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

Row reduction gives rank $= 2$ (two pivots). So nullity $= 3 - 2 = 1$, confirmed by $\ker(T_A) = \text{span}\{(1, -2, 1)\}$.

$A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}$. The third row is the sum of the first two, so rank $= 2$ and nullity $= 3 - 2 = 1$.

$\ker(T_A) = \text{span}\{(1, 1, -1)\}$ (verify: $A(1, 1, -1)^T = (0, 0, 0)^T$).

$D : P_3(\mathbb{R}) \to P_2(\mathbb{R})$, $D(p) = p'$.

$\dim P_3(\mathbb{R}) = 4$. $\ker(D) = \{$ constants $\}$, nullity $= 1$. $\text{im}(D) = P_2(\mathbb{R})$, rank $= 3$.

$4 = 3 + 1$. Check.

$\text{tr} : M_{n \times n}(F) \to F$.

$\dim M_{n \times n} = n^2$. $\text{im}(\text{tr}) = F$, rank $= 1$. So nullity $= n^2 - 1$.

The kernel (trace-zero matrices) has dimension $n^2 - 1$, as expected.

$\pi : \mathbb{R}^4 \to \mathbb{R}^2$, $\pi(x_1, x_2, x_3, x_4) = (x_1, x_2)$.

rank $= 2$, nullity $= 4 - 2 = 2$. $\ker(\pi) = \{(0, 0, x_3, x_4)\}$.

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

rank $= 0$, nullity $= \dim V$. $\dim V = 0 + \dim V$. Check.

$\text{id}_V : V \to V$.

rank $= \dim V$, nullity $= 0$. $\dim V = \dim V + 0$. Check.

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

$\ker(T) = \{0\}$, nullity $= 0$. So rank $= 2 - 0 = 2$. The image is the plane $z = x + y$ in $\mathbb{R}^3$.

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

$\text{im}(T) = \mathbb{R}^2$ (surjective), rank $= 2$. So nullity $= 3 - 2 = 1$.

$\ker(T)$: $x + z = 0$ and $y - z = 0$, so $x = -z$, $y = z$. $\ker(T) = \text{span}\{(-1, 1, 1)\}$.

The system $Ax = b$ (where $A$ is $m \times n$) has a solution iff $b \in \text{im}(T_A) = C(A)$ (the column space). If $x_0$ is a particular solution, then the general solution is

$x = x_0 + \ker(T_A)$

i.e., the solution set is a coset (translate) of the null space. The number of free variables equals nullity$(T_A) = n - \text{rank}(A)$.

For $T : F^n \to F^n$ (equivalently, an $n \times n$ matrix $A$):

$T \text{ injective} \iff \ker(T) = \{0\} \iff \text{rank}(T) = n \iff T \text{ surjective} \iff T \text{ isomorphism}.$

This equivalence is specific to maps between spaces of the same finite dimension.

For an $m \times n$ matrix $A$ with $m > n$ (more equations than unknowns): rank$(A) \leq n$. If rank$(A) = n$, then nullity $= 0$ ($T_A$ is injective) but $\text{im}(T_A) \subsetneq F^m$ (not surjective). Most $b \in F^m$ give no solution.

For $m < n$ (fewer equations than unknowns): rank$(A) \leq m < n$, so nullity $= n - \text{rank}(A) \geq n - m > 0$. The null space is nontrivial, so the homogeneous system $Ax = 0$ always has nonzero solutions.