(a) To prove this statement, we can write out the components of the vectors $\mathbf{u}$ and $\mathbf{v}$:

$$\mathbf{u} = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix} \quad \mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ \vdots \ v_n \end{pmatrix}$$

Then, the vector $\mathbf{u} + \mathbf{v}$ is given by:

$$\mathbf{u} + \mathbf{v} = \begin{pmatrix} u_1 + v_1 \ u_2 + v_2 \ \vdots \ u_n + v_n \end{pmatrix}$$

Similarly, the vector $\mathbf{v} + \mathbf{u}$ is given by:

$$\mathbf{v} + \mathbf{u} = \begin{pmatrix} v_1 + u_1 \ v_2 + u_2 \ \vdots \ v_n + u_n \end{pmatrix}$$

Since $u_i + v_i = v_i + u_i$ for all $i$, it follows that $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$.

(b) To prove this statement, we can again write out the components of the vectors $\mathbf{u}, \mathbf{v}$, and $\mathbf{w}$:

$$\mathbf{u} = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix} \quad \mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ \vdots \ v_n \end{pmatrix} \quad \mathbf{w} = \begin{pmatrix} w_1 \ w_2 \ \vdots \ w_n \end{pmatrix}$$

Then, the vector $(\mathbf{u} + \mathbf{v}) + \mathbf{w}$ is given by:

$$(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \begin{pmatrix} u_1 + v_1 \ u_2 + v_2 \ \vdots \ u_n + v_n \end{pmatrix} + \begin{pmatrix} w_1 \ w_2 \ \vdots \ w_n \end{pmatrix} = \begin{pmatrix} u_1 + v_1 + w_1 \ u_2 + v_2 + w_2 \ \vdots \ u_n + v_n + w_n \end{pmatrix}$$

Similarly, the vector $\mathbf{u} + (\mathbf{v} + \mathbf{w})$ is given by:

$$\mathbf{u} + (\mathbf{v} + \mathbf{w}) = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix} + \begin{pmatrix}

v_1 + w_1 \ v_2 + w_2 \ \vdots \ v_n + w_n \end{pmatrix} = \begin{pmatrix} u_1 + v_1 + w_1 \ u_2 + v_2 + w_2 \ \vdots \ u_n + v_n + w_n \end{pmatrix}$$

Since $u_i + v_i + w_i = u_i + (v_i + w_i)$ for all $i$, it follows that $(\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})$.

(c) To prove this statement, we can again write out the components of the vector $\mathbf{u}$:

$$\mathbf{u} = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix}$$

Then, the vector $\mathbf{u} + \mathbf{0}$ is given by:

$$\mathbf{u} + \mathbf{0} = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix} + \begin{pmatrix} 0 \ 0 \ \vdots \ 0 \end{pmatrix} = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix}$$

Similarly, the vector $\mathbf{0} + \mathbf{u}$ is given by:

$$\mathbf{0} + \mathbf{u} = \begin{pmatrix} 0 \ 0 \ \vdots \ 0 \end{pmatrix} + \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix} = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix}$$

Since $\mathbf{u} + \mathbf{0} = \mathbf{0} + \mathbf{u}$, it follows that $\mathbf{u} + \mathbf{0} = \mathbf{0} + \mathbf{u} = \mathbf{u}$.

(d) To prove this statement, we can again write out the components of the vector $\mathbf{u}$:

$$\mathbf{u} = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix}$$

Then, the vector $\mathbf{u} + (-\mathbf{u})$ is given by:

$$\mathbf{u} + (-\mathbf{u}) = \begin{pmatrix} u_1 \ u_2 \ \vdots

\ u_n \end{pmatrix} + \begin{pmatrix} -u_1 \ -u_2 \ \vdots \ -u_n \end{pmatrix} = \begin{pmatrix} u_1 - u_1 \ u_2 - u_2 \ \vdots \ u_n - u_n \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ \vdots \ 0 \end{pmatrix} = \mathbf{0}$$

Thus, $\mathbf{u} + (-\mathbf{u}) = \mathbf{0}$.

(e) To prove this statement, we can again write out the components of the vectors $\mathbf{u}$ and $\mathbf{v}$:

$$\mathbf{u} = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix} \quad \mathbf{v} = \begin{pmatrix} v_1 \ v_2 \ \vdots \ v_n \end{pmatrix}$$

Then, the vector $k(\mathbf{u} + \mathbf{v})$ is given

$$k(\mathbf{u} + \mathbf{v}) = k \begin{pmatrix} u_1 + v_1 \ u_2 + v_2 \ \vdots \ u_n + v_n \end{pmatrix} = \begin{pmatrix} k(u_1 + v_1) \ k(u_2 + v_2) \ \vdots \ k(u_n + v_n) \end{pmatrix} = \begin{pmatrix} ku_1 + kv_1 \ ku_2 + kv_2 \ \vdots \ ku_n + kv_n \end{pmatrix} = k\mathbf{u} + k\mathbf{v}$$

Thus, $k(\mathbf{u} + \mathbf{v}) = k \mathbf{u}+k \mathbf{v}$.

(f) To prove this statement, we can again write out the components of the vector $\mathbf{u}$:

$$\mathbf{u} = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix}$$

Then, the vector $(k+m)\mathbf{u}$ is given by:

$$(k+m)\mathbf{u} = (k+m) \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix} = \begin{pmatrix} (k+m)u_1 \ (k+m)u_2 \ \vdots \ (k+m)u_n \end{pmatrix} = \begin{pmatrix} ku_1 + mu_1 \ ku_2 + mu_2 \ \vdots \ ku_n + mu_n \end{pmatrix} = k\mathbf{u} + m\mathbf{u}$$

Thus, $k(m\mathbf{u}) = (km)\mathbf{u}$.

(h) To prove this statement, we can again write out the components of the vector $\mathbf{u}$:

$$\mathbf{u} = \begin{pmatrix} u_1 \ u_2 \ \vdots \ u_n \end{pmatrix}$$

Then, the vector $\mathbf{0}$ is given by:

$$\mathbf{0} = \begin{pmatrix} 0 \ 0 \ \vdots \ 0 \end{pmatrix}$$

Since $u_i = 0$ for all $i$, it follows that $\mathbf{u} = \mathbf{0}$.

Therefore, all of the statements are true.