

4a

\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}

4b

\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}

4c

\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}

5a

\begin{matrix} Let T : R^{3} \to R^{3}, T (\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} y + z \\ z \\ 0 \end{matrix}) \\ Prove: T is nilpotent \\ Proof: \\ T ((\begin{matrix} x \\ y \\ z \end{matrix})) = 0 \\ T^{2} (\begin{matrix} x \\ y \\ z \end{matrix}) = T (\begin{matrix} y + z \\ z \\ 0 \end{matrix}) = (\begin{matrix} z \\ 0 \\ 0 \end{matrix}) \\ T^{3} (\begin{matrix} x \\ y \\ z \end{matrix}) = T (T^{2} (\begin{matrix} x \\ y \\ z \end{matrix})) = T (\begin{matrix} z \\ 0 \\ 0 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) \\ ⟹ T^{3} = 0 \end{matrix}

5b

\begin{matrix} Let V be a vector space over F, d i m (V) = n \\ Let T : V \to V be a nilpotent linear transformation \\ Let α \neq 0 \in F \\ Let S = T - α I \\ Prove: S is invertible \\ Proof: \\ Let v \in k e r (S) \\ S (v) = 0 ⟹ (T - α I) (v) = 0 ⟹ T (v) = α v \\ Let k \in N \cup {0} : T^{k} = 0 \\ T^{k} (v) = α^{k} v = 0 \\ α \neq 0 ⟹ α^{k} \neq 0 ⟹ v = 0 ⟹ k e r (S) = {0} ⟹ S is surjective \\ S = (T - α I) : V \to V ⟹ S is injective \\ ⟹ S is bijective ⟹ S is invertible \end{matrix}

5c

\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}