2. Nullspace and column space

a)

$A^{2} = (v v^{⊤}) (v v^{⊤}) = v (v^{⊤} v) v^{⊤} = v ∥ v ∥^{2} v^{⊤} = v v^{⊤} = A$ $P^{2} = (I_{3} - A) (I_{3} - A) = I_{3}^{2} - 2 I_{3} A + A^{2} = I_{3} - 2 A + A = I_{3} - A = P$

b)

$w is orthogonal to v ⟹ v^{⊤} w = 0 ⟹ v v^{⊤} w = v 0 = 0 ⟹ A w = 0$

c)

$A w = 0 ⟹ v v^{⊤} w = 0 ⟹ v^{⊤} v v^{⊤} w = v^{⊤} 0 = 0 ⟹ ∥ v ∥^{2} v^{⊤} w = 0 ⟹ v^{⊤} w = 0$ $⟹ v \cdot w = 0 ⟹ w \cdot v = 0$

d)

$N (A) = {w \in R^{3} ∣ v^{⊤} w = 0}$

e)

the rank of $A$ is 1 $A$ is not invertible

f)

$C (A) = {A w ∣ w \in R^{3}} = {v v^{⊤} w ∣ w \in R^{3}} = {(v^{⊤} w) v ∣ w \in R^{3}} = {α v ∣ α \in R} = s p an (v)$

g)

Let $y \in C (A)$ $y = A x$ for some $x \in R^{3}$ $⟹ A y = A^{2} x ⟹ A y = A x = y$ $y \in {w \in R^{3} ∣ A w = w}$

Conversely, Let $A w = w$ , then $w = A w \in C (A)$ Thus $C (A) = {w \in R^{3} ∣ A w = w}$

h)

Let $w \in N (P)$ $⟺ P w = 0$ $⟺ (I_{3} - A) w = 0$ $⟺ I_{3} w - A w = 0$ $⟺ w - A w = 0$ $⟺ w = A w$ $⟹ N (P) = {w \in R^{3} ∣ A w = w} = C (A)$

i)

First we prove $C (P) \subseteq N (A)$ Let $w \in C (P)$ $⟹ w = P x$ for some $x \in R^{3}$ $⟹ w = (I_{3} - A) x$ $⟹ A w = A (I_{3} - A) x = (A - A^{2}) x = (A - A) x = 0 x = 0$ Thus $w \in N (A)$

Then we prove $N (A) \subseteq C (P)$ Let $w \in N (A)$ $⟹ A w = 0$ $⟹ P w = (I_{3} - A) w = w - A w = w - 0 = w$ $⟹ w \in C (P)$

Since $C (P) \subseteq N (A)$ and $N (A) \subseteq C (P)$ , $C (P) = N (A)$

Thomas Second Brain

Explorer

LinArg assignment 7

2. Nullspace and column space

a)

b)

c)

d)

e)

f)

g)

h)

i)

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