Exam 2024 (C)

1a

\begin{matrix} Let V be a vector space over F \\ Let S \subseteq V \\ 1. S is a linear dependence \\ 2. \exists v \in S : v \in s p (S ∖ {v}) \\ 3. \exists v \in S : s p (S) = s p (S ∖ {v}) \\ 1 ⟺ 2 ⟺ 3 \\ Proof: \\ Let S = {v_{1}, v_{2}, \dots, v_{n}} \\ Let 1 \\ ⟹ \exists α_{i} \neq 0 : \sum_{i = 1}^{n} α_{i} v_{i} = 0 \\ Let α_{1} \neq 0 (WLOG) ⟹ v_{1} = \sum_{i = 2}^{n} (\frac{- α_{i}}{α_{i - 1}}) v_{i} \\ \forall i \in [2, n] : v_{i} \neq v_{1} ⟹ v_{1} \in s p (S ∖ {v_{1}}) ⟹ 1 ⟹ 2 \\ Let 2 \\ Let v_{1} \in s p (S ∖ {v_{1}}) (WLOG) \\ S ∖ {v_{1}} \subseteq S ⟹ s p (S ∖ {v_{1}}) \subseteq s p (S) \\ Let u \in s p (S) \\ u = \sum_{i = 1}^{n} α_{i} v_{i} = α_{1} v_{1} + \sum_{i = 2}^{n} α_{i} v_{i} \\ v_{1} \in s p (S ∖ {v_{1}}) ⟹ v_{1} = \sum_{j = 2}^{n} β_{j} v_{j} \\ ⟹ u = α_{1} \sum_{j = 2}^{n} β_{j} v_{j} + \sum_{i = 2}^{n} α_{i} v_{i} = \sum_{i = 2}^{n} (α_{1} β_{i} + α_{i}) v_{i} ⟹ u \in s p (S ∖ {v_{1}}) \\ ⟹ s p (S) \subseteq s p (S ∖ {v_{1}}) ⟹ s p (S) = s p (S ∖ {v_{1}}) ⟹ 2 ⟹ 3 \\ Let 3 \\ Let s p (S ∖ {v_{1}}) = s p (S) (WLOG) \\ v_{1} \in S ⟹ v_{1} \in s p (S) ⟹ v_{1} \in s p (S ∖ {v_{1}}) \\ ⟹ v_{1} = \sum_{i = 2}^{n} α_{i} v_{i} ⟹ v_{1} - \sum_{i = 2}^{n} α_{i} v_{i} = 0 \\ ⟹ {1, - α_{2}, - α_{3}, \dots, - α_{n}} is a non-trivial linear combination of S \\ ⟹ S is a linear dependence ⟹ 3 ⟹ 1 \\ ⟹ [1 ⟹ 2 ⟹ 3 ⟹ 1] ⟹ 1 ⟺ 2 ⟺ 3 \end{matrix}

1b

\begin{matrix} Let T : R^{2} \to R^{2} \\ T ((\begin{matrix} x \\ y \end{matrix})) = (\begin{matrix} 2 x \\ x + y \end{matrix}) \\ Let B = {(\begin{matrix} 1 \\ 3 \end{matrix}), (\begin{matrix} 1 \\ 1 \end{matrix})} basis of R^{2} \\ Find [T]_{B}^{B} \\ Solution: \\ [T]_{B}^{B} = (\begin{matrix} [T (\begin{matrix} 1 \\ 3 \end{matrix})]_{B} & [T (\begin{matrix} 1 \\ 1 \end{matrix})]_{B} \end{matrix}) = (\begin{matrix} [(\begin{matrix} 2 \\ 4 \end{matrix})]_{B} & [(\begin{matrix} 2 \\ 2 \end{matrix})]_{B} \end{matrix}) = (\begin{matrix} 1 & 0 \\ 1 & 2 \end{matrix}) \end{matrix}

2

\begin{matrix} U = {p (x) \in R_{2} [x] | \begin{matrix} p^{'} (1) = 0 \\ p (1) = 0 \end{matrix}} \\ W = s p ({1 + x + x_{2}, x + x^{2}}) \\ Find basis and dimension of U + W, U \cap W \\ Solution: \\ Let p (x) \in R_{2} [x] \\ p (x) = a + b x + c x^{2} ⟹ p^{'} (x) = b + 2 c x \\ p^{'} (1) = b + 2 c \\ p (1) = a + b + c \\ {\begin{matrix} b + 2 c = 0 \\ a + b + c = 0 \end{matrix} ⟹ {\begin{matrix} b = - 2 c \\ a = c \end{matrix} \\ ⟹ U = s p ({1 - 2 x + x^{2}}) \\ U + W = s p (U \cup W) = s p ({1 + x + x^{2}, x + x^{2}, 1 - 2 x + x^{2}}) \\ (\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 1 & 1 & - 2 & 0 \\ 1 & 1 & 1 & 0 \end{array}) \to (\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & - 3 & 0 \\ 0 & 1 & 0 & 0 \end{array}) \to (\begin{array}{cccc} 1 & 0 & 1 & 0 \\ 0 & 1 & - 3 & 0 \\ 0 & 0 & 3 & 0 \end{array}) \\ ⟹ {1 + x + x^{2}, x + x^{2}, 1 - 2 x + x^{2}} is a linear independence and is a basis of U + W \\ ⟹ d i m (U + W) = 3 \\ \underset{3}{\underset{⏟}{d i m (U + W)}} = \underset{1}{\underset{⏟}{d i m (U)}} + \underset{2}{\underset{⏟}{d i m (W)}} - d i m (U \cap W) \\ ⟹ d i m (U \cap W) = 0 ⟹ {0} is a basis of U \cap W \end{matrix}

3a

\begin{matrix} Let V be a vector space, d i m (V) = n \\ Let T, S : V \to V linear transformations such that \\ T S = T \\ Prove: d i m (I m (T)) + d i m (I m (S - I)) \leq n \\ Proof: \\ T S = T ⟹ T S - T = 0 \\ \forall v \in V : T (S (v)) - T (v) = T (S (v) - v) = T (S (v) - I (v)) = T (S - I) (v) = 0 \\ ⟹ \forall u \in I m (S - I) : T (u) = 0 ⟹ I m (S - I) \subseteq k e r (T) \\ ⟹ d i m (I m (S - I)) \leq d i m (k e r (T)) ⟹ d i m (I m (S - I)) \leq n - d i m (I m (T)) \\ ⟹ d i m (I m (T)) + d i m (I m (S - I)) \leq n \end{matrix}

3b

\begin{matrix} Find T, S : R^{2} \to R^{2} : T \neq 0, S \neq I, T S = T \\ Solution: \\ T (S (v)) = T (v) \\ Let S (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x \\ 0 \end{matrix}) \\ T ((\begin{matrix} x \\ y \end{matrix})) = T S (\begin{matrix} x \\ y \end{matrix}) = T (\begin{matrix} x \\ 0 \end{matrix}) ⟹ T (\begin{matrix} x - x \\ y - 0 \end{matrix}) = T (\begin{matrix} 0 \\ y \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) \\ Let T (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x \\ 0 \end{matrix}) \\ T S = T \end{matrix}

4a

\begin{matrix} Let A \in R^{n \times m}, B \in R^{n \times k} such that \\ \exists! C : A = B C \\ Prove or disprove: r a n k (B) = k \\ Proof: \\ Let S = {C_{1} (B), C_{2} (B), \dots, C (k) (B)} be a linear dependence \\ ⟹ \exists i \in [1, k] : C_{i} (B) \in s p (S ∖ {C_{i} (B)}) \\ Let C_{1} (B) \in s p (S ∖ {C_{1} (B)}) (WLOG) \\ ⟹ \exists {β_{j}}_{j \in [2, k]} \subseteq R : C_{1} (B) = \sum_{j = 2}^{k} β_{j} C_{j} (B) \\ A = B C ⟹ \forall i \in [1, m] : C_{i} (A) = C_{i} (B C) = B C_{i} (C) \\ C_{1} (A) = C_{1} (B C) = B C_{1} (C) = \sum_{i = 1}^{k} C_{1 i} C_{i} (B) = C_{11} C_{1} (B) + \sum_{i = 2}^{k} C_{1 i} C_{i} (B) = \\ = C_{11} \sum_{j = 2}^{k} β_{j} C_{j} (B) + \sum_{i = 2}^{k} C_{1 i} C_{i} (B) = \sum_{i = 2}^{k} (C_{11} β_{i} + C_{1 i}) C_{i} (B) \\ Let v = (\begin{matrix} C_{11} - 1 \\ C_{12} + β_{2} \\ ⋮ \\ C_{1 k} + β_{k} \end{matrix}) \\ Let \hat{C} = (\begin{matrix} \overset{|}{\underset{|}{v}} & \overset{|}{\underset{|}{C_{2} (C)}} & \dots & \overset{|}{\underset{|}{C_{m} (C)}} \end{matrix}) \\ C_{1} (B \hat{C}) = B C_{1} (\hat{C}) = \dots = \sum_{i = 2}^{k} (C_{11} β_{i} - β_{i} + C_{1 i} + β_{i}) C_{i} (B) = \\ = \sum_{i = 2}^{k} (C_{11} β_{i} + C_{1 i}) C_{i} (B) = C_{1} (A) \\ \forall i \in [2, m] : C_{i} (B \hat{C}) = B C_{i} (\hat{C}) = B C_{i} (C) = C_{i} (A) \\ ⟹ A = B \hat{C} \land \hat{C} \neq C - Contradiction! \\ ⟹ S is a linear independence ⟹ d i m (C (B)) = k \\ ⟹ r a n k (B) = k \end{matrix}

\begin{matrix} Alternative proof (might be wrong): \\ C_{i} (A) = C_{i} (B C) = B C_{i} (C) \\ \exists! C ⟹ C_{i} (C) is a unique solution of B x = C_{i} (A) \\ Let k > n \\ ⟹ C F (B) has k - n free variables \\ ⟹ B x = C_{i} (A) has more than one solution \\ ⟹ k \leq n \\ Let r a n k (B) < k \\ ⟹ d i m (R (B)) < k ⟹ d i m (R (B)) < n ⟹ \exists i \in [1, n] : R_{i} (C F (B)) = 0 \\ ⟹ B x = 0 has infinitely many solutions \\ ⟹ B x = C_{i} (A) has zero or infinitely many solutions - Contradiction! \\ ⟹ r a n k (B) = k \end{matrix}

4b

\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 V = R^{2} \\ Let T (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x \\ 0 \end{matrix}) \\ Let S (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ x \end{matrix}) \\ I m (T) = s p {(\begin{matrix} 1 \\ 0 \end{matrix})} \\ I m (S) = s p {(\begin{matrix} 0 \\ 1 \end{matrix})} \\ I m (T) \cap I m (S) = {0}, I m (T) \oplus I m (S) = R^{2} \\ I m (T + S) = s p {(\begin{matrix} 1 \\ 1 \end{matrix})} \neq R^{2} ⟹ T + S is not surjective \\ d i m (V) = d i m (V) ⟹ T + S is not injective \end{matrix}

4c

\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: \\ Let T + S be injective \\ \forall v \in V : (T + S) (v) = T (v) + S (v) \in I m (T) + I m (S) ⟹ I m (T + S) \subseteq I m (T) + I m (S) \\ d i m (V) = d i m (V) ⟹ T + S is surjective ⟹ I m (T + S) = V \\ V = I m (T + S) \subseteq I m (T) + I m (S) \subseteq V ⟹ I m (T) + I m (S) = V \end{matrix}

5a

\begin{matrix} Let A \in R^{n \times n} \\ E_{i j} = {\begin{cases} 1 & i, j \\ 0 & otherwise \end{cases} \end{matrix}

5a.i

\begin{matrix} \forall 1 \leq k \leq n : define C_{k} (A E_{i j}) via A, i, j \\ Solution: \\ Let 1 \leq k \leq n \\ Let k \neq j \\ ⟹ C_{k} (A E_{i j}) = A C_{k} (E_{i j}) = A \cdot 0 = 0 \\ Let k = j \\ ⟹ C_{k} (A E_{i j}) = C_{j} (A E_{i j}) = A C_{j} (E_{i j}) = A e_{i} = C_{i} (A) \\ ⟹ C_{k} (A E_{i j}) = {\begin{cases} C_{i} (A) & k = j \\ 0 & otherwise \end{cases} \end{matrix}

5a.ii

\begin{matrix} \begin{matrix} \forall 1 \leq k \leq n : define R_{k} (E_{i j} A) via A, i, j \\ Solution: \\ Let 1 \leq k \leq n \\ Let k \neq i \\ ⟹ R_{k} (E_{i j} A) = R_{k} (E_{i j}) A = 0 \cdot A = 0 \\ Let k = i \\ ⟹ R_{k} (E_{i j} A) = R_{i} (E_{i j} A) = R_{i} (E_{i j}) A = e_{j}^{T} A = R_{j} (A) \\ ⟹ R_{k} (E_{i j} A) = {\begin{cases} R_{j} (A) & k = i \\ 0 & otherwise \end{cases} \end{matrix} \end{matrix}

5b

\begin{matrix} Let A \in R^{n \times n} : \forall B \in R^{n \times n} : A B = B A \\ Prove: A is diagonal \\ Proof: \\ \forall i \in [1, n] : E_{i i} A = (\begin{matrix} 0 \\ ⋮ \\ 0 \\ R_{i} (A) \\ 0 \\ ⋮ \\ 0 \end{matrix}) = A E_{i i} = (\begin{matrix} \overset{|}{\underset{|}{0}} & \dots & \overset{|}{\underset{|}{0}} & \overset{|}{\underset{|}{C_{i} (A)}} & \overset{|}{\underset{|}{0}} & \dots & \overset{|}{\underset{|}{0}} \end{matrix}) \\ For example: \\ E_{11} A = (\begin{matrix} A_{11} & A_{12} & A_{13} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) = A E_{11} = (\begin{matrix} A_{11} & 0 & 0 \\ A_{21} & 0 & 0 \\ A_{31} & 0 & 0 \end{matrix}) ⟹ {\begin{matrix} A_{12} = 0 \\ A_{13} = 0 \\ A_{21} = 0 \\ A_{31} = 0 \end{matrix} \\ ⟹ \forall i \in [1, n] : \forall j \in [1, n] : [i \neq j ⟹ A_{i j} = 0] ⟹ A is diagonal \end{matrix}

5c

\begin{matrix} Let A \in R^{n \times n} : \forall B \in R^{n \times n} : A B = B A \\ Prove: A is scalar (\exists α \in R : A = α I) \\ Proof: \\ \forall i \in [2, n] : E_{1 i} A = (\begin{matrix} R_{i} (A) \\ 0 \\ ⋮ \\ 0 \end{matrix}) = A E_{1 i} = (\begin{matrix} \overset{|}{\underset{|}{0}} & \dots & \overset{|}{\underset{|}{0}} & \overset{|}{\underset{|}{C_{1} (A)}} & \overset{|}{\underset{|}{0}} & \dots & \overset{|}{\underset{|}{0}} \end{matrix}) \\ For example: \\ E_{12} A = (\begin{matrix} A_{21} & A_{22} & A_{23} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) = A E_{12} = (\begin{matrix} 0 & A_{11} & 0 \\ 0 & A_{12} & 0 \\ 0 & A_{13} & 0 \end{matrix}) ⟹ A_{22} = A_{11} \\ ⟹ \forall i \in [1, n] : A_{i i} = A_{11} ⟹ A_{11} \in R, A = A_{11} I \end{matrix}

5d

\begin{matrix} Let A \in R^{n \times n} \\ Prove: [\forall B \in R^{n \times n} : A B = B A] ⟺ \exists α \in R : A = α I \\ Proof: \\ One direction is proved in 5c \\ Let \exists α \in R : A = α I \\ Let B \in R^{n \times n} \\ A B = α I B = α (I B) = α B = (B I) α = B α I = B A \\ ⟹ \forall B \in R^{n \times n} : A B = B A \end{matrix}