Linear-Algebra-Done-Right · Unit 5 · Video 1 · Interactive Practice

Linear Maps and the Linear Map Lemma: Values on a Basis

IKey Formulas

Formula	Name	What it says
$T(u + v) = Tu + Tv$ and $T(\lambda v) = \lambda (Tv)$	Linear map $T \colon V \to W$	Additivity and homogeneity for all $u, v \in V$ , $\lambda \in \mathbf{F}$ ; the sum and scalar multiple on the left are taken in $V$ , on the right in $W$
$\mathcal{L}(V, W)$ , $\ \mathcal{L}(V) = \mathcal{L}(V, V)$	Sets of linear maps	Examples: $0v = 0$ , $Iv = v$ , $Dp = p'$ , $Tp = \int_0^1 p$ , $(Tp)(x) = x^2 p(x)$ , $(Tp)(x) = p(q(x))$ , the backward shift
$T(x_1, \dots, x_n) = \bigl(A_{1,1} x_1 + \cdots + A_{1,n} x_n,\ \dots,\ A_{m,1} x_1 + \cdots + A_{m,n} x_n\bigr)$	Linear maps $\mathbf{F}^n \to \mathbf{F}^m$	$A_{j,k}$ is coordinate $j$ of $T$ at the $k$ -th standard basis vector; e.g. $T(x, y, z) = (2x - y + 3z,\ 7x + 5y - 6z)$
$T(c_1 v_1 + \cdots + c_n v_n) = c_1 w_1 + \cdots + c_n w_n$	Linear map lemma	$v_1, \dots, v_n$ a basis of $V$ and any $w_1, \dots, w_n \in W$ $\Rightarrow$ exactly one $T \in \mathcal{L}(V, W)$ with $Tv_k = w_k$

Key Insight: On a basis, the values of a linear map can be chosen freely (existence), and once chosen they determine the map at every vector (uniqueness).

IIVisualization 1 — Additivity and Homogeneity

Linearity demands $T(u + v) = Tu + Tv$ and $T(\lambda u) = \lambda\, Tu$ for every $u$ , $v$ , $\lambda$ , not just for a few.

Map Property λ 1.5

IIIVisualization 2 — The Linear Map Lemma

Any choice of $w_1 = Tv_1$ and $w_2 = Tv_2$ on a basis gives exactly one linear map: $Tv = c_1 w_1 + c_2 w_2$ .

IVVisualization 3 — Coefficients Are Values on the Standard Basis

The coefficients of $x$ , $y$ , $z$ in $T(x, y, z)$ are exactly the coordinates of $T(1, 0, 0)$ , $T(0, 1, 0)$ , $T(0, 0, 1)$ .

Values on the standard basis x 1 y 2 z −1

VQuiz Questions

Problem 1 · Spot the Linear Map

Given: four maps $\mathbf{R}^2 \to \mathbf{R}^2$ — find the one that is linear.

✅ Correct! Each coordinate of

(5y,\ x - 4y)

is a sum of constants times

x

and

y

, with no constant term and no products, so additivity and homogeneity both hold.

❌ The constant term breaks it. Every linear map sends

0

0

, because

T(0) = T(0 + 0) = T(0) + T(0)

. Here

T(0, 0) = (2, 0) \neq (0, 0)

❌ The product breaks homogeneity.

T(2 \cdot (1, 1)) = T(2, 2) = (4, 2)

, but

2\, T(1, 1) = 2(1, 1) = (2, 2)

❌ The absolute value breaks it. With

\lambda = -1

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

, but

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

. Additivity fails too:

T\bigl((1, 0) + (-1, 0)\bigr) = (0, 0)

while

T(1, 0) + T(-1, 0) = (2, 0)

Show solution

For (c), take $u = (x, y)$ , $u' = (x', y')$ and $\lambda \in \mathbf{R}$ :

T(u + u') = \bigl(5(y + y'),\ (x + x') - 4(y + y')\bigr) = (5y,\ x - 4y) + (5y',\ x' - 4y') = Tu + Tu'

T(\lambda u) = (5\lambda y,\ \lambda x - 4\lambda y) = \lambda (5y,\ x - 4y) = \lambda\, Tu

The other three each fail for a specific pair or scalar:

(a) $T(0, 0) = (2, 0)$ , but a linear map must send $0$ to $0$ .
(b) $T(2, 2) = (4, 2) \neq (2, 2) = 2\,T(1, 1)$ .
(d) $T(-1, 0) = (1, 0) \neq (-1, 0) = -T(1, 0)$ .

One failing instance is enough to show a map is not linear.

Problem 2 · Can the Values Be Chosen?

Given: the vectors $(1, 2)$ and $(2, 4)$ in $\mathbf{R}^2$ — decide whether some $T \in \mathcal{L}(\mathbf{R}^2)$ has $T(1, 2) = (3, -1)$ and $T(2, 4) = (6, 1)$ .

✅ Correct!

(2, 4) = 2(1, 2)

, so the value at

(2, 4)

is already decided by the value at

(1, 2)

, and

(6, 1) \neq (6, -2)

❌ The lemma needs a basis. Its hypothesis is that

v_1, \dots, v_n

is a basis;

(1, 2), (2, 4)

is linearly dependent, since

(2, 4) = 2(1, 2)

❌ Not even one exists. Infinitely many maps would exist only if the two values were compatible, i.e. if

T(2, 4)

were

2\,T(1, 2) = (6, -2)

❌ Check the second value. This map gives

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

, but

T(2, 4) = (6, 10 - 12) = (6, -2)

, not

(6, 1)

Show solution

Notice that $(2, 4) = 2(1, 2)$ . If $T$ is linear, homogeneity gives

T(2, 4) = T\bigl(2 \cdot (1, 2)\bigr) = 2\, T(1, 2) = 2(3, -1) = (6, -2).

Since $(6, -2) \neq (6, 1)$ , no linear map has both values. The linear map lemma does not apply: it lets the values be chosen freely only on a basis, and $(1, 2), (2, 4)$ is linearly dependent — exactly the video's warning that $v_2 = 2v_1$ forces $Tv_2 = 2\,Tv_1$ .

Problem 3 · Build the Map from a Basis

Given: the basis $v_1 = (1, 1)$ , $v_2 = (1, -1)$ of $\mathbf{R}^2$ and the unique $T \in \mathcal{L}(\mathbf{R}^2)$ with $Tv_1 = (3, 1)$ and $Tv_2 = (1, 5)$ — find $T(4, 2)$ and the formula for $T(x, y)$ .

What is T(4, 2)?

Which formula is T?

✅ Correct!

T(x, y) = (2x + y,\ 3x - 2y)

gives

T(1, 1) = (3, 1)

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

and

T(4, 2) = (10, 8)

❌ Not quite. First write

(4, 2) = c_1 (1, 1) + c_2 (1, -1)

, then use

T(4, 2) = c_1\, Tv_1 + c_2\, Tv_2

❌ Those are standard coordinates, not basis coefficients.

4(3, 1) + 2(1, 5)

treats

(4, 2)

4v_1 + 2v_2

, but

4v_1 + 2v_2 = (6, 2)

. Solve

c_1 + c_2 = 4

c_1 - c_2 = 2

first.

❌ The coefficients are swapped.

(4, 2) = 3v_1 + 1v_2

, so

T(4, 2) = 3(3, 1) + 1(1, 5)

, not

1(3, 1) + 3(1, 5)

❌ The coefficients are missing.

Tv_1 + Tv_2 = (4, 6)

T(v_1 + v_2) = T(2, 0)

; the vector

(4, 2)

3v_1 + v_2

❌ Check the formula on the basis. The right formula must give

T(1, 1) = (3, 1)

and

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

❌ That map uses the standard basis. It sends

(1, 0) \mapsto (3, 1)

and

(0, 1) \mapsto (1, 5)

, so

T(1, 1) = (4, 6) \neq (3, 1)

❌ The images are swapped. This map gives

T(1, 1) = (1, 5)

and

T(1, -1) = (3, 1)

— it sends

v_1

w_2

and

v_2

w_1

❌ The coefficient array is transposed. This map gives

T(1, 1) = (5, -1) \neq (3, 1)

Show solution

Step 1: Coefficients. Write $(x, y) = c_1 (1, 1) + c_2 (1, -1)$ , so $c_1 + c_2 = x$ and $c_1 - c_2 = y$ :

c_1 = \frac{x + y}{2}, \qquad c_2 = \frac{x - y}{2}.

For $(4, 2)$ : $c_1 = 3$ , $c_2 = 1$ ; indeed $3(1, 1) + 1(1, -1) = (4, 2)$ .

Step 2: Value. By the lemma's formula,

T(4, 2) = 3\, Tv_1 + 1\, Tv_2 = 3(3, 1) + (1, 5) = (10, 8).

Step 3: Formula.

T(x, y) = \frac{x + y}{2}(3, 1) + \frac{x - y}{2}(1, 5) = \Bigl(\frac{4x + 2y}{2},\ \frac{6x - 4y}{2}\Bigr) = (2x + y,\ 3x - 2y).

Check: $T(1, 1) = (3, 1)$ and $T(1, -1) = (1, 5)$ . By uniqueness, this is the only linear map with these values.

Problem 4 · Uniqueness on Polynomials

Given: $T \in \mathcal{L}(\mathcal{P}_2(\mathbf{R}), \mathbf{R})$ with $T(1) = 1$ , $T(x) = \tfrac{1}{2}$ and $T(x^2) = \tfrac{1}{3}$ — identify $T$ and find $T(6x^2 - 2x + 5)$ .

Which map is T?

What is T(6x² − 2x + 5)?

✅ Correct!

p \mapsto \int_0^1 p

is linear with the same values on the basis

1, x, x^2

, so by uniqueness it is

T

; and

\int_0^1 (6x^2 - 2x + 5)\,dx = 2 - 1 + 5 = 6

❌ Not quite. Compare the candidate's values at

1

x

x^2

with

1

\tfrac{1}{2}

\tfrac{1}{3}

❌ Evaluation at 1 gives $1, 1, 1$ . It is linear, but its value at

x

1

, not

\tfrac{1}{2}

❌ Evaluation at ½ agrees at $1$ and $x$ but not at $x^2$ . It gives

\bigl(\tfrac{1}{2}\bigr)^2 = \tfrac{1}{4} \neq \tfrac{1}{3}

❌ The derivative at 1 gives $0, 1, 2$ . Already

T(1) = 0 \neq 1

❌ Not quite. Use

T(6x^2 - 2x + 5) = 6\,T(x^2) - 2\,T(x) + 5\,T(1)

❌ That is the value at $x = 1$ .

6 - 2 + 5 = 9

evaluates the polynomial;

T

weights

x^2

\tfrac{1}{3}

and

x

\tfrac{1}{2}

❌ The constant term counts.

5 = 5 \cdot 1

is a multiple of the basis vector

1

, so it contributes

5\,T(1) = 5

to the total.

❌ The coefficients are paired with the wrong basis vectors.

6 \cdot 1 - 2 \cdot \tfrac{1}{2} + 5 \cdot \tfrac{1}{3}

multiplies the

x^2

-coefficient by

T(1)

; it belongs with

T(x^2) = \tfrac{1}{3}

Show solution

Identify $T$ . The list $1, x, x^2$ is a basis of $\mathcal{P}_2(\mathbf{R})$ . The map $p \mapsto \int_0^1 p$ is linear (integration of a sum and of a constant multiple), and

\int_0^1 1\,dx = 1, \qquad \int_0^1 x\,dx = \tfrac{1}{2}, \qquad \int_0^1 x^2\,dx = \tfrac{1}{3}.

It agrees with $T$ on a basis, so by the uniqueness half of the linear map lemma, $Tp = \int_0^1 p$ . The other candidates differ at some basis vector: $p(1)$ gives $1$ at $x$ , $p(\tfrac{1}{2})$ gives $\tfrac{1}{4}$ at $x^2$ , and $p'(1)$ gives $0$ at $1$ .

Evaluate. By linearity,

T(6x^2 - 2x + 5) = 6 \cdot \tfrac{1}{3} - 2 \cdot \tfrac{1}{2} + 5 \cdot 1 = 2 - 1 + 5 = 6,

which matches $\int_0^1 (6x^2 - 2x + 5)\,dx = \bigl[2x^3 - x^2 + 5x\bigr]_0^1 = 6$ .

Solved: 0 / 4