Proof:

Let some $mxn$ matrix $A = [\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_n], \vec{x} = \left[\begin{array}{c} x_1 \\x_2 \\\vdots \\x_n\end{array}\right] \in \mathbb{R}^{n}$ and let the matrix equation $A\vec{x}=\vec{b}$ have a valid solution.

Since $A\vec{x}$ has a valid solution $\vec{b}$ we can write $\vec{b}$ as a linear combination of the columns of $A$ and the weights in $\vec{x}$.

$$x_1\vec{a}_1 + x_2\vec{a}_2 + \cdots + x_n\vec{a}_n = \vec{b}$$ since the matrix equation has a valid solution it must follow that the corresponding vector equation has the same solution set.

Now, we can construct an augmented matrix from this:

$$\left[ \begin{array}{cccc|c} a_{11} & a_{21} & \cdots & a_{n1} & b_1 \\ a_{12} & a_{22} & \cdots & a_{n2} & b_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ a_{1m} & a_{2m} & \cdots & a_{nm} & b_n \end{array} \right]$$ From this we can now construct a system of linear equations $$\left\{ \begin{array}{l} x_{1}a_{11} + x_{2}a_{21} + \dots + x_{n}a_{n1} = b_1 \\\\ x_{2}a_{12} + x_{2}a_{22} + \dots + x_{n}a_{n2} = b_2 \\\\ \dots \\\\ x_{n}a_{1m} + x_{2}a_{2m} + \dots + x_{n}a_{nm} = b_n \\\\ \end{array} \right.$$

Since the solution ($x_1, x_2,\cdots,x_n$) is valid for the matrix equation it must also follow that it is valid for the assocaited vector equation and for the associated system of equations since we showed that they are all equivelant.

Proof $\rightarrow$:

Suppose the matrix equation $A\vec{x}=\vec{b}$ has a solution. Let $A$ be an $mxn$ matrix where $A = [\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_{n}]$ and let $x = \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right]$.

We can rewrite this as the following vector equation: $$x_1\vec{a}_1 + x_2\vec{a}_2 + \cdots + x_n\vec{a}_n = \vec{b}$$ $\vec{b}$ is generated as a linear comibination of the columns of $A$ and the weights $(x_1, x_2,\cdots,x_n)$. Since the matrix equation has a solution, $\vec{b}$ can be generated as a linear combination of the columns of $A$.

Proof $\leftarrow$:

Suppose $\vec{b}$ can be generated by a linear combination of the columns of some matrix $A$. Let $A = [\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_{n}]$, then there exist some weights $(x_1, x_2, \cdots, x_n)$ that satisfy the following vector equation: $$x_1\vec{a}_1 + x_2\vec{a}_2 + \cdots + x_n\vec{a}_n = \vec{b}$$ We can rewrite this as: $$[\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_n]\left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right]=\vec{b}$$ By definition this is a matrix equation and can be written in the following form: $$A\vec{x}=\vec{b}$$ Since $\vec{b}$ can be genereated by a linear combination of the columns of A it must also follow that the matrix equation has a solution because we showed that they are equivelant.

Proof (statement 1 given statement 3):

Given the columns of some $m x n$ matrix $A$ span $\mathbb{R}^{m}$ lets assume the matrix equation $A\vec{x}=\vec{b}$ does not have a solution. Lets denote the columns of $A$ as [$\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_n$]. By definiton Span{$\vec{a}_1, \vec{a}_2, \cdots, \vec{a}_n$} is the set of all linear combinations of $\vec{a}_1, \vec{a}_2, \cdots, \vec{a}_n$ and since the columns of $A$ span $\mathbb{R}^{m}$ it must be true that we can generate a vector $\vec{b}$ as a linear combination of the columns of $A$. More specifically, there exists a solution $x_1, x_2,\cdots,x_n$ for the vector equation $x_1\vec{a}_1 + x_2\vec{a}_{2} + \cdots + x_n\vec{a}_n = \vec{b}$. Since we previously showed that a vector equation has the same solution set to its associated matrix equation then $x_1\vec{a}_1 + x_2\vec{a}_{2} + \cdots + x_n\vec{a}_n = \vec{b}$ has the same solution set as $A\vec{x}=\vec{b}$ but this is a contradiction to our original assumption that the matrix equation does not have a solution.

Proof (statement 3 given statement 1):

Given that for each $\vec{b} \in \mathbb{R}^{m}$, $A\vec{x}=\vec{b}$ has a solution lets denote the columns of some $mxn$ matrix $A$ as $\vec{a}_1 \ \vec{a}_2 \ \cdots \vec{a}_n \in \mathbb{R}^m$ and let $x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$. We can rewrite this equation now in the following form: $$[\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_n] \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}=\vec{b}$$ Which in turn can now be written in the following form: $$x_1\vec{a}_1 + x_2\vec{a}_2 + \cdots + x_n\vec{a}_n = \vec{b}$$ Since the solution holds for the intial matrix equation, it must also hold for the vector equation because we have shown that they are equivelant. Furthermore, since we can generate some $\vec{b}$ as a linear combination of the columns of $A$ it must also be true that the columns of $A$ Span $\mathbb{R}^{m}$.

Proof (statement 1 given statement 4):

Given that some $mxn$ matrix $A$ has a pivot in each row, lets denote the columns of $A$ as $\vec{a}_1 \ \vec{a}_2 \ \cdots \vec{a}_n$. Since we have a pivot in each row of the coefficient matrix then it must also be true that the associated augmented matrix of $A$ will never have a pivot in the augmented column. Since no contradictions can arise from this, the system will always have a consistent solution. Then there exists weights $x_1, x_2, \cdots, x_n \in \mathbb{R}$ such that $x_1\vec{a}_1 + x_2\vec{a}_2 + \cdots + x_n\vec{a}_n = \vec{b}$ will always have a solution for each $\vec{b} \in \mathbb{R}^m$ because we have m pivot columns. Since the vector equation will always have a valid solution and since we have already shown that a vector equation has the same solution set as the associated matrix equation, then it must follow that the matrix equation has a solution for each $\vec{b} \in \mathbb{R}^m$

Proof (statement 4 given statement 1):

Given that for each $\vec{b} \in \mathbb{R}^{m}$, $A\vec{x}=\vec{b}$ has a solution lets assume that $A$ does not have a pivot position in each row. Then there exists a row in the form of [$0 \ 0 \ \cdots \ 0 \ b$] where b is the augmented column. Now, we are given that for each $\vec{b} \in \mathbb{R}^{m}$, $A\vec{x}=\vec{b}$ has a solution this means that there exists a solution where b is non-zero but this is a contradiction to our original assumption because if b is non-zero then our system is inconsistent and we already stated that the matrix equation has a solution for each $\vec{b} \in \mathbb{R}^m$

Proof(Statement one):

Let $A = [\vec{a}_1 \ \vec{a}_2 \ \cdots \ \vec{a}_n]$, $\vec{u} = (u_1 \ u_2 \ \cdots \ u_n)$ and $\vec{v} = (v_1 \ v_2\ \cdots \ \_n)$

Proof:

Suppose $\vec{x} \in \mathbb{R}^n$ lets assume that $I_n \vec{x} \ne \vec{x}$, where $I_n$ is the $n \times n$ identity matrix. This implies that there is at least one component of $I_n \vec{x}$ that is different from the corresponding component of $\vec{x}$. By definition, $I_n$ has 1s on the main diagonal and 0s elsewhere. Therefore, each component of $I_n \vec{x}$ is simply the corresponding component of $\vec{x}$. There cannot exist a value in the main diagonal of $I_n$ that is not 1, nor can there be a value elsewhere that affects $\vec{x}$ due to the 0s. Then it must be true that $I_n \vec{x} = \vec{x}$, which is a contradiction to our assumption that $I_n \vec{x} \ne \vec{x}$. It must be true that $I_n \vec{x} = \vec{x}$ for all $\vec{x} \in \mathbb{R}^n$.