$T = \text{tr} : M_{2 \times 2}(\mathbb{R}) \to \mathbb{R}$, $\dim V = 4$.

Step 1: $\ker(T) = \{A \mid a_{11} + a_{22} = 0\}$. Basis for kernel:

$u_1 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\; u_2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\; u_3 = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}.$

So $k = 3$.

Step 2: Extend to a basis of $M_{2 \times 2}(\mathbb{R})$ by adding $w_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$.

Step 3: $\{T(w_1)\} = \{1\}$ is a basis for $\text{im}(T) = \mathbb{R}$.

So rank $= 1 = 4 - 3$. Check: $4 = 1 + 3$.

$D : P_2(\mathbb{R}) \to P_1(\mathbb{R})$, $\dim V = 3$.

$\ker(D) = \text{span}\{1\}$, $k = 1$.

Extend: $\{1, x, x^2\}$ is a basis for $P_2(\mathbb{R})$, with $u_1 = 1$, $w_1 = x$, $w_2 = x^2$.

$\{D(x), D(x^2)\} = \{1, 2x\}$ is a basis for $P_1(\mathbb{R}) = \text{im}(D)$.

rank $= 2 = 3 - 1$. Check: $3 = 2 + 1$.

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

$\ker(T) = \text{span}\{e_3\}$, $k = 1$.

Extend: $\{e_3, e_1, e_2\}$ is a basis for $\mathbb{R}^3$.

$\{T(e_1), T(e_2)\} = \{(1,0), (0,1)\}$ is a basis for $\mathbb{R}^2 = \text{im}(T)$.

rank $= 2 = 3 - 1$. Check: $3 = 2 + 1$.

$T : \mathbb{R}^2 \to \mathbb{R}^3$, $T(x, y) = (x, y, 0)$, $\dim V = 2$.

$\ker(T) = \{0\}$, $k = 0$. No kernel basis to extend.

$\{T(e_1), T(e_2)\} = \{(1,0,0), (0,1,0)\}$ is a basis for $\text{im}(T)$.

rank $= 2 = 2 - 0$. Check: $2 = 2 + 0$.

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

$\ker(T_A) = \text{span}\{(-2, 1)\}$, $k = 1$.

Extend: $\{(-2, 1), (1, 0)\}$ is a basis for $\mathbb{R}^2$.

$\{T_A(1, 0)\} = \{(1, 2)\}$ is a basis for $\text{im}(T_A) = \text{span}\{(1, 2)\}$ (the column space).

rank $= 1 = 2 - 1$. Check: $2 = 1 + 1$.

The proof strategy is beautifully simple:

1. Find a basis for $\ker(T)$ (size $k$).
2. Extend to a basis for all of $V$ (total size $n$, added $n - k$ vectors).
3. Show the images of the added vectors form a basis for $\text{im}(T)$.

The kernel vectors contribute nothing to the image (they map to $0$), while the remaining vectors map to an independent spanning set.

The Rank-Nullity Theorem says the sequence

$0 \to \ker(T) \xrightarrow{\iota} V \xrightarrow{T} \text{im}(T) \to 0$

is a short exact sequence of vector spaces. The dimension formula $\dim V = \dim \ker(T) + \dim \text{im}(T)$ is the statement that dimensions are additive across short exact sequences. This generalizes to the theory of exact sequences in abelian categories.

The first isomorphism theorem gives $V/\ker(T) \cong \text{im}(T)$. Therefore:

$\dim V = \dim \ker(T) + \dim(V/\ker(T)) = \dim \ker(T) + \dim \text{im}(T).$

This is perhaps the most conceptual proof of the Rank-Nullity Theorem.

Does $T(x, y, z) = (x + y, y + z, x + z)$ satisfy $T(v) = (1, 2, 3)$ for some $v$?

$\text{rank}(T)$: the matrix is $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{pmatrix}$ with $\det = 2 \neq 0$. So rank $= 3 = \dim \mathbb{R}^3$, meaning $T$ is surjective. Yes, a solution exists (and it is unique since nullity $= 0$).

For a matrix $A \in M_{m \times n}(F)$, the Rank-Nullity Theorem applied to $T_A$ gives rank$(A) = n -$ nullity$(A)$. Here rank$(A)$ is the column rank. A separate argument shows row rank = column rank, giving a single notion of "rank."

If $T : F^5 \to F^3$ is linear, then nullity$(T) \geq 5 - 3 = 2$. So $\ker(T)$ is at least 2-dimensional. This "pigeonhole principle for linear maps" is a consequence of Rank-Nullity.

For $Ax = b$ with $A \in M_{m \times n}(F)$:

• Existence: $b \in C(A)$ iff rank$(A) =$ rank$(A | b)$.
• Uniqueness: the solution is unique iff nullity$(A) = 0$, i.e., rank$(A) = n$.
• Number of free variables: $n -$ rank$(A)$.