Matrix representation of linear transformations: Exercises

Notation reminder, so nothing here is a surprise:

$[v]_{B}$ means "the list of numbers you multiply the basis vectors of $B$ by to build $v$ ." If $B = {b_{1}, b_{2}}$ and $v = 3 b_{1} - 2 b_{2}$ , then $[v]_{B} = (- 2 3)$ .
$[T]_{C}^{B}$ means "the table that converts $B$ -coordinates of an input into $C$ -coordinates of the output." Read it right-to-left: bottom label $B$ is what goes in, top label $C$ is what comes out.
$e_{1}, e_{2}, \dots$ are the standard basis arrows: $e_{1} = (1, 0, \dots)$ , $e_{2} = (0, 1, 0, \dots)$ , and so on.

Level 1 — Recognition

Recall Solution 1.1

WHAT we do: hit each standard basis arrow with $T$ , read off coordinates.

$T (e_{1}) = 2 (1, 0) = (2, 0)$ → first column $(0 2)$ .
$T (e_{2}) = 2 (0, 1) = (0, 2)$ → second column $(2 0)$ .

WHY: each column catches where a basis vector goes (the mnemonic "Columns Catch Outputs in $C$ "). Here $C$ is the standard basis, so the output already is its coordinates. $[T] = (2002) .$

Recall Solution 1.2

Size: $3 \times 3$ (three-dimensional in, three-dimensional out).

$T (e_{1}) = (1, 0, 0)$ → column $(1, 0, 0)^{⊤}$ .
$T (e_{2}) = (0, 1, 0)$ → column $(0, 1, 0)^{⊤}$ .
$T (e_{3}) = (0, 0, 0)$ → column $(0, 0, 0)^{⊤}$ . Why zero? the $z$ -axis is flattened away. $[T] = 100010000 .$ The all-zero third column is the fingerprint of a degenerate direction — see Kernel and image: that column being zero means $e_{3}$ lives in the kernel.

Recall Solution 1.3

Read the columns back out — they tell you where the basis arrows land.

Column 1 = $(1 0)$ : so $T (e_{1}) = (0, 1)$ . The arrow pointing right now points up.
Column 2 = $(0 1)$ : so $T (e_{2}) = (1, 0)$ . The arrow pointing up now points right.

Right ↔ up swap means every point $(x, y)$ goes to $(y, x)$ : this is the ==reflection across the line $y = x$ ==. ✓

Level 2 — Application

Recall Solution 2.1

A $180°$ turn sends every arrow to its exact opposite: $(x, y) \mapsto (- x, - y)$ .

$T (e_{1}) = (- 1, 0)$ → column $(- 1, 0)^{⊤}$ .
$T (e_{2}) = (0, - 1)$ → column $(0, - 1)^{⊤}$ . $[T] = (- 1 0 0 - 1) .$ Verify: $[T] (- 3 4) = (3 - 4)$ , and indeed $(4, - 3)$ flipped through the origin is $(- 4, 3)$ . ✓

Recall Solution 2.2

Now the output basis is the full ${1, x, x^{2}}$ , so every column has three entries.

$T (1) = 0$ → $(0, 0, 0)^{⊤}$ .
$T (x) = 1 = 1 \cdot 1$ → $(1, 0, 0)^{⊤}$ .
$T (x^{2}) = 2 x = 2 \cdot x$ → $(0, 2, 0)^{⊤}$ . $[T] = 000100020 .$ Apply: $[p]_{B} = (7, 2, 5)^{⊤}$ , so $[T] [p]_{B} = (2, 10, 0)^{⊤}$ , meaning $2 + 10 x$ . And $p^{'} = 2 + 10 x$ . ✓ Note the bottom row is all zeros: the derivative can never produce an $x^{2}$ term, so it can't hit that basis direction — a preview of Rank and nullity.

Recall Solution 2.3

Size forecast: $3 \times 2$ (three out, two in).

$T (e_{1}) = T (1, 0) = (1, 1, 0)$ → column $(1, 1, 0)^{⊤}$ .
$T (e_{2}) = T (0, 1) = (- 1, 1, 2)$ → column $(- 1, 1, 2)^{⊤}$ . $[T] = 110 - 1 12 .$ Test: $[T] (1 3) = (3 - 1, 3 + 1, 0 + 2)^{⊤} = (2, 4, 2)^{⊤}$ , and $T (3, 1) = (3 - 1, 3 + 1, 2) = (2, 4, 2)$ . ✓

Level 3 — Analysis

Figure — Matrix representation of linear transformations

Recall Solution 3.1

WHAT: apply $T$ to each new basis arrow, then express the answer back in $B^{'}$ . First, in standard coordinates the map is $T (x, y) = (x + 2 y, y)$ .

$T (1, 0) = (1, 0)$ . Write $(1, 0) = a (1, 0) + b (1, 1)$ : from the second slot $b = 0$ , then $a = 1$ . → column $(1, 0)^{⊤}$ .
$T (1, 1) = (1 + 2, 1) = (3, 1)$ . Write $(3, 1) = a (1, 0) + b (1, 1)$ : second slot gives $b = 1$ , then $a + 1 = 3 \Rightarrow a = 2$ . → column $(2, 1)^{⊤}$ . $[T]_{B^{'}} = (1021) .$ Surprise: it's the same matrix! WHY: $b_{1}^{'} = (1, 0) = e_{1}$ lies along the shear's fixed direction, and the shear moves $b_{2}^{'}$ by exactly one copy of $b_{1}^{'}$ — the geometry the new basis measures happens to match the old. This is the Change of basis formula $P^{- 1} A P$ landing on the same matrix because $P$ commutes with this particular $A$ .

Recall Solution 3.2

$T (e_{2}) = 3 e_{2}$ . In $B^{'} = {e_{2}, e_{1}}$ this is $3 \cdot (first B^{'} vector)$ → column $(3, 0)^{⊤}$ .
$T (e_{1}) = 2 e_{1}$ . In $B^{'}$ this is $2 \cdot (second B^{'} vector)$ → column $(0, 2)^{⊤}$ . $[T]_{B^{'}} = (3002) .$ Reordering the basis just swaps the diagonal entries. This is exactly what Eigenvalues and diagonalization uses: when the basis is made of eigenvectors, the matrix is diagonal and the entries are the stretch factors (eigenvalues).

Recall Solution 3.3

Matrices. $[S] = (01 - 1 0)$ , $[T] = (10 0 - 1)$ .

Way 1 — multiply (composition rule $[T \circ S] = [T] [S]$ ):

=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}.$$ **Way 2 — track basis vectors** (apply $S$ then $T$): - $e_1=(1,0)\xrightarrow{S}(0,1)\xrightarrow{T}(0,-1)$ → column $(0,-1)^\top$. - $e_2=(0,1)\xrightarrow{S}(-1,0)\xrightarrow{T}(-1,0)$ → column $(-1,0)^\top$. $$[T\circ S]=\begin{pmatrix}0&-1\\-1&0\end{pmatrix}. \checkmark$$ Both agree — this is [[Composition of linear maps]]: matrix multiplication is *defined* to make these two routes match.

Level 4 — Synthesis

Recall Solution 4.1

Apply the derivative, then express each answer in $C$ by solving.

$T (1) = 0 = 0 \cdot 1 + 0 \cdot (1 + x)$ → column $(0, 0)^{⊤}$ .
$T (1 + x) = 1$ . Write $1 = a \cdot 1 + b \cdot (1 + x)$ : match $x$ -term $b = 0$ , then constant $a = 1$ . → column $(1, 0)^{⊤}$ .
$T (1 + x + x^{2}) = 1 + 2 x$ . Write $1 + 2 x = a \cdot 1 + b (1 + x)$ : match $x$ -term $b = 2$ , then constant $a + b = 1 \Rightarrow a = - 1$ . → column $(- 1, 2)^{⊤}$ . $[T]_{C}^{B} = (0010 - 1 2) .$ Sanity test. Take $p = 1 + x + x^{2}$ ; its $B$ -coords are $(0, 0, 1)^{⊤}$ (it's the third basis vector). Then $[T]_{C}^{B} (0, 0, 1)^{⊤} = (- 1, 2)^{⊤}$ , meaning $- 1 \cdot 1 + 2 \cdot (1 + x) = 1 + 2 x$ . And directly $p^{'} = (1 + x + x^{2})^{'} = 1 + 2 x$ . ✓

Recall Solution 4.2

The given vectors ${(1, 2), (0, 1)}$ form a basis, so this data determines $T$ . But we want the matrix in the standard basis, so we need $T (e_{1})$ and $T (e_{2})$ .

$e_{2} = (0, 1)$ is already given: $T (e_{2}) = (1, 1)$ .
$e_{1} = (1, 0)$ : write $e_{1} = 1 \cdot (1, 2) - 2 \cdot (0, 1)$ . By linearity $T (e_{1}) = 1 \cdot T (1, 2) - 2 \cdot T (0, 1) = (3, 0) - 2 (1, 1) = (1, - 2)$ . $[T] = (1 - 2 11) .$ Check both conditions: $[T] (2 1) = (- 2 + 2 1 + 2) = (0 3)$ ✓ and $[T] (1 0) = (1 1)$ ✓.

The plane's unit normal is $n = \frac{1}{3} (1, 1, 1)$ . Projection onto the plane removes the component along $n$ : $T (v) = v - (v \cdot n) n = v - \frac{( v \cdot ( 1 , 1 , 1 ))}{3} (1, 1, 1) .$ Why subtract the normal part? whatever sticks out of the plane along $n$ is exactly the shadow being removed; the rest lies flat in the plane.

$T (e_{1}) = (1, 0, 0) - \frac{1}{3} (1, 1, 1) = (\frac{2}{3}, - \frac{1}{3}, - \frac{1}{3})$ .
$T (e_{2}) = (0, 1, 0) - \frac{1}{3} (1, 1, 1) = (- \frac{1}{3}, \frac{2}{3}, - \frac{1}{3})$ .
$T (e_{3}) = (0, 0, 1) - \frac{1}{3} (1, 1, 1) = (- \frac{1}{3}, - \frac{1}{3}, \frac{2}{3})$ . $[T] = \frac{1}{3} 2 - 1 - 1 - 1 2 - 1 - 1 - 1 2 .$ (b) The image is the whole plane (2-dimensional) so rank $= 2$ . Everything along the normal $(1, 1, 1)$ collapses to $0$ , so the kernel is the line spanned by $(1, 1, 1)$ , nullity $= 1$ . Check: $[T] (1, 1, 1)^{⊤} = \frac{1}{3} (0, 0, 0)^{⊤} = 0$ ✓, and rank $+$ nullity $= 2 + 1 = 3 = dim R^{3}$ , the Rank and nullity theorem. See Kernel and image.

Recall Solution 5.2

(a) Apply $T$ to each basis vector.

$T (1) = 1 + 0 = 1$ → column $(1, 0, 0)^{⊤}$ .
$T (x) = x + 1 = 1 + x$ → column $(1, 1, 0)^{⊤}$ .
$T (x^{2}) = x^{2} + 2 x$ → column $(0, 2, 1)^{⊤}$ . $[T] = 100110021 .$ (b) It's upper-triangular with all diagonal entries $1$ , so $det = 1 \neq = 0$ : invertible. Invert (back-substitute the triangular system): $[T]^{- 1} = 100 - 1 10 2 - 2 1 .$ Check: $[T] [T]^{- 1} = I$ . As a map, $T^{- 1} (p) = p - p^{'} + p^{''}$ (a finite version of $\sum (- 1)^{k} p^{(k)}$ , which terminates because third and higher derivatives vanish on $P_{2}$ ). Quick test: $T^{- 1} (x^{2})$ has coords $[T]^{- 1} (0, 0, 1)^{⊤} = (2, - 2, 1)^{⊤}$ , i.e. $2 - 2 x + x^{2}$ ; apply $T$ : $(2 - 2 x + x^{2}) + (- 2 + 2 x) = x^{2}$ ✓.

Recall Self-check: the one-line recipe behind every problem

To build $[T]_{C}^{B}$ ::: apply $T$ to each vector of $B$ , write each output in the basis $C$ , and make those coordinate lists the columns. Matrix size for $T : V \to W$ ::: $dim W$ rows by $dim V$ columns. When is $[T]$ not invertible ::: exactly when $T$ has a nonzero kernel (some direction maps to $0$ ), equivalently $det = 0$ .

Connections

Coordinate vectors and bases — every "write it in basis $C$ " step lives here.
Change of basis — Exercises 3.1 and 3.2 are $P^{- 1} A P$ in disguise.
Composition of linear maps — Exercise 3.3's two-route check.
Rank and nullity — read off from zero columns/rows (2.2, 5.1).
Kernel and image — the flattened direction in 5.1.
Eigenvalues and diagonalization — the eigenbasis matrix in 3.2.

Exercises — Matrix representation of linear transformations