Linear-Algebra-Done-Right · Unit 28 · Video 4 · Interactive Practice

The Determinant of an Operator: αT = (det T)α

IKey Formulas

Formula	Name	What it says
$\alpha_T(v_1, \dots, v_m) = \alpha(Tv_1, \dots, Tv_m)$	Transforming a form by $T$	$\alpha \mapsto \alpha_T$ is linear on $V_{\text{alt}}^{(m)}$
$\alpha_T = (\det T)\,\alpha$ for all $\alpha \in V_{\text{alt}}^{(\dim V)}$	Determinant of $T$	Defined with no basis and no matrix
$\det I = 1, \quad \det(\lambda T) = \lambda^n \det T$	First properties ( $n = \dim V$ )	One factor of $\lambda$ per slot
$\det T = \lambda_1 \cdots \lambda_n$	Eigenvector basis, $Te_k = \lambda_k e_k$	Diagonal matrix: product of diagonal entries

Key Insight: $V_{\text{alt}}^{(n)}$ is one-dimensional, and a linear map on a one-dimensional space is multiplication by one scalar. That scalar is $\det T$ .

IIOne Scalar for Every Input List

For a fixed $T$ on $\mathbf{R}^2$ , the ratio $\alpha(Tv_1, Tv_2) / \alpha(v_1, v_2)$ is the same for every independent $v_1, v_2$ .

Te₁ = (3, 0) Te₂ = (1, 1) α

IIIA Factor of λ in Every Slot

$\lambda T$ puts a factor $\lambda$ into each of the $n$ slots, so $\det(\lambda T) = \lambda^n \det T$ .

n = dim V 3 λ 2 det T 1

IVQuiz Questions

Problem 1 · Doubling Every Vector

Given: $\dim V = 3$ and the operator $2I$ on $V$ . Find $\det(2I)$ .

✅ Correct!

\det(\lambda I) = \lambda^n

, here

2^3 = 8

❌ Not quite. The determinant is not linear in the operator: each of the

3

slots contributes its own factor of

2

❌ Not quite. The factors multiply, they do not add:

2 \cdot 2 \cdot 2

❌ Not quite. Pull one factor of

2

out of each slot of

\alpha(2v_1, 2v_2, 2v_3)

Show solution

For $\alpha \in V_{\text{alt}}^{(3)}$ , linearity in each slot gives

\alpha_{2I}(v_1, v_2, v_3) = \alpha(2v_1, 2v_2, 2v_3) = 2^3\,\alpha(v_1, v_2, v_3).

So $\alpha_{2I} = 8\alpha$ for every $\alpha$ , and $\det(2I) = 8$ .

Problem 2 · Scaling an Operator

Given: $\dim V = 4$ and $T \in \mathcal{L}(V)$ with $\det T = 3$ . Find $\det(-2T)$ .

✅ Correct!

(-2)^4 \cdot 3 = 16 \cdot 3 = 48

❌ Not quite.

-2 \cdot 3

treats the determinant as linear; the factor

-2

appears once per slot, four times.

❌ Not quite.

(-2)^4

is positive: an even number of negative factors.

❌ Not quite. That uses

(-2)^3

; the exponent is

n = \dim V = 4

❌ Not quite. Use

\det(\lambda T) = \lambda^n \det T

Show solution

For $\alpha \in V_{\text{alt}}^{(4)}$ , pulling $\lambda = -2$ out of each of the four slots,

\alpha_{-2T}(v_1, \dots, v_4) = \alpha(-2Tv_1, \dots, -2Tv_4) = (-2)^4 \alpha_T(v_1, \dots, v_4) = 16 \cdot 3\,\alpha(v_1, \dots, v_4).

So $\det(-2T) = 48$ .

Problem 3 · An Eigenvector Basis

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ and a basis $u_1, u_2, u_3$ of $\mathbf{R}^3$ with $Tu_1 = 2u_1$ , $Tu_2 = -u_2$ , $Tu_3 = 5u_3$ . Find $\det T$ .

✅ Correct! With an eigenvector basis,

\det T

is the product of the eigenvalues:

2 \cdot (-1) \cdot 5 = -10

❌ Not quite.

6

is the sum of the eigenvalues; the determinant is their product.

❌ Not quite. Keep the sign of the eigenvalue

-1

❌ Not quite. The eigenvector basis is enough: evaluate

\alpha_T

u_1, u_2, u_3

for a nonzero

\alpha

❌ Not quite. Pull each eigenvalue out of its slot.

Show solution

Take a nonzero $\alpha \in (\mathbf{R}^3)_{\text{alt}}^{(3)}$ . Then

\alpha_T(u_1, u_2, u_3) = \alpha(2u_1, -u_2, 5u_3) = 2 \cdot (-1) \cdot 5\,\alpha(u_1, u_2, u_3),

and also $\alpha_T(u_1, u_2, u_3) = (\det T)\,\alpha(u_1, u_2, u_3)$ . Since $u_1, u_2, u_3$ is a basis, $\alpha(u_1, u_2, u_3) \ne 0$ , so $\det T = -10$ .

Problem 4 · Reading det T from One List

Given: $\dim V = 3$ , $T \in \mathcal{L}(V)$ , a nonzero $\alpha \in V_{\text{alt}}^{(3)}$ , and a linearly independent list $u, v, w$ with $\alpha(Tu, Tv, Tw) = 4\,\alpha(u, v, w)$ . Find $\det(2T)$ .

✅ Correct!

\det T = 4

, so

\det(2T) = 2^3 \cdot 4 = 32

❌ Not quite.

2 \cdot 4 = 8

forgets that the factor

2

appears in all three slots.

❌ Not quite. The exponent is

\dim V = 3

, not

2

❌ Not quite.

\alpha_T = (\det T)\alpha

holds on every list, and

\alpha(u, v, w) \ne 0

because the list is independent of length

\dim V

, so you may divide by it.

❌ Not quite. First find

\det T

from the given list.

Show solution

By definition $\alpha(Tu, Tv, Tw) = \alpha_T(u, v, w) = (\det T)\,\alpha(u, v, w)$ . The list $u, v, w$ is linearly independent of length $3 = \dim V$ and $\alpha \ne 0$ , so $\alpha(u, v, w) \ne 0$ (the independence test from the previous video). Dividing, $\det T = 4$ .

Then $\det(2T) = 2^3 \det T = 8 \cdot 4 = 32$ .

Solved: 0 / 4