The Matrix of a Transformation, Kernel, Range & the Dimension Theorem
How to read off a transformation’s matrix from where it sends a basis, and the two subspaces — kernel and range — that measure everything a transformation destroys and everything it can reach
§1Finding A From Where the Standard Basis Lands
Lecture 20 told us every linear transformation $T:\mathbb{R}^2\to\mathbb{R}^2$ has some matrix $A$ with $T(v)=Av$. Here is the practical question: given a transformation, how do you actually find $A$?
The columns of $A$ are exactly the images of the standard basis vectors: $$A = \begin{pmatrix} | & | & & | \\ T(\mathbf{e}_1) & T(\mathbf{e}_2) & \cdots & T(\mathbf{e}_n) \\ | & | & & | \end{pmatrix}.$$ Nothing more is needed — knowing where $T$ sends $\mathbf{e}_1,\ldots,\mathbf{e}_n$ pins down $A$ completely, because those columns are the matrix.
Suppose $T(1,0)=(2,1)$ and $T(0,1)=(0,1)$. Reading these off as columns:
$$A = \begin{pmatrix}2&0\\1&1\end{pmatrix}.$$
Check: $A\binom{1}{0}=\binom{2}{1}$ ✓ and $A\binom{0}{1}=\binom{0}{1}$ ✓ — both match exactly.
§2Part B — Generalizing to Non-Standard Bases
The standard basis is convenient, but nothing about the underlying idea requires it. The real principle is broader:
If we know the image of every vector in a basis — any basis, not just the standard one — then we can find the image of any vector at all. Write the target vector as a combination of the basis, then apply linearity term by term.
§3Fully Worked Example — Using a Non-Standard Basis
§4Kernel and Range — Definitions
Every linear transformation $T:V\to W$ carries two natural subspaces along with it — one living inside $V$, one living inside $W$.
$$\operatorname{Ker}(T) := \{\,v\in V : T(v)=\mathbf{0}\,\} \subseteq V$$ — every vector $T$ crushes down to zero. Ker(T) is a subspace of $V$.
$$\operatorname{Range}(T) := \{\,w\in W : \exists\, v \text{ with } T(v)=w\,\} \subseteq W$$ — every output $T$ can actually produce. Range(T) is a subspace of $W$.
For both sets, check the three subspace criteria: contains the zero vector, closed under addition, closed under scalar multiplication.
§5The Dimension (Rank–Nullity) Theorem
$$\dim(\operatorname{Ker}(T)) + \dim(\operatorname{Range}(T)) = \dim(V).$$
$$\operatorname{Nullity}(A) + \operatorname{Rank}(A) = \#\,\text{columns of } A.$$
§6Fully Worked Example — T : ℝ³ → ℝ⁴
§7Why Do We Study Ker(T) and Range(T)?
$$\dim(\operatorname{Ker}(T)) = 0 \iff T \text{ is one-to-one (injective)}.$$
$$\dim(\operatorname{Range}(T)) = \dim(W) \iff T \text{ is onto (surjective)}.$$
§8The General Shortcut
Given the matrix $A$ of $T:V\to W$, to find $\dim(\operatorname{Ker}\,T)$ and $\dim(\operatorname{Range}\,T)$: just find $\operatorname{rank}(A)$. If $\operatorname{rank}(A)=r$, then by the dimension theorem:
$$\dim(\operatorname{Ker}(T)) = \dim(V) - r.$$
One row reduction answers both questions at once — $\operatorname{rank}(A)$ is $\dim(\operatorname{Range}(T))$ directly, and subtracting from $\dim(V)$ hands you $\dim(\operatorname{Ker}(T))$ for free.
§9Fully Worked Example — T : M₂₂ → M₂₂
§10Two Examples Not in the Book
Everything above used $\mathbb{R}^n$ or a matrix space as domain and codomain. The Ker/Range machinery does not care — it works exactly the same way when the vectors are polynomials.
Define $T:P_3\to\mathbb{R}^4$ by $$T(a+bx+cx^2+dx^3) = (a,b,c,d).$$ This is exactly the coordinate-vector isomorphism — it says $P_3$ and $\mathbb{R}^4$ are structurally identical, differing only in notation.
Define $T:M_{22}\to P_3$ by
$$T\begin{pmatrix}a&b\\c&d\end{pmatrix} = (a+b) + (b+c)x + (c+d)x^2 + (d+a)x^3.$$
This uses exactly the same coefficient pattern as the $M_{22}\to M_{22}$ example in §9 — only now the four output numbers are read off as polynomial coefficients instead of matrix entries. So, relative to the standard bases $\{E_{11},E_{12},E_{21},E_{22}\}$ of $M_{22}$ and $\{1,x,x^2,x^3\}$ of $P_3$, the matrix of $T$ is identical to the one in §9:
$$A = \begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\1&0&0&1\end{pmatrix}, \qquad \operatorname{rank}(A)=3 \;\text{ (identical row reduction as before).}$$
Where a transformation sends a basis tells you everything; one rank computation tells you the rest.
The matrix of a transformation is built from basis images, its kernel is a null space, its range is a column space, and the dimension theorem ties the two together with no room for anything to go missing or get double-counted. Every worked example today — $\mathbb{R}^2$, $\mathbb{R}^3\to\mathbb{R}^4$, $M_{22}$, $P_3$ — was the exact same three-step recipe, just wearing different notation.
For the upcoming Quiz and Exam: sections §7.1, §7.2.
There is a double quiz on Monday, covering §8.1, §8.2, §7.1, and §7.2.
A policy referred to in passing as the "n−2 policy" was also mentioned — see the official course policy document for the exact details rather than relying on this summary.