Linear-Algebra-Done-Right · Unit 6 · Video 1 · Interactive Practice

The Matrix of a Linear Map: Each Column Is Tvₖ

IKey Formulas

Formula	Name	What it says
$A = \begin{pmatrix} A_{1,1} & \cdots & A_{1,n} \\ \vdots & & \vdots \\ A_{m,1} & \cdots & A_{m,n} \end{pmatrix}$	$m$ -by- $n$ matrix	$m$ rows, $n$ columns, entries in $\mathbf{F}$ ; $A_{j,k}$ is the entry in row $j$ , column $k$ (first index = row)
$Tv_k = A_{1,k} w_1 + \cdots + A_{m,k} w_m = \displaystyle\sum_{j=1}^{m} A_{j,k} w_j$	Matrix of $T$ : $\mathcal{M}(T)$	$T \in \mathcal{L}(V, W)$ , bases $v_1, \dots, v_n$ of $V$ and $w_1, \dots, w_m$ of $W$ ; column $k$ holds the coordinates of $Tv_k$ ; full notation $\mathcal{M}\big(T, (v_1, \dots, v_n), (w_1, \dots, w_m)\big)$
column $k$ of $\mathcal{M}(T)$ $=$ $T$ applied to the $k$ -th standard basis vector	Standard bases	Default for $T \in \mathcal{L}(\mathbf{F}^n, \mathbf{F}^m)$ ; e.g. $T(x, y) = (x + 3y,\ 2x + 5y,\ 7x + 9y)$ has $\mathcal{M}(T) = \begin{pmatrix} 1 & 3 \\ 2 & 5 \\ 7 & 9 \end{pmatrix}$
$\dim V = n$ , $\dim W = m$ $\Rightarrow$ $\mathcal{M}(T)$ is $m$ -by- $n$	Size	Columns match the basis of the domain, rows the basis of the target; e.g. $D \colon \mathcal{P}_3(\mathbf{R}) \to \mathcal{P}_2(\mathbf{R})$ gives a 3-by-4 matrix

Key Insight: Together with the two bases, the $mn$ entries of $\mathcal{M}(T)$ determine $T$ : they determine every $Tv_k$ , and by the linear map lemma the values $Tv_1, \dots, Tv_n$ determine $T$ .

IIVisualization 1 — Rows First, Then Columns

In $A_{j,k}$ the first index $j$ picks the row and the second index $k$ picks the column.

IIIVisualization 2 — Column by Column: a Map from $\mathbf{F}^2$ to $\mathbf{F}^3$

With standard bases, column $k$ of $\mathcal{M}(T)$ is $T$ of the $k$ -th standard basis vector.

Map Step 0

IVVisualization 3 — Differentiation, and Why the Bases Matter

For $D \in \mathcal{L}(\mathcal{P}_3(\mathbf{R}), \mathcal{P}_2(\mathbf{R}))$ , column $k$ lists the coordinates of $Dv_k$ ; other bases give another matrix.

Basis of P₃(ℝ) Basis of P₂(ℝ) Column k 3

VQuiz Questions

Problem 1 · Build the Matrix

Given: $T \in \mathcal{L}(\mathbf{F}^2, \mathbf{F}^3)$ , $T(x, y) = (3x - y,\ 2x + 4y,\ 5y)$ , standard bases — find $\mathcal{M}(T)$ .

✅ Correct! Column 1 is

T(1, 0) = (3, 2, 0)

and column 2 is

T(0, 1) = (-1, 4, 5)

❌ Wrong size. The domain

\mathbf{F}^2

has dimension 2 and the target

\mathbf{F}^3

has dimension 3, so

\mathcal{M}(T)

is 3-by-2: the values

T(1,0)

and

T(0,1)

stand as columns, not rows.

❌ Not quite. The third coordinate

5y

has

x

-coefficient

0

, so

T(1, 0)

has third coordinate

0

and

T(0, 1)

has third coordinate

5

❌ Columns in the wrong order. Column 1 belongs to the first basis vector

(1, 0)

, so it is

T(1, 0) = (3, 2, 0)

Show solution

$\dim \mathbf{F}^2 = 2$ and $\dim \mathbf{F}^3 = 3$ , so $\mathcal{M}(T)$ is 3-by-2. With the standard bases, column $k$ is $T$ of the $k$ -th standard basis vector:

T(1, 0) = (3 - 0,\ 2 + 0,\ 0) = (3, 2, 0), \qquad T(0, 1) = (0 - 1,\ 0 + 4,\ 5) = (-1, 4, 5).

\mathcal{M}(T) = \begin{pmatrix} 3 & -1 \\ 2 & 4 \\ 0 & 5 \end{pmatrix}.

Check by rows: row $j$ holds the coefficients of $x$ and $y$ in coordinate $j$ — row 3 is $(0, 5)$ because coordinate 3 is $0x + 5y$ .

Problem 2 · Count the Basis, Not the Degree

Given: $T \in \mathcal{L}(\mathcal{P}_4(\mathbf{R}), \mathbf{R}^2)$ , $Tp = \big(p(1),\ p(2)\big)$ , with the standard bases $1, x, x^2, x^3, x^4$ and $(1, 0), (0, 1)$ — find the size of $\mathcal{M}(T)$ and its third column.

What size is $\mathcal{M}(T)$ ?

What is the third column of $\mathcal{M}(T)$ ?

✅ Correct!

\dim \mathcal{P}_4(\mathbf{R}) = 5

columns,

\dim \mathbf{R}^2 = 2

rows, and column 3 is

T(x^2) = (1^2, 2^2) = (1, 4)

❌ Rows and columns swapped. The rows match the basis of the target

\mathbf{R}^2

, so the target's dimension comes first:

m

-by-

n

with

m = \dim W

❌ The degree is not the dimension.

\mathcal{P}_4(\mathbf{R})

has basis

1, x, x^2, x^3, x^4

— five vectors, so five columns.

❌ Two slips.

\mathcal{P}_4(\mathbf{R})

has basis

1, x, x^2, x^3, x^4

— five vectors, not four — and the rows match the basis of the target

\mathbf{R}^2

, so the size is

\dim \mathbf{R}^2

-by-

\dim \mathcal{P}_4(\mathbf{R})

❌ Check the size. Columns match the basis of the domain, rows the basis of the target.

❌ That is column 2. The basis starts with

1

, so the third basis vector is

x^2

, not

x

;

(1, 2) = T(x)

❌ Off by one.

(1, 8) = T(x^3)

is column 4; the third basis vector is

x^2

because the list starts at

x^0 = 1

❌ Coordinates out of order.

Tp = (p(1), p(2))

, so the first coordinate of

T(x^2)

1^2 = 1

and the second is

2^2 = 4

❌ Check the column. Column 3 is the coordinates of

T

of the third basis vector.

Show solution

The domain basis $1, x, x^2, x^3, x^4$ has 5 vectors and the target basis has 2, so $\mathcal{M}(T)$ is 2-by-5.

The third basis vector of $\mathcal{P}_4(\mathbf{R})$ is $x^2$ , and in the standard basis of $\mathbf{R}^2$ the coordinates of a vector are its entries:

T(x^2) = (1^2,\ 2^2) = (1, 4).

The whole matrix is $\mathcal{M}(T) = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 & 16 \end{pmatrix}$ , column $k$ being $T(x^{k-1}) = (1, 2^{k-1})$ .

Problem 3 · Multiplication by $x + 1$

Given: $T \in \mathcal{L}(\mathcal{P}_2(\mathbf{R}), \mathcal{P}_3(\mathbf{R}))$ , $(Tp)(x) = (x + 1)\,p(x)$ , standard bases $1, x, x^2$ and $1, x, x^2, x^3$ — find $\mathcal{M}(T)$ .

✅ Correct! The columns are the coordinates of

T(1) = 1 + x

T(x) = x + x^2

T(x^2) = x^2 + x^3

1, x, x^2, x^3

❌ Coordinates written as rows. The coordinates of

T(1) = 1 + x

form column 1, not row 1; the matrix is 4-by-3 because

\dim \mathcal{P}_3(\mathbf{R}) = 4

and

\dim \mathcal{P}_2(\mathbf{R}) = 3

❌ The target is $\mathcal{P}_3(\mathbf{R})$ .

T(x^2) = x^2 + x^3

has an

x^3

-coordinate, so the matrix needs a fourth row for the basis vector

x^3

❌ Rows in the wrong order. Row

j

belongs to the

j

-th basis vector of the target: row 1 to

1

, row 4 to

x^3

. Reading coefficients from the highest power down reverses the rows.

Show solution

$\dim \mathcal{P}_2(\mathbf{R}) = 3$ and $\dim \mathcal{P}_3(\mathbf{R}) = 4$ , so $\mathcal{M}(T)$ is 4-by-3. Apply $T$ to each basis vector and write the result in $1, x, x^2, x^3$ :

T(1) = 1 + x = 1 \cdot 1 + 1 \cdot x + 0 \cdot x^2 + 0 \cdot x^3

T(x) = x + x^2 = 0 \cdot 1 + 1 \cdot x + 1 \cdot x^2 + 0 \cdot x^3

T(x^2) = x^2 + x^3 = 0 \cdot 1 + 0 \cdot x + 1 \cdot x^2 + 1 \cdot x^3

Those coordinate lists are the three columns:

\mathcal{M}(T) = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}.

Problem 4 · A Basis That Is Not Standard

Given: $T \in \mathcal{L}(\mathbf{R}^2, \mathbf{R}^2)$ , $T(x, y) = (3x + y,\ x - y)$ , the basis $v_1 = (1, 1)$ , $v_2 = (1, -1)$ of the domain and the standard basis of the target — find $\mathcal{M}\big(T, (v_1, v_2), ((1, 0), (0, 1))\big)$ .

✅ Correct! Column 1 is

Tv_1 = (4, 0)

and column 2 is

Tv_2 = (2, 2)

— the same map has a different matrix than with the standard bases.

❌ That is the matrix with the standard bases. Here the columns record

Tv_1

and

Tv_2

for the given

v_1 = (1, 1)

v_2 = (1, -1)

; the matrix depends on the basis of the domain too.

❌ Values written as rows.

Tv_1 = (4, 0)

is column 1, read top to bottom.

❌ Columns in the wrong order. Column 1 belongs to

v_1 = (1, 1)

, so it is

Tv_1 = (4, 0)

Show solution

Column $k$ holds the coordinates of $Tv_k$ in the target basis; with the standard basis of $\mathbf{R}^2$ these are just its entries.

Tv_1 = T(1, 1) = (3 + 1,\ 1 - 1) = (4, 0) = 4\,(1, 0) + 0\,(0, 1)

Tv_2 = T(1, -1) = (3 - 1,\ 1 + 1) = (2, 2) = 2\,(1, 0) + 2\,(0, 1)

\mathcal{M}\big(T, (v_1, v_2), ((1, 0), (0, 1))\big) = \begin{pmatrix} 4 & 2 \\ 0 & 2 \end{pmatrix}.

With the standard bases on both sides the matrix of the same $T$ is $\begin{pmatrix} 3 & 1 \\ 1 & -1 \end{pmatrix}$ : one map, two bases of the domain, two different matrices.

Solved: 0 / 4

The Matrix of a Linear Map: Each Column Is Tvₖ

IKey Formulas

IIVisualization 1 — Rows First, Then Columns

IIIVisualization 2 — Column by Column: a Map from F2\mathbf{F}^2F2 to F3\mathbf{F}^3F3

IVVisualization 3 — Differentiation, and Why the Bases Matter

VQuiz Questions

IIIVisualization 2 — Column by Column: a Map from $\mathbf{F}^2$ to $\mathbf{F}^3$