Linear-Algebra-Done-Right · Unit 7 · Video 4 · Interactive Practice

The Matrix of a Vector: $\mathcal{M}(Tv) = \mathcal{M}(T)\mathcal{M}(v)$

IKey Formulas

Formula	Name	Setting
$\mathcal{M}(v) = \begin{pmatrix} b_1 \\ \vdots \\ b_n \end{pmatrix}$ where $v = b_1 v_1 + \cdots + b_n v_n$	Matrix of a vector	Basis $v_1, \dots, v_n$ of $V$ ; $\mathcal{M} : V \to \mathbf{F}^{n,1}$ is an isomorphism
$\mathcal{M}(T)_{\cdot,k} = \mathcal{M}(Tv_k)$	Column $k$ of $\mathcal{M}(T)$	$T \in \mathcal{L}(V,W)$ ; coordinates with respect to $w_1, \dots, w_m$
$\mathcal{M}(Tv) = \mathcal{M}(T)\,\mathcal{M}(v)$	Linear map = matrix times column	Same bases on both sides
$\dim \operatorname{range} T = \text{column rank of } \mathcal{M}(T)$	Rank from the matrix	$V, W$ finite-dimensional; any bases

Key Insight: Once bases are chosen, every vector becomes a column of coordinates, and applying $T$ becomes multiplying that column by $\mathcal{M}(T)$ .

IICoordinates Depend on the Basis

One vector $v \in \mathbf{R}^2$ generally has a different column $\mathcal{M}(v)$ for each basis: the unique weights in $v = b_1 v_1 + b_2 v_2$ .

Basis v: first entry 3 v: second entry 2

IIITwo Routes to the Same Column

For $D : \mathcal{P}_4(\mathbf{R}) \to \mathcal{P}_3(\mathbf{R})$ , differentiating then taking coordinates equals multiplying $\mathcal{M}(p)$ by $\mathcal{M}(D)$ .

a₀ 2 a₁ −7 a₂ 0 a₃ 5 a₄ 1 Row of M(D) 3

IVBuilding $\mathcal{M}(T)$ Column by Column

Column $k$ of $\mathcal{M}(T)$ is $\mathcal{M}(Tv_k)$ , and the dimension of their span is $\dim \operatorname{range} T$ .

Map Columns 0

💡 $\operatorname{range} T$ is defined without any basis, so although $\mathcal{M}(T)$ can change when the bases of $\mathcal{P}_3(\mathbf{R})$ and $\mathbf{R}^3$ change, its column rank never does.

VQuiz Questions

Problem 1 · Coordinates of a Polynomial

Given: $p = 4x - x^3 + 6 \in \mathcal{P}_3(\mathbf{R})$ with the standard basis $1, x, x^2, x^3$ — find $\mathcal{M}(p)$ .

✅ Correct! One entry per basis vector, in the order

1, x, x^2, x^3

, with

0

for the missing

x^2

term.

❌ Not quite. The entries follow the basis

1, x, x^2, x^3

, not the order in which the terms are written — and

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

, so

\mathcal{M}(p)

4

-by-

1

❌ Not quite. That column lists the coefficients from

x^3

down to

1

; entry

k

must be the coefficient of the

k

-th basis vector, starting with

1

❌ Close, but… there is no

x^2

term, yet

x^2

is still a basis vector: its coefficient

0

keeps its place, so

\mathcal{M}(p)

4

-by-

1

Show solution

Write $p$ as a linear combination of the basis $1, x, x^2, x^3$ :

p = 6 \cdot 1 + 4 \cdot x + 0 \cdot x^2 + (-1) \cdot x^3.

The coefficients, in basis order, form the column

\mathcal{M}(p) = \begin{pmatrix} 6 \\ 4 \\ 0 \\ -1 \end{pmatrix}.

Problem 2 · Same Vector, Other Basis

Given: the basis $v_1 = (2, 1)$ , $v_2 = (3, 2)$ of $\mathbf{R}^2$ and $v = (4, 3)$ — find $\mathcal{M}(v)$ with respect to $v_1, v_2$ .

✅ Correct!

-1 \cdot (2, 1) + 2(3, 2) = (4, 3)

, so the coordinates of

v

in this basis are

-1

and

2

❌ Not quite.

(4, 3)

\mathcal{M}(v)

for the standard basis.

\mathcal{M}(v)

depends on the basis: you need

b_1, b_2

with

b_1 (2, 1) + b_2 (3, 2) = (4, 3)

❌ Not quite. Those weights solve

2b_1 + b_2 = 4

3b_1 + 2b_2 = 3

, which treats

v_1, v_2

as the rows of the system; check:

5(2, 1) - 6(3, 2) = (-8, -7) \neq (4, 3)

❌ Close, but… the order matters: entry

1

is the coefficient of

v_1

. Check:

2(2, 1) - 1 \cdot (3, 2) = (1, 0)

Show solution

Find the unique $b_1, b_2$ with $b_1 (2, 1) + b_2 (3, 2) = (4, 3)$ :

2b_1 + 3b_2 = 4, \qquad b_1 + 2b_2 = 3.

Subtract twice the second equation from the first: $-b_2 = 4 - 6 = -2$ , so $b_2 = 2$ and $b_1 = 3 - 4 = -1$ . Check: $-1 \cdot (2, 1) + 2(3, 2) = (-2 + 6,\ -1 + 4) = (4, 3)$ . Hence

\mathcal{M}(v) = \begin{pmatrix} -1 \\ 2 \end{pmatrix}.

Problem 3 · Columns, Then the Product

Given: $T \in \mathcal{L}\big(\mathcal{P}_2(\mathbf{R}), \mathbf{R}^2\big)$ defined by $Tp = \big(p(0) + p(1),\ p'(1)\big)$ , with the standard bases $1, x, x^2$ and $e_1, e_2$ — find column $2$ of $\mathcal{M}(T)$ , then $\mathcal{M}(Tp)$ for $p = 3 - x + 2x^2$ .

Column 2 of M(T)?

M(Tp)?

✅ Correct! Column

2

\mathcal{M}(Tx) = (1, 1)

, and

\mathcal{M}(T)\mathcal{M}(p) = (7, 3)

— the same column as computing

Tp = (3 + 4,\ 3)

directly.

❌ Not quite. That is row

2

\mathcal{M}(T)

. Column

k

\mathcal{M}(Tv_k)

, an

m

-by-

1

column — here

2

-by-

1

❌ Not quite.

(2, 0) = \mathcal{M}(T1)

is column

1

. The second basis vector is

v_2 = x

, so column

2

\mathcal{M}(Tx)

❌ Check column 2. Column

2

\mathcal{M}(Tv_2)

with

v_2 = x

: compute

x(0) + x(1)

and the derivative of

x

1

❌ Not quite. That multiplies

\mathcal{M}(T)

(2, -1, 3)

, the coefficients of

p

read from

x^2

down. In the basis

1, x, x^2

\mathcal{M}(p) = (3, -1, 2)

❌ Check the product.

\mathcal{M}(Tp) = \mathcal{M}(T)\mathcal{M}(p)

with

\mathcal{M}(p) = (3, -1, 2)

; both entries of

Tp

must be included:

p(0) + p(1)

, then

p'(1)

Show solution

The columns. Column $k$ of $\mathcal{M}(T)$ is $\mathcal{M}(Tv_k)$ :

T1 = (1 + 1,\ 0) = (2, 0), \quad Tx = (0 + 1,\ 1) = (1, 1), \quad Tx^2 = (0 + 1,\ 2) = (1, 2),

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

So column $2$ is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ .

The product. $\mathcal{M}(p) = (3, -1, 2)$ as a column, so

\mathcal{M}(Tp) = \mathcal{M}(T)\mathcal{M}(p) = 3\begin{pmatrix} 2 \\ 0 \end{pmatrix} - 1\begin{pmatrix} 1 \\ 1 \end{pmatrix} + 2\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 7 \\ 3 \end{pmatrix}.

Check directly: $p(0) = 3$ , $p(1) = 3 - 1 + 2 = 4$ , $p'(x) = -1 + 4x$ so $p'(1) = 3$ . Thus $Tp = (7, 3)$ , whose coordinates in $e_1, e_2$ are $(7, 3)$ . ✓

Problem 4 · Range From the Matrix

Given: $T \in \mathcal{L}(V, W)$ with $\dim V = 4$ , $\dim W = 3$ , and with respect to some bases of $V$ and $W$

\mathcal{M}(T) = \begin{pmatrix} 1 & 2 & 0 & 3 \\ 2 & 4 & 1 & 7 \\ -1 & -2 & 1 & -2 \end{pmatrix}.

Find $\dim \operatorname{range} T$ and $\dim \operatorname{null} T$ .

dim range T?

dim null T?

✅ Correct! Only two columns are independent, so

\dim \operatorname{range} T = 2

, and

\dim \operatorname{null} T = 4 - 2 = 2

— whatever bases were used.

❌ Not quite.

4

counts the columns, but the column rank counts only linearly independent ones, and column

2

is twice column

1

❌ Not quite.

3 = \dim W

would mean

T

is surjective; the columns are dependent — column

2 = 2 \cdot

column

1

and column

4 = 3 \cdot

column

1 +

column

3

❌ Check the column rank.

\dim \operatorname{range} T

is the column rank: find how many columns are linearly independent.

❌ Not quite. The fundamental theorem of linear maps uses the domain:

\dim V = \dim \operatorname{null} T + \dim \operatorname{range} T

, and

\dim V = 4

, not

3

❌ Check the null space. Use

\dim V = \dim \operatorname{null} T + \dim \operatorname{range} T

with

\dim V = 4

Show solution

Call the columns $c_1, c_2, c_3, c_4$ . Then $c_2 = 2c_1$ , and

3c_1 + c_3 = \begin{pmatrix} 3 \\ 6 \\ -3 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 7 \\ -2 \end{pmatrix} = c_4.

So the span of the columns is $\operatorname{span}(c_1, c_3)$ , and $c_1, c_3$ are linearly independent (neither is a scalar multiple of the other: $c_3$ has first entry $0$ but is not zero). The column rank is $2$ , so

\dim \operatorname{range} T = \text{column rank of } \mathcal{M}(T) = 2.

By the fundamental theorem of linear maps, $\dim \operatorname{null} T = \dim V - \dim \operatorname{range} T = 4 - 2 = 2$ . Neither answer depends on which bases produced $\mathcal{M}(T)$ .

Solved: 0 / 4

The Matrix of a Vector: M(Tv)=M(T)M(v)\mathcal{M}(Tv) = \mathcal{M}(T)\mathcal{M}(v)M(Tv)=M(T)M(v)

IKey Formulas

IICoordinates Depend on the Basis

IIITwo Routes to the Same Column

IVBuilding M(T)\mathcal{M}(T)M(T) Column by Column

VQuiz Questions

The Matrix of a Vector: $\mathcal{M}(Tv) = \mathcal{M}(T)\mathcal{M}(v)$

IVBuilding $\mathcal{M}(T)$ Column by Column