Matrix representation of linear transformations: Worked examples

Before anything: a coordinate vector $[v]_{B}$ just means "the list of mixing amounts you need to build $v$ out of the basis $B$ ." If $B = {b_{1}, b_{2}}$ and $v = 3 b_{1} - 2 b_{2}$ , then $[v]_{B} = (- 2 3)$ . That is all the notation $[\cdot]_{B}$ means — see Coordinate vectors and bases. A basis is a smallest set of "ingredient arrows" that can build every vector by mixing.

The scenario matrix

Every representation problem falls into one of these cells. The examples that follow are tagged with the cell they cover.

Cell	What makes it different	Example
A. Square, standard basis	$n = m$ , both bases standard — easiest	Ex 1
B. Non-square (more inputs)	$dim V > dim W$ , matrix is "wide"	Ex 2
C. Non-square (more outputs)	$dim V < dim W$ , matrix is "tall"	Ex 3
D. Non-standard codomain basis	must SOLVE for coordinates in $C$	Ex 4
E. Degenerate / collapsing map	$T$ crushes a dimension (zero column)	Ex 5
F. Zero / limiting input	the origin and scaling behaviour	inside Ex 5
G. Composition	two machines chained → matrix product	Ex 6
H. Real-world word problem	translate a story into a matrix	Ex 7
I. Exam twist (recover $T$ from matrix)	matrix given, find the rule/action	Ex 8

Sign and quadrant coverage is built into the geometric examples (Ex 1 sweeps all four quadrants of a rotation; Ex 5 shows the sign of a projected component in every quadrant).

A · Square, standard basis (all four quadrants)

Figure — Matrix representation of linear transformations

Find $T (e_{1})$ . Why this step? Columns catch where basis vectors go. The point $(1, 0)$ sits on the horizontal axis; reflecting across $y = x$ sends it to $(0, 1)$ . So the first column is $(1 0)$ .
Find $T (e_{2})$ . Why? Same recipe for the second ingredient. $(0, 1)$ reflects to $(1, 0)$ , giving second column $(0 1)$ .
Assemble. Why columns not rows? Because $[T] e_{j}$ must pick out column $j$ (parent's boxed formula): $[T] = (0110) .$
Sweep all quadrants (look at the coloured dots in the figure). Reflection swaps coordinates, so $(x, y) \mapsto (y, x)$ :
- Q1: $(3, 1) \mapsto (1, 3)$
- Q2: $(- 2, 4) \mapsto (4, - 2)$
- Q3: $(- 3, - 2) \mapsto (- 2, - 3)$
- Q4: $(2, - 5) \mapsto (- 5, 2)$

Verify: $[T] (4 - 2) = (- 2 4)$ ✓. Reflecting twice must return the original, so $[T]^{2}$ should be the identity: $(0110)^{2} = (1001)$ ✓.

B · Wide matrix (more inputs than outputs)

Shape. Why? rows $= dim W = 2$ , columns $= dim V = 3$ . So $[T]$ is $2 \times 3$ — "wide."
Column for $e_{1} = (1, 0, 0)$ . Why? Feed the first ingredient in. $T (1, 0, 0) = (2 \cdot 1 - 0, 1 + 0 + 0) = (2, 1)$ → $(1 2)$ .
Column for $e_{2} = (0, 1, 0)$ . $T (0, 1, 0) = (0 - 0, 0 + 1 + 0) = (0, 1)$ → $(1 0)$ .
Column for $e_{3} = (0, 0, 1)$ . $T (0, 0, 1) = (0 - 1, 0 + 0 + 1) = (- 1, 1)$ → $(1 - 1)$ .
Assemble: $[T] = (2101 - 1 1) .$

Verify: take $(3, - 1, 2)$ . Directly $T (3, - 1, 2) = (2 \cdot 3 - 2, 3 - 1 + 2) = (4, 4)$ . Via matrix: $(2101 - 1 1) 3 - 1 2 = (3 - 1 + 2 6 - 2) = (4 4)$ ✓. A wide matrix means inputs can collapse together — related to Rank and nullity and Kernel and image.

C · Tall matrix (more outputs than inputs)

Shape. $dim W = 3$ rows, $dim V = 2$ columns → $3 \times 2$ , "tall."
Column for $e_{1} = (1, 0)$ . $T (1, 0) = (1, 1 + 0, 0) = (1, 1, 0)$ → first column.
Column for $e_{2} = (0, 1)$ . $T (0, 1) = (0, 0 + 1, 1) = (0, 1, 1)$ → second column.
Assemble: $[T] = 110011 .$

Verify: $(5, 2) \mapsto (5, 7, 2)$ directly; matrix gives $110011 (2 5) = (5, 7, 2)^{⊤}$ ✓. A tall matrix can never be onto $R^{3}$ — its image is at most a 2-D plane (Kernel and image).

D · Non-standard codomain basis (you must SOLVE)

Compute raw outputs. $T (e_{1}) = (1 \cdot 1 + 0, 0) = (1, 0)$ ; $T (e_{2}) = (0 + 1, 2) = (1, 2)$ .
Express $T (e_{1})$ in $C$ . Why this step? The parent's second "mistake" callout: you MUST solve $(1, 0) = a c_{1} + b c_{2} = a (1, 1) + b (1, - 1)$ . That gives $a + b = 1$ and $a - b = 0$ , so $a = b = \frac{1}{2}$ . First column $= (1/2 1/2)$ .
Express $T (e_{2})$ in $C$ . Solve $(1, 2) = a (1, 1) + b (1, - 1)$ : $a + b = 1, a - b = 2 \Rightarrow a = \frac{3}{2}, b = - \frac{1}{2}$ . Second column $= (- 1/2 3/2)$ .
Assemble: $[T]_{C}^{E} = (\frac{1}{2} \frac{1}{2} \frac{3}{2} - \frac{1}{2}) .$

Verify: test $v = e_{1} = (1, 0)$ , so $[v]_{E} = (0 1)$ . Then $[T]_{C}^{E} (0 1) = (1/2 1/2)$ , claiming $T (v) = \frac{1}{2} c_{1} + \frac{1}{2} c_{2} = \frac{1}{2} (1, 1) + \frac{1}{2} (1, - 1) = (1, 0)$ . And directly $T (1, 0) = (1, 0)$ ✓. Skipping step 2 (dumping raw $(1, 0)$ into the column) would give a different, wrong map — see Change of basis.

E · Degenerate map + F · Zero/limiting input

Column for $e_{1}$ . $T (1, 0) = (1, 0)$ → $(0 1)$ . The horizontal ingredient survives untouched.
Column for $e_{2}$ . $T (0, 1) = (0, 0)$ → $(0 0)$ . Why a zero column? The vertical ingredient is entirely flattened; nothing of it survives. That zero column is the fingerprint of a degenerate (non-invertible) map.
Assemble: $[T] = (1000) .$
Sign in every quadrant (look at the coloured projection arrows in the figure): the output keeps the sign of $x$ and kills $y$ .
- Q1 $(4, 3) \mapsto (4, 0)$ — positive $x$
- Q2 $(- 4, 3) \mapsto (- 4, 0)$ — negative $x$
- Q3 $(- 4, - 3) \mapsto (- 4, 0)$ — negative $x$
- Q4 $(4, - 3) \mapsto (4, 0)$ — positive $x$
Cell F, zero/limiting input. Why examine the origin? Every linear map sends $0 \mapsto 0$ (set $c = 0$ in $T (c v) = c T (v)$ ): $T (0, 0) = (0, 0)$ ✓. And scaling the input scales the output: $T (t \cdot (4, 3)) = (4 t, 0) = t \cdot T (4, 3)$ , so as $t \to 0$ the output smoothly shrinks to the origin — no jumps.

Verify: the kernel (what maps to zero) is the whole $y$ -axis: $T (0, c) = (0, 0)$ for all $c$ ✓. Applying $T$ twice changes nothing (already flat): $[T]^{2} = (1000)$ ✓ — a projection. A zero column ⟹ non-zero kernel ⟹ not invertible.

G · Composition = matrix product

Recall the pieces. From the parent note $[S] = (01 - 1 0)$ (rotation), and from Ex 1 $[T] = (0110)$ (reflection).
Order of the product. Why $[T] [S]$ and not $[S] [T]$ ? The parent's composition rule: $[T \circ S] = [T] [S]$ — the last machine applied sits on the left, because $[T (S (v))] = [T] ([S] [v])$ .
Multiply: $=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$$
Interpret. What does it look like? $(10 0 - 1)$ keeps $x$ , flips $y$ : that's a reflection across the $x$ -axis. Rotate-then-reflect $=$ a single reflection.

Verify: track $(2, 0)$ . $S (2, 0) = (0, 2)$ (rotated up); then $T (0, 2) = (2, 0)$ (swapped back). Net $(2, 0) \mapsto (2, 0)$ . Matrix: $(10 0 - 1) (0 2) = (0 2)$ ✓. Check $(0, 3)$ : $S \to (- 3, 0)$ , $T \to (0, - 3)$ ; matrix $(10 0 - 1) (3 0) = (- 3 0)$ ✓. See Composition of linear maps.

H · Real-world word problem

Column for $b_{1} =$ one Applejoy. Why? Feed one unit of the first "ingredient" (one bottle of Applejoy) and record the output. It needs $(2 apples, 1 orange)$ → $(1 2)$ .
Column for $b_{2} =$ one Sunburst. Needs $(1 apple, 3 oranges)$ → $(3 1)$ .
Assemble ( $M$ takes bottles → fruit): $M = (2113) .$
Apply to the order $(4 5)$ : $M (4 5) = (1 \cdot 5 + 3 \cdot 4 2 \cdot 5 + 1 \cdot 4) = (17 14) .$

Verify (with units): apples $= 2 (5) + 1 (4) = 14$ apples; oranges $= 1 (5) + 3 (4) = 17$ oranges ✓. Units are consistent: (fruit per bottle) $\times$ (bottles) $=$ fruit. The matrix is exactly the mixing rule.

I · Exam twist — recover the map from the matrix

Read the formula off the matrix. Why? Because $T (x, y) = [T] (y x)$ : $(30 - 1 2) (y x) = (2 y 3 x - y) \Rightarrow T (x, y) = (3 x - y, 2 y) .$
Set up the fixed-point condition $T (v) = v$ . Why? A "fixed vector" satisfies $T (x, y) = (x, y)$ : $3 x - y = x and 2 y = y .$
Solve. From $2 y = y$ we get $y = 0$ . Substitute into $3 x - y = x$ : $3 x = x \Rightarrow x = 0$ . Wait — is there more? Check the diagonal entries: $2 \neq = 1$ and $3 \neq = 1$ , so no eigenvalue equals $1$ except at the origin. The only fixed vector is $(0 0)$ .

Verify: $T (0, 0) = (3 \cdot 0 - 0, 2 \cdot 0) = (0, 0) = (0, 0)$ ✓. Spot-check the formula on $(1, 1)$ : $T (1, 1) = (3 - 1, 2) = (2, 2)$ , and $[T] (1 1) = (2 2)$ ✓. "Fixed vectors" are eigenvectors with eigenvalue $1$ — here there are none but the origin, see Eigenvalues and diagonalization.

Recall Quick self-test

A map $R^{4} \to R^{2}$ has a matrix of what shape? ::: $2 \times 4$ (rows $= dim W = 2$ , columns $= dim V = 4$ ). Which cell forces you to solve a linear system for each column? ::: Cell D — a non-standard codomain basis. What in the matrix signals a collapsing (degenerate) map? ::: A zero column (its basis vector maps to $0$ ), giving a non-trivial kernel. Compose $T$ after $S$ : which order is the product? ::: $[T] [S]$ — the last-applied map is on the left.

Connections

Coordinate vectors and bases — every column is a coordinate vector $[T (b_{j})]_{C}$ .
Change of basis — why Ex 4's non-standard basis reshuffles the matrix.
Composition of linear maps — the engine behind Ex 6.
Rank and nullity / Kernel and image — read from the wide/tall/degenerate matrices (Ex 2, 3, 5).
Eigenvalues and diagonalization — the fixed-vector question in Ex 8.

Worked examples — Matrix representation of linear transformations