Let's solve this step-by-step:

1. Verify Linearity of $f$:

   To prove $f$ is linear, we need to check if $f$ satisfies the following properties:

   • Additivity: $f(A + B) = f(A) + f(B)$
   • Homogeneity: $f(\lambda A) = \lambda f(A)$

   For $A = \begin{pmatrix} x & y \\ 0 & z \end{pmatrix}$ and $B = \begin{pmatrix} x' & y' \\ 0 & z' \end{pmatrix}$ in $V$: $f(A + B) = f\left( \begin{pmatrix} x + x' & y + y' \\ 0 & z + z' \end{pmatrix} \right) = \begin{pmatrix} (x + x') + (y + y') & z + z' \\ z + z' & (x + x') + (y + y') \end{pmatrix}$ On the other hand: $f(A) + f(B) = \begin{pmatrix} x + y & z \\ z & x + y \end{pmatrix} + \begin{pmatrix} x' + y' & z' \\ z' & x' + y' \end{pmatrix} = \begin{pmatrix} (x + x') + (y + y') & z + z' \\ z + z' & (x + x') + (y + y') \end{pmatrix}$ Thus, $f(A + B) = f(A) + f(B)$, satisfying additivity.

   For homogeneity, let $\lambda \in \mathbb{R}$: $f(\lambda A) = f\left( \begin{pmatrix} \lambda x & \lambda y \\ 0 & \lambda z \end{pmatrix} \right) = \begin{pmatrix} \lambda x + \lambda y & \lambda z \\ \lambda z & \lambda x + \lambda y \end{pmatrix} = \lambda \begin{pmatrix} x + y & z \\ z & x + y \end{pmatrix} = \lambda f(A)$ Hence, $f$ is linear.

2. Find a Basis for $V + W$ and $V \cap W$:

   To determine $V + W$, we combine the bases of $V$ and $W$ and identify linearly independent vectors.

   • Basis for $V$: The given basis for $V$ is: $\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\}$
   • Basis for $W$: The given basis for $W$ is: $\left\{ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right\}$

   To form a basis for $V + W$, we look at these five matrices and find the set of linearly independent ones.

   For $V \cap W$, find matrices that can be expressed both as a linear combination of $V$'s basis and $W$'s basis.

3. Find a Basis for $\ker(f)$ and $\text{im}(f)$:

   • Kernel: To find $\ker(f)$, solve $f(A) = 0$ for $A = \begin{pmatrix} x & y \\ 0 & z \end{pmatrix}$. This implies: $\begin{pmatrix} x + y & z \\ z & x + y \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ Thus, $x + y = 0$ and $z = 0$. So, $\ker(f)$ consists of matrices of the form $\begin{pmatrix} x & -x \\ 0 & 0 \end{pmatrix}$.

   • Image: The image $\text{im}(f)$ is spanned by the form of matrices produced by $f$. Any $f(A)$ is of the form $\begin{pmatrix} x + y & z \\ z & x + y \end{pmatrix}$, implying the image matrices have equal diagonal entries and are symmetric.

Would you like further details on any step?

Here are related questions to expand on this topic:

1. How do we confirm if a set of matrices forms a basis for a given subspace?
2. What are the implications of a linear map's kernel and image in linear algebra?
3. How can we generalize this process to higher dimensions or different types of matrices?
4. How would we compute the rank of the map $f$ in this example?
5. What are other methods to determine if a map is linear besides verifying additivity and homogeneity?

Tip: For matrix spaces, analyzing dimensions of subspaces like $V$, $W$, $V \cap W$, and $V + W$ can often simplify finding bases and understanding the structure of the space.