For $A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix}$, $T_A : \mathbb{R}^3 \to \mathbb{R}^2$ with $T_A(x) = Ax$.

Row reducing: $\begin{pmatrix} 1 & 2 & 3 \\ 0 & -3 & -6 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{pmatrix}$.

So $\ker(T_A) = \{t(1, -2, 1) \mid t \in \mathbb{R}\}$. Nullity $= 1$, rank $= 2$.

$D : P_n(\mathbb{R}) \to P_{n-1}(\mathbb{R})$, $D(p) = p'$.

$\ker(D) = \{p \in P_n(\mathbb{R}) \mid p' = 0\} = \{c \mid c \in \mathbb{R}\}$ (the constant polynomials).

So $\text{nullity}(D) = 1$ and $\text{im}(D) = P_{n-1}(\mathbb{R})$, giving $\text{rank}(D) = n$.

$\pi : \mathbb{R}^3 \to \mathbb{R}^2$, $\pi(x, y, z) = (x, y)$.

$\ker(\pi) = \{(0, 0, z) \mid z \in \mathbb{R}\}$ (the $z$-axis), nullity $= 1$.

$\text{im}(\pi) = \mathbb{R}^2$, rank $= 2$.

$\text{tr} : M_{2 \times 2}(F) \to F$, $\text{tr}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = a + d$.

$\ker(\text{tr}) = \{A \mid a + d = 0\}$, the trace-zero matrices. This has dimension $3$ (nullity $= 3$).

$\text{im}(\text{tr}) = F$, rank $= 1$.

Check: $3 + 1 = 4 = \dim M_{2 \times 2}(F)$.

$R_\theta : \mathbb{R}^2 \to \mathbb{R}^2$ (rotation by $\theta \neq 0$).

$\ker(R_\theta) = \{0\}$ (only the origin is fixed), nullity $= 0$.

$\text{im}(R_\theta) = \mathbb{R}^2$ (rotation is surjective), rank $= 2$.

$T_0 : V \to W$, $T_0(v) = 0$ for all $v$.

$\ker(T_0) = V$, nullity $= \dim V$.

$\text{im}(T_0) = \{0\}$, rank $= 0$.

$\text{id}_V : V \to V$, $\text{id}_V(v) = v$.

$\ker(\text{id}_V) = \{0\}$, nullity $= 0$.

$\text{im}(\text{id}_V) = V$, rank $= \dim V$.

$T : M_{2 \times 2}(F) \to M_{2 \times 2}(F)$, $T(A) = A^T$.

$\ker(T) = \{0\}$ (since $A^T = 0$ implies $A = 0$), nullity $= 0$.

$\text{im}(T) = M_{2 \times 2}(F)$ (since $(B^T)^T = B$), rank $= 4$.

For any $A \in M_{m \times n}(F)$, $N(A) = \ker(T_A)$ is the solution set of $Ax = 0$. The column space $C(A) = \text{im}(T_A)$ is spanned by the columns of $A$.

$T : C[-1, 1] \to \mathbb{R}$, $T(f) = \int_{-1}^1 f(x)\,dx$.

$\ker(T)$ is the set of continuous functions with zero integral over $[-1, 1]$. For instance, all odd functions belong to $\ker(T)$. This kernel is infinite-dimensional.

$\text{ev}_c : P_n(F) \to F$, $\text{ev}_c(p) = p(c)$.

$\ker(\text{ev}_c) = \{p \in P_n(F) \mid p(c) = 0\}$, i.e., polynomials with root at $c$. These are of the form $(x - c)q(x)$ with $\deg q \leq n - 1$, so nullity $= n$.

$L : F^\infty \to F^\infty$, $L(a_1, a_2, a_3, \ldots) = (a_2, a_3, a_4, \ldots)$.

$\ker(L) = \{(a_1, 0, 0, \ldots) \mid a_1 \in F\}$, which is 1-dimensional.

$\text{im}(L) = F^\infty$ (surjective: for any $(b_1, b_2, \ldots)$, take $(0, b_1, b_2, \ldots)$).