$3$. Let $W$ be the set of all $2 \times 2$ complex Hermitian matrices, that is, the set of $2 \times 2$ complex matrices $A$ such that $A_{ij} = \overline{A_{ji}}$ (the bar denoting complex conjugation). As we pointed out in Example $6$ of Chapter $2$, $W$ is a vector space over the field of real numbers, under the usual operations. Verify that $$(x, y, z, t) \to \begin{bmatrix} t+x & y+iz \\ y-iz & t-x \end{bmatrix}$$ is an isomorphism of $\Bbb R^4$ onto $W$.
I can't understand how the $4 \times 1$ matrix $(x,y,z,t)$ became the $2 \times 2$ matrix $\begin{bmatrix} t+x & y+iz \\ y-iz & t-x \end{bmatrix}$.
Assuming that the matrix of transformation is $m \times n$. Thus, $n$ should be $4$, then the resulting matrix is $m \times 1$, not $2 \times 2$.
The problem asks you about the linear transformation $$ (x,y,z,t)\in\mathbb R^4 \mapsto \begin{bmatrix} t+x & y+iz \\ y-iz & t-x \end{bmatrix} \in W$$ This is a function that produces a new matrix for each real 4-vector you give it.
The question does not ask for, or even speak about, a matrix that represents the entire linear transformation. We know that every linear transformation between finite-dimensional vector spaces can be represented with a matrix, but that doesn't mean we always should. And here there's no reason to.
In fact, another way of stating roughly the same thing you're asked to prove here would be,
Once we have a basis for $W$, we would be able to write down a matrix for the above mapping (which in this case would be the $4\times 4$ identity matrix, so that is not even terribly interesting).
Aficionados will recognize the first three basis matrices in my formulation -- which are the images of $\mathbf e_1$, $\mathbf e_2$, $\mathbf e_3$ under your linear transformation -- as the Pauli matrices. They're important in the mathematics behind quantum mechanics.