Proof (a):

Let $A = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix}$, $B = \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix}$

$A + B = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix} + \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix}$

$= \begin{bmatrix} \vec{a}_1 + \vec{b}_1 & \vec{a}_2 + \vec{b}_2 & \dots & \vec{a}_n + \vec{b}_n \end{bmatrix}$

$= \begin{bmatrix} \vec{b}_1 + \vec{a}_1 & \vec{b}_2 + \vec{a}_2 & \dots & \vec{b}_n + \vec{a}_n \end{bmatrix}$

$= B + A$

Proof (b):

Let $A = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix}$, $B = \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix}$, and $C = \begin{bmatrix} \vec{c}_1 & \vec{c}_2 & \dots & \vec{c}_n \end{bmatrix}$

$(A + B) + C = \left( \begin{bmatrix} \vec{a}_1 + \vec{b}_1 & \vec{a}_2 + \vec{b}_2 & \dots & \vec{a}_n + \vec{b}_n \end{bmatrix} \right) + \begin{bmatrix} \vec{c}_1 & \vec{c}_2 & \dots & \vec{c}_n \end{bmatrix}$

$= \begin{bmatrix} (\vec{a}_1 + \vec{b}_1) + \vec{c}_1 & (\vec{a}_2 + \vec{b}_2) + \vec{c}_2 & \dots & (\vec{a}_n + \vec{b}_n) + \vec{c}_n \end{bmatrix}$

$= \begin{bmatrix} \vec{a}_1 + (\vec{b}_1 + \vec{c}_1) & \vec{a}_2 + (\vec{b}_2 + \vec{c}_2) & \dots & \vec{a}_n + (\vec{b}_n + \vec{c}_n) \end{bmatrix}$

$= \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix} + \left( \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix} + \begin{bmatrix} \vec{c}_1 & \vec{c}_2 & \dots & \vec{c}_n \end{bmatrix} \right)$

$= A + (B + C)$

Proof (c):

Let $A = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix}$.

$A + 0 = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix} + \begin{bmatrix} \vec{0} & \vec{0} & \dots & \vec{0} \end{bmatrix}$

$= \begin{bmatrix} \vec{a}_1 + \vec{0} & \vec{a}_2 + \vec{0} & \dots & \vec{a}_n + \vec{0} \end{bmatrix}$

$= \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix}$

$= A$.

Proof (d):

Let $A = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix}$ and $B = \begin{bmatrix} \vec{b}_1 & \vec{b}_2 & \dots & \vec{b}_n \end{bmatrix}$.

$r(A + B) = r\left( \begin{bmatrix} \vec{a}_1 + \vec{b}_1 & \vec{a}_2 + \vec{b}_2 & \dots & \vec{a}_n + \vec{b}_n \end{bmatrix} \right)$

$= \begin{bmatrix} r(\vec{a}_1 + \vec{b}_1) & r(\vec{a}_2 + \vec{b}_2) & \dots & r(\vec{a}_n + \vec{b}_n) \end{bmatrix}$

$= \begin{bmatrix} r\vec{a}_1 + r\vec{b}_1 & r\vec{a}_2 + r\vec{b}_2 & \dots & r\vec{a}_n + r\vec{b}_n \end{bmatrix}$

$= rA + rB$.

Proof (e):

Let $A = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix}$.

$(r + s)A = (r + s)\begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix}$

$= \begin{bmatrix} (r + s)\vec{a}_1 & (r + s)\vec{a}_2 & \dots & (r + s)\vec{a}_n \end{bmatrix}$

$= \begin{bmatrix} r\vec{a}_1 + s\vec{a}_1 & r\vec{a}_2 + s\vec{a}_2 & \dots & r\vec{a}_n + s\vec{a}_n \end{bmatrix}$

$= \begin{bmatrix} r\vec{a}_1 & r\vec{a}_2 & \dots & r\vec{a}_n \end{bmatrix} + \begin{bmatrix} s\vec{a}_1 & s\vec{a}_2 & \dots & s\vec{a}_n \end{bmatrix}$

$= rA + sA$.

Proof (f):

Let $A = \begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix}$.

$r(sA) = r\left(s\begin{bmatrix} \vec{a}_1 & \vec{a}_2 & \dots & \vec{a}_n \end{bmatrix} \right)$

$= r\begin{bmatrix} s\vec{a}_1 & s\vec{a}_2 & \dots & s\vec{a}_n \end{bmatrix}$

$= \begin{bmatrix} r(s\vec{a}_1) & r(s\vec{a}_2) & \dots & r(s\vec{a}_n) \end{bmatrix}$

$= \begin{bmatrix} (rs)\vec{a}_1 & (rs)\vec{a}_2 & \dots & (rs)\vec{a}_n \end{bmatrix}$

$= (rs)A$.

Proof (a):

Let $A$ be an $m \times n$ matrix. By the definition of the transpose, the $(i, j)$-th entry of $A^T$ is $(A^T)_{ij} = A_{ji}$.

Taking the transpose again, the $(i, j)$-th entry of $(A^T)^T$ becomes $((A^T)^T)_{ij} = (A^T)_{ji} = A_{ij}$.

Then $(A^T)^T = A$.

Proof (b):

Let $A$ and $B$ be $m \times n$ matrices. The $(i, j)$-th entry of $A + B$ is $(A + B)_{ij} = A_{ij} + B_{ij}$.

Taking the transpose, the $(i, j)$-th entry of $(A + B)^T$ is $((A + B)^T)_{ij} = (A + B)_{ji} = A_{ji} + B_{ji}$.

The $(i, j)$-th entry of $A^T + B^T$ is also $(A^T + B^T)_{ij} = A_{ji} + B_{ji}$.

So $(A + B)^T = A^T + B^T$.

Proof (c):

Let $A$ be an $m \times n$ matrix and $r$ be a scalar. The $(i, j)$-th entry of $rA$ is $(rA)_{ij} = r(A_{ij})$.

Taking the transpose, the $(i, j)$-th entry of $(rA)^T$ is $((rA)^T)_{ij} = (rA)_{ji} = r(A_{ji})$.

The $(i, j)$-th entry of $rA^T$ is $(rA^T)_{ij} = r(A^T)_{ij} = r(A_{ji})$.

Then $(rA)^T = rA^T$.

Proof (d):

Let $A$ be an $m \times n$ matrix and $B$ be an $n \times p$ matrix. The $(i, j)$-th entry of $AB$ is $(AB)_{ij} = \sum_{k=1}^n A_{ik}B_{kj}$.

Taking the transpose, the $(i, j)$-th entry of $(AB)^T$ is $((AB)^T)_{ij} = (AB)_{ji} = \sum_{k=1}^n A_{jk}B_{ki}$.

The $(i, j)$-th entry of $B^T A^T$ is $(B^T A^T)_{ij} = \sum_{k=1}^n (B^T)_{ik}(A^T)_{kj}$.

By the definition of the transpose, $(B^T)_{ik} = B_{ki}$ and $(A^T)_{kj} = A_{jk}$.

Then $(B^T A^T)_{ij} = \sum_{k=1}^n B_{ki}A_{jk}$.

So $(AB)^T = B^T A^T$.

Proof (a):

Let $A$ be an $m \times n$ matrix, $B$ an $n \times p$ matrix, and $C$ a $p \times q$ matrix.

Define the product $BC$ as: $$(BC)_{ij} = \sum_{k=1}^p b_{ik} c_{kj}.$$ Now, multiply $A$ with $BC$. The $(i, j)$-th entry of $A(BC)$ is: $$(A(BC))_{ij} = \sum_{k=1}^n a_{ik} (BC)_{kj} = \sum_{k=1}^n a_{ik} \left( \sum_{l=1}^p b_{kl} c_{lj} \right). \\= \sum_{k=1}^n \sum_{l=1}^p a_{ik} b_{kl} c_{lj}.$$ Next, compute $AB$ first. The $(i, k)$-th entry of $AB$ is: $$(AB)_{il} = \sum_{k=1}^n a_{ik} b_{kl}.$$ Now multiply $AB$ with $C$. The $(i, j)$-th entry of $(AB)C$ is: $$((AB)C)_{ij} = \sum_{l=1}^p (AB)_{il} c_{lj} = \sum_{l=1}^p \left( \sum_{k=1}^n a_{ik} b_{kl} \right) c_{lj}. \\= \sum_{k=1}^n \sum_{l=1}^p a_{ik} b_{kl} c_{lj}.$$ Since the entries of $A(BC)$ and $(AB)C$ are identical, we have: $$A(BC) = (AB)C.$$

Proof (b):

Let $A$, $B$, and $C$ be matrices such that $A(B + C)$ is defined.

By definition of matrix addition: $$B + C = \begin{bmatrix} b_{ij} + c_{ij} \end{bmatrix}.$$ $$A(B + C) = A\begin{bmatrix} b_{ij} + c_{ij} \end{bmatrix} = \begin{bmatrix} \sum_{j=1}^n a_{ij}(b_{jk} + c_{jk}) \end{bmatrix}.$$ Distributing: $$A(B + C) = \begin{bmatrix} \sum_{j=1}^n a_{ij}b_{jk} \end{bmatrix} + \begin{bmatrix} \sum_{j=1}^n a_{ij}c_{jk} \end{bmatrix} \\= AB + AC.$$

Proof (c):

Let $A$, $B$, and $C$ be matrices such that $(B + C)A$ is defined.

$$B + C = \begin{bmatrix} b_{ij} + c_{ij} \end{bmatrix}.$$ $$(B + C)A = \begin{bmatrix} b_{ij} + c_{ij} \end{bmatrix}A = \begin{bmatrix} \sum_{j=1}^n (b_{ij} + c_{ij})a_{jk} \end{bmatrix}.$$ Distributing: $$(B + C)A = \begin{bmatrix} \sum_{j=1}^n b_{ij}a_{jk} \end{bmatrix} + \begin{bmatrix} \sum_{j=1}^n c_{ij}a_{jk} \end{bmatrix} \\= BA + CA.$$

Proof (d):

Let $A$ and $B$ be matrices such that $AB$ is defined, and let $r$ be a scalar.

By definition of scalar multiplication: $$r(AB) = r\begin{bmatrix} \sum_{j=1}^n a_{ij}b_{jk} \end{bmatrix} = \begin{bmatrix} r\sum_{j=1}^n a_{ij}b_{jk} \end{bmatrix}.$$ Distributing $r$: $$r(AB) = \begin{bmatrix} \sum_{j=1}^n (ra_{ij})b_{jk} \end{bmatrix} = (rA)B,$$ and similarly: $$r(AB) = \begin{bmatrix} \sum_{j=1}^n a_{ij}(rb_{jk}) \end{bmatrix} = A(rB).$$ Thus, $r(AB) = (rA)B = A(rB)$.

Proof (e):

Let $A$ be an $m \times n$ matrix, and let $I_m$ and $I_n$ be the identity matrices of size $m \times m$ and $n \times n$, respectively.

The columns of $A$ can be written as $A\vec{e}_j$, where $\vec{e}_j$ is the $j$-th standard basis vector.

Multiplying $I_m$ with $A$: $$I_m A = \begin{bmatrix} \vec{e}_1 & \cdots & \vec{e}_m \end{bmatrix} A = A.$$ Similarly: $$A I_n = A\begin{bmatrix} \vec{e}_1 & \cdots & \vec{e}_n \end{bmatrix} = A.$$ Thus, $I_m A = A$ and $A I_n = A$.