Proof:

We want to prove that the column space of an $m \times n$ matrix $A$ is all of $\mathbb{R}^m$ if and only if the equation $A \vec{x} = \vec{b}$ has a solution for every $\vec{b} \in \mathbb{R}^m$.

$\text{Col}(A) = \mathbb{R}^m \\ \Leftrightarrow \text{For every } \vec{b} \in \mathbb{R}^m, \vec{b} \text{ can be expressed as a linear combination of the columns of } A. \\ \Leftrightarrow \text{For every } \vec{b} \in \mathbb{R}^m, \vec{b} \in \text{Col}(A) \\ \Leftrightarrow \text{For every } \vec{b} \in \mathbb{R}^m, \exists \, \vec{x} \in \mathbb{R}^n \text{ such that } A \vec{x} = \vec{b}.$

Thus, the column space of $A$ is all of $\mathbb{R}^m$ if and only if the equation $A \vec{x} = \vec{b}$ has a solution for every $\vec{b} \in \mathbb{R}^m$.

Proof:

Let $A$ be an $m \times n$ matrix. The null space of $A$, denoted as $\text{Nul}(A)$, is defined as: $$\text{Nul}(A) = \{ \vec{x} \in \mathbb{R}^n \mid A \vec{x} = \vec{0} \}.$$ We will show that $\text{Nul}(A)$ is a subspace of $\mathbb{R}^n$ by verifying the three subspace properties:

1. The zero vector is in $\text{Nul}(A)$:
Let $\vec{x} = \vec{0}$. Then: $$A \vec{0} = \vec{0}.$$ Hence, $\vec{0} \in \text{Nul}(A)$.

2. Closed under addition:
Let $\vec{u}, \vec{w} \in \text{Nul}(A)$. Then: $$A \vec{u} = \vec{0} \quad \text{and} \quad A \vec{w} = \vec{0}.$$ Adding these equations gives: $$A (\vec{u} + \vec{w}) = A \vec{u} + A \vec{w} = \vec{0} + \vec{0} = \vec{0}.$$ Hence, $\vec{u} + \vec{w} \in \text{Nul}(A)$.

3. Closed under scalar multiplication:
Let $\vec{u} \in \text{Nul}(A)$ and let $c \in \mathbb{R}$. Then: $$A \vec{u} = \vec{0}.$$ Multiplying by $c$ gives: $$A (c \vec{u}) = c (A \vec{u}) = c \vec{0} = \vec{0}.$$ Hence, $c \vec{u} \in \text{Nul}(A)$.

Thus, $\text{Nul}(A)$ satisfies the conditions for a subspace of $\mathbb{R}^n$.

Proof:

Let $A$ be an $m \times n$ matrix. The column space of $A$, denoted by $\text{Col}(A)$, is defined as: $$\text{Col}(A) = \{ A \vec{x} \mid \vec{x} \in \mathbb{R}^n \}.$$ This is the set of all linear combinations of the columns of $A$. We will show that $\text{Col}(A)$ is a subspace of $\mathbb{R}^m$ by verifying the three subspace properties:

1. The zero vector is in $\text{Col}(A)$:
Let $\vec{x} = \vec{0} \in \mathbb{R}^n$. Then: $$A \vec{x} = A \vec{0} = \vec{0}.$$ Hence, $\vec{0} \in \text{Col}(A)$.

2. Closed under addition:
Let $\vec{u}, \vec{w} \in \text{Col}(A)$. Then there exist $\vec{x}_1, \vec{x}_2 \in \mathbb{R}^n$ such that: $$\vec{u} = A \vec{x}_1 \quad \text{and} \quad \vec{w} = A \vec{x}_2.$$ Adding $\vec{u}$ and $\vec{w}$ gives: $$\vec{u} + \vec{w} = A \vec{x}_1 + A \vec{x}_2 = A (\vec{x}_1 + \vec{x}_2).$$ Since $\vec{x}_1 + \vec{x}_2 \in \mathbb{R}^n$, $\vec{u} + \vec{w} \in \text{Col}(A)$.

3. Closed under scalar multiplication:
Let $\vec{u} \in \text{Col}(A)$ and $c \in \mathbb{R}$. Then there exists $\vec{x} \in \mathbb{R}^n$ such that: $$\vec{u} = A \vec{x}.$$ Multiplying by $c$ gives: $$c \vec{u} = c (A \vec{x}) = A (c \vec{x}).$$ Since $c \vec{x} \in \mathbb{R}^n$, $c \vec{u} \in \text{Col}(A)$.

Thus, $\text{Col}(A)$ satisfies the conditions for a subspace of $\mathbb{R}^m$.