Linear-1 10

1

\begin{matrix} Let A \in R^{n \times n} \\ Let T : R^{n \times n} \to R^{n \times n}, T (B) = A B - B A \\ Let S : R^{n \times n} \to R^{n \times n}, S (B) = A B + B A \end{matrix}

1a

\begin{matrix} Prove: T, S are linear transformations \\ Proof: \\ Let α \in R, B_{1}, B_{2} \in R^{n \times n} \\ T (B_{1} + α B_{2}) = A (B_{1} + α B_{2}) - (B_{1} + α B_{2}) A = A B_{1} - B_{1} A + α A B_{2} - α B_{2} A = \\ = T (B_{1}) + α T (B_{2}) \\ ⟹ T is a linear transformation \\ S (B_{1} + α B_{2}) = A (B_{1} + α B_{2}) + (B_{1} + α B_{2}) A = A B_{1} + B_{1} A + α A B_{2} + α B_{2} A = \\ = S (B_{1}) + α S (B_{2}) \\ ⟹ S is a linear transformation \end{matrix}

1b

\begin{matrix} Let A = (\begin{matrix} 2 & 2 \\ 2 & 2 \end{matrix}) \\ Find basis and dimension of I m (T), k e r (T) \\ Prove: I m (T) \oplus k e r (T) = R^{2 \times 2} \\ Solution: \\ I m (T) = {B \in R^{2 \times 2} | \exists C \in R^{2 \times 2} : T (B) = C} = {A B - B A | B \in R^{2 \times 2}} \\ A B = (\begin{matrix} 2 & 2 \\ 2 & 2 \end{matrix}) (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} 2 a + 2 c & 2 b + 2 d \\ 2 a + 2 c & 2 b + 2 d \end{matrix}) \\ B A = (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} 2 & 2 \\ 2 & 2 \end{matrix}) = (\begin{matrix} 2 a + 2 b & 2 a + 2 b \\ 2 c + 2 d & 2 c + 2 d \end{matrix}) \\ A B - B A = (\begin{matrix} 2 c - 2 b & 2 d - 2 a \\ 2 a - 2 d & 2 b - 2 c \end{matrix}) = (2 b - 2 c) (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}) + (2 a - 2 d) (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}) \\ I m (T) = {(2 b - 2 c) (\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}) + (2 a - 2 d) (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}) | a, b, c, d \in R} = \\ = s p ({(\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})}) \\ {(\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})} is a linear independence \\ ⟹ {(\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})} is a basis of I m (T) and d i m (I m (T)) = 2 \end{matrix}

\begin{matrix} k e r (T) = {B \in R^{2 \times 2} | T (B) = 0} = {(\begin{matrix} a & b \\ c & d \end{matrix}) | \begin{matrix} a - d = 0 \\ b - c = 0 \end{matrix}} = {(\begin{matrix} a & b \\ b & a \end{matrix}) | a, b \in R} = \\ = s p ({(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})}) \\ {(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})} is a linear independence \\ ⟹ {(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})} is a basis of k e r (T) and d i m (k e r (T)) = 2 \\ k e r (T) \cap I m (T) = s p ({(\begin{matrix} - 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})}) \cap s p ({(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})}) = {0} \\ ⟹ d i m (k e r (T) \cap I m (T)) = 0 \\ k e r (T), I m (T) \subseteq R^{2 \times 2} ⟹ k e r (T) + I m (T) \subseteq R^{2 \times 2} \\ d i m (k e r (T) + I m (T)) = d i m (k e r (T)) + d i m (I m (T)) - d i m (k e r (T) \cap I m (T)) \\ ⟹ d i m (k e r (T) + I m (T)) = 2 + 2 - 0 = 4 = d i m (R^{2 \times 2}) \\ ⟹ k e r (T) + I m (T) = R^{2 \times 2} \land k e r (T) \cap I m (T) = {0} \\ ⟹ k e r (T) \oplus I m (T) = R^{2 \times 2} \end{matrix}

1c

\begin{matrix} Let A = (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) \\ Find basis of k e r (S T), I m (S T) \\ Solution: \\ S T (B) = S (T (B)) = S (A B - B A) = A (A B - B A) + (A B - B A) A = \\ = A A B - A B A + A B A - B A A = A^{2} B - B A^{2} \\ A^{2} = (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}) = (\begin{matrix} 2 & 2 \\ 2 & 2 \end{matrix}) \\ Let A^{'} = A^{2} \\ ⟹ S T (B) = A^{'} B - B A^{'} = T_{A^{'}} (B) \\ ⟹ I m (S T), k e r (S T) are equal to I m (T), k e r (T) from the previous task \end{matrix}

1d

\begin{matrix} Let C_{A} = {B \in R^{n \times n} | A B = B A} \\ Prove: C_{A} is a vector subspace of R^{n \times n} \\ Proof: \\ C_{A} \subseteq R^{n \times n} \\ C_{A} = {B \in R^{n \times n} | A B - B A = 0} = k e r (T) \\ Kernel of a linear transformation is always a vector subspace of the domain \\ ⟹ C_{A} is a vector subspace of R^{n \times n} \end{matrix}

2

\begin{matrix} Let n \in N : 1 \leq m \leq n \\ Let f^{(m)} be m ’th derivative of f \\ T : R_{n} [x] \to R_{n} [x], T (f) = x^{m - 1} f^{(m)} \end{matrix}

2a

\begin{matrix} Prove: T is a linear transformation \\ Proof: \\ Let f, g \in R_{n} [x], α \in R \\ T (f + α g) = x^{m - 1} (f + α g)^{(m)} = x^{m - 1} (f^{(m)} + α (g^{(m)})) = x^{m - 1} f^{(m)} + α x^{m - 1} g^{(m)} = \\ = T (f) + α T (g) \\ ⟹ T is a linear transformation \end{matrix}

2b

\begin{matrix} Find basis and dimension of I m (T), k e r (T) \\ Solution: \\ Let f \in R_{n} [x] \\ f (x) = \sum_{k = 0}^{n} a_{k} x^{k} \\ (a_{k} x^{k})^{(m)} = {\begin{cases} 0 & k < m \\ \frac{k!}{(k - m)!} a_{k} x^{k - m} & k \geq m \end{cases} \\ f^{(m)} (x) = {(\sum_{k = 0}^{n} a_{k} x^{k})}^{(m)} = \sum_{k = 0}^{n} (a_{k} x^{k})^{(m)} = \sum_{k = m}^{n} \frac{k!}{(k - m)!} a_{k} x^{k - m} \\ ⟹ T (f) = x^{m - 1} (\sum_{k = m}^{n} \frac{k!}{(k - m)!} a_{k} x^{k - m}) = \sum_{k = m}^{n} \frac{k!}{(k - m)!} a_{k} x^{k - 1} = \\ = m! a_{m} x^{m - 1} + (m + 1)! a_{m + 1} x^{m} + \dots + \frac{n!}{(n - m)!} a_{n} x^{n - 1} \\ ⟹ T (f) \in s p ({x^{m - 1}, x^{m}, \dots, x^{n - 1}}) \\ ⟹ I m (T) \subseteq s p ({x^{m - 1}, x^{m}, \dots, x^{n - 1}}) \end{matrix}

\begin{matrix} Let m - 1 \leq k \leq n - 1 \\ x^{k} \in s p ({x^{m - 1}, x^{m}, \dots, x^{n - 1}}) \\ x^{k + 1} \in R_{n} [x] \\ T (x^{k + 1}) = x^{m - 1} \cdot \frac{(k + 1)!}{(k + 1 - m)!} \cdot x^{k + 1 - m} = \frac{(k + 1)!}{(k + 1 - m)!} x^{k} \\ ⟹ T (\frac{(k + 1 - m)!}{(k + 1)!} x^{k + 1}) = x^{k} \\ ⟹ x^{k} \in I m (T) ⟹ {x^{m - 1}, \dots, x^{n - 1}} \subseteq I m (T) \\ s p ({x^{m - 1}, \dots, x^{n - 1}}) is a minimal vector space containing {x^{m - 1}, \dots, x^{n - 1}} \\ ⟹ s p ({x^{m - 1}, \dots, x^{n - 1}}) \subseteq I m (T) \\ ⟹ I m (T) = s p ({x^{m - 1}, \dots, x^{n - 1}}) \\ {x^{m - 1}, \dots, x^{n - 1}} is a linear independence \\ ⟹ {x^{m - 1}, \dots, x^{n - 1}} is a basis of I m (T) and d i m (I m (T)) = n - m + 1 \\ d i m (k e r (T)) = d i m (R_{n} [x]) - d i m (I m (T)) = m \\ 1. {1, x, \dots, x^{m - 1}} is a linear independence \\ 2. d i m (s p ({1, x, \dots, x^{m - 1}})) = m = d i m (k e r (T)) \\ ⟹ {1, x, \dots, x^{m - 1}} is a basis of k e r (T) and d i m (k e r (T)) = m \end{matrix}

3

\begin{matrix} Let V, W be finitely generated vector spaces over F \\ Let S, T : V \to W be linear transformations \end{matrix}

3a

\begin{matrix} Prove or disprove: k e r (S) \cap k e r (T) = {0_{V}} ⟹ I m (S) \cap I m (T) = {0_{W}} \\ Disproof: \\ Let W \neq {0} \\ Let S = T be invertible \\ ⟹ \begin{matrix} 1. & d i m (V) = d i m (W) \\ 2. & k e r (T) = k e r (S) = k e r (S) \cap k e r (T) = {0_{V}} \\ 3. & I m (T) = I m (S) = I m (S) \cap I m (T) = W \neq {0_{W}} \end{matrix} \\ For example: \\ V = W = R^{3} \\ T = S = I d_{R^{3}} \\ k e r (S) \cap k e r (T) = {0} \\ I m (S) \cap I m (T) = R^{3} \end{matrix}

3b

\begin{matrix} Prove: I m (T) \subseteq I m (S) ⟺ \exists a linear transformation R : V \to V, T = S R \\ Proof: \\ Let I m (T) \subseteq I m (S) \\ ⟹ \forall v \in V : \exists v_{1} \in V : T (v) = S (v_{1}) \\ Let u, v \in V, α \in F \\ ⟹ \exists u_{1}, v_{1} : S (u_{1}) = T (u), S (v_{1}) = T (v) \\ ⟹ S (v_{1} + α u_{1}) = S (v_{1}) + α S (u_{1}) = T (v) + α T (u) = T (v + α u) \\ Let R : V \to V, R (v) = v_{1} \\ R (v + α u) = w \\ S (w) = T (v + α u) ⟹ w = v_{1} + α u_{1} = R (v) + α R (u) \\ ⟹ R (v + α u) = R (v) + α R (u) ⟹ R is a linear transformation \\ \forall v \in V : \exists v_{1} : T (v) = S (v_{1}) = S (R (v)) ⟹ T = S R & (1) \\ Let \exists a linear transformation R : V \to V, T = S R \\ ⟹ \forall v \in V : T (v) = S (R (v)) \\ ⟹ \forall v \in V : \exists v_{1} \in V : R (v) = v_{1} \\ ⟹ \forall v \in V : \exists v_{1} \in V : T (v) = S (v_{1}) \\ ⟹ I m (T) \subseteq I m (S) & (2) \\ (1) and (2) ⟹ \\ ⟹ I m (T) \subseteq I m (S) ⟺ \exists a linear transformation R : V \to V, T = S R \end{matrix}

4

\begin{matrix} Let T : V \to V be a linear transformation \\ 1. I m (T) = I m (T^{2}) \\ 2. k e r (T) = k e r (T^{2}) \\ 3. k e r (T) \oplus I m (T) = V \\ Prove: 1 ⟺ 2 ⟺ 3 \\ Proof: \\ Let I m (T) = I m (T^{2}) \\ ⟹ d i m (I m (T)) = d i m (I m (T^{2})) \\ ⟹ d i m (k e r (T)) = d i m (V) - d i m (I m (T)) = d i m (V) - d i m (I m (T^{2})) = d i m (k e r (T^{2})) \\ Let v \in k e r (T) \\ ⟹ T (v) = 0 ⟹ T (T (v)) = T (0) = 0 \\ ⟹ v \in k e r (T^{2}) ⟹ k e r (T) \subseteq k e r (T^{2}) \\ ⟹ k e r (T) = k e r (T^{2}) ⟹ 1 ⟹ 2 \\ Let k e r (T) = k e r (T^{2}) \\ ⟹ d i m (k e r (T)) = d i m (k e r (T^{2})) \\ ⟹ d i m (I m (T)) = d i m (V) - d i m (k e r (T)) = d i m (V) - d i m (k e r (T^{2})) = d i m (I m (T^{2})) \\ Let v \in I m (T^{2}) \\ ⟹ \exists u \in V : T (T (u)) = v \\ T (u) \in V ⟹ v \in I m (T) ⟹ I m (T^{2}) \subseteq I m (T) \\ ⟹ I m (T) = I m (T^{2}) ⟹ 2 ⟹ 1 \\ ⟹ 1 ⟺ 2 \end{matrix}

\begin{matrix} Let k e r (T) = k e r (T^{2}) \\ k e r (T), I m (T) \subseteq V ⟹ k e r (T) + I m (T) \subseteq V \\ Let v \in k e r (T) \cap I m (T) \\ ⟹ T (v) = 0 \land \exists u \in V : T (u) = v \\ ⟹ T (T (u)) = 0 ⟹ u \in k e r (T^{2}) ⟹ u \in k e r (T) \\ ⟹ T (u) = 0 ⟹ v = 0 ⟹ k e r (T) \cap I m (T) = {0} \\ ⟹ d i m (k e r (T) \cap I m (T)) = 0 \\ ⟹ d i m (k e r (T) + I m (T)) = d i m (k e r (T)) + d i m (I m (T)) = d i m (V) \\ ⟹ k e r (T) + I m (T) = V \land k e r (T) \cap I m (T) = {0} \\ ⟹ k e r (T) \oplus I m (T) = V ⟹ 2 ⟹ 3 \\ Let k e r (T) \oplus I m (T) = V \\ Let v \in k e r (T^{2}) \\ ⟹ T^{2} (v) = T (T (v)) = 0 \\ Let w = T (v) \\ ⟹ w \in I m (T) \\ T (w) = 0 ⟹ w \in k e r (T) ⟹ w \in k e r (T) \cap I m (T) ⟹ w = 0 \\ ⟹ T (v) = 0 ⟹ v \in k e r (T) ⟹ k e r (T^{2}) \subseteq k e r (T) \\ As shown earlier, k e r (T) \subseteq k e r (T^{2} ⟹ k e r (T) = k e r (T^{2}) ⟹ 3 ⟹ 2 \\ ⟹ 2 ⟺ 3 \\ ⟹ 1 ⟺ 2 ⟺ 3 \end{matrix}

5a

\begin{matrix} Prove or disprove: \exists different non-invertible linear transformations S, T : R^{3} \to R^{3} \\ such that: I m (S) = I m (T), k e r (S) = k e r (T) \\ Proof: \\ A = (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}), B = (\begin{matrix} 0 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 1 \end{matrix}) \\ T (v) = A v, S (v) = B v \\ ⟹ T ((\begin{matrix} a \\ b \\ c \end{matrix})) = (\begin{matrix} 0 \\ b \\ c \end{matrix}), S ((\begin{matrix} a \\ b \\ c \end{matrix})) = (\begin{matrix} 0 \\ - b \\ c \end{matrix}) \\ I m (T) = I m (S) = s p ({(\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}) \\ k e r (T) = k e r (S) = s p ({(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})}) \end{matrix}

5b

\begin{matrix} Prove or disprove: \exists non-invertible linear transformations S, T : R^{3} \to R^{3} such that: \\ k e r (S) = I m (T), k e r (T) = I m (S) \\ Proof: \\ A = (\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}), B = (\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) \\ T (v) = A v, S (v) = B v \\ ⟹ T ((\begin{matrix} a \\ b \\ c \end{matrix})) = (\begin{matrix} 0 \\ b \\ c \end{matrix}), S ((\begin{matrix} a \\ b \\ c \end{matrix})) = (\begin{matrix} a \\ 0 \\ 0 \end{matrix}) \\ I m (T) = k e r (S) = s p ({(\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix})}) \\ I m (S) = k e r (T) = s p ({(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix})}) \end{matrix}