LINEAR-ALGEBRA · Unit 29 · Video 2

The Basis Trick: Interactive Practice

IKey Formulas

Formula	Name	Meaning
$v = c_1 v_1 + c_2 v_2 + \dots + c_n v_n$	Basis decomposition	Every vector is a unique combination of basis vectors
$T(v) = c_1 T(v_1) + c_2 T(v_2) + \dots + c_n T(v_n)$	The Basis Trick	Linearity transports the decomposition through $T$
$T(au + bv) = a\,T(u) + b\,T(v)$	Linearity	Additivity + homogeneity
$M = [\,T(v_1)\;\;T(v_2)\;\;\dots\;\;T(v_n)\,]$	Matrix of $T$	Columns are images of basis vectors

IIVisualization 1 — A Vector as Its Basis Components

A vector is $c_1 e_1 + c_2 e_2$ — its horizontal and vertical basis pieces added tip-to-tail.

c₁ 3 c₂ 2

IIIVisualization 2 — The Basis Trick in Action

For the 90° rotation $T$ , the output $T(v)=c_1T(e_1)+c_2T(e_2)$ rebuilds from the same coefficients on the rotated basis.

c₁ 3 c₂ 2

Input space

Output space — apply T

IVVisualization 3 — Same Arrow, Different Coordinates

Same arrow, two dictionaries: the standard basis $\{e_1,e_2\}$ and the skew basis $\{v_1=(1,1),\,v_2=(1,-1)\}$ .

x 3 y 2

Standard basis {e₁, e₂}

Basis {v₁, v₂}

VPractice Questions

Question 1

A linear transformation $T$ on $\mathbb{R}^2$ satisfies $T(e_1) = (2, 3)$ and $T(e_2) = (-1, 4)$ .

Using the basis trick, what is $T(v)$ for $v = (5, 2)$ ?

✅ Correct! You used the coefficients of $v$ together with the images of the basis vectors.

❌ Not quite. Remember: $T(v) = c_1 T(e_1) + c_2 T(e_2)$ , where $(c_1, c_2)$ are the coordinates of $v$ .

Show solution

Solution:

Decompose $v$ in the standard basis: $v = (5, 2) = 5\,e_1 + 2\,e_2$ .

Apply the basis trick: $T(v) = 5\,T(e_1) + 2\,T(e_2) = 5(2, 3) + 2(-1, 4) = (10, 15) + (-2, 8) = (8, 23).$

Question 2

For a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ , the standard matrix of $T$ is

$M = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.$

What do the columns of $M$ represent?

✅ Correct! A matrix is the record of where the basis vectors are sent.

❌ Not quite. Recall the slogan: 'A matrix = where the basis goes.'

Show solution

Solution:

A matrix is just a table recording where the basis vectors go. The first column of $M$ is $T(e_1)$ and the second column is $T(e_2)$ .

For this matrix:

$T(e_1) = (0, 1)$ — first column
$T(e_2) = (-1, 0)$ — second column

That's the entire content of the matrix — once you know where the basis goes, the basis trick fills in $T(v)$ for every other vector.

Question 3

True or False: The coordinates $(3, 2)$ are an intrinsic property of a vector — they describe the vector no matter what basis you use.

✅ Correct! Coordinates are an invoice written in the currency of a chosen basis.

❌ Not quite. Remember Visualization 3 — the same arrow had two different coordinate representations.

Show solution

Solution:

False. Coordinates are not intrinsic to a vector — they depend on the chosen basis.

Writing a vector as $(3, 2)$ silently assumes the standard basis, meaning $3\,e_1 + 2\,e_2$ . Switch to a different basis like $v_1 = (1,1)$ , $v_2 = (1,-1)$ , and the same arrow gets a different coordinate label, namely $(2.5, 0.5)$ .

The arrow itself is geometric — a length and direction in space. The numbers are bookkeeping in a chosen language.

Question 4

A linear transformation $T$ on $\mathbb{R}^2$ sends $T(e_1) = (1, 0)$ and $T(e_2) = (0, 0)$ .

Which of the following best describes $T$ geometrically, and what is $T(v)$ for $v = (4, 7)$ ?

✅ Correct! Knowing $T(e_1)$ and $T(e_2)$ tells you everything: the second basis vector gets sent to zero, so the $y$ -component is annihilated.

❌ Not quite. Use the basis trick: $T(v) = 4\,T(e_1) + 7\,T(e_2)$ . What happens when $T(e_2)$ is the zero vector?

Show solution

Solution:

Apply the basis trick. For $v = (4, 7) = 4\,e_1 + 7\,e_2$ : $T(v) = 4\,T(e_1) + 7\,T(e_2) = 4(1, 0) + 7(0, 0) = (4, 0).$

The $y$ -component is annihilated and the $x$ -component is preserved — this is exactly the projection onto the x-axis. The corresponding matrix is

$M = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},$

whose columns are $T(e_1) = (1, 0)$ and $T(e_2) = (0, 0)$ .

Solved: 0 / 4