Linear-Algebra-Done-Right · Unit 29 · Video 1 · Interactive Practice

det(ST) = det S · det T: Invertibility and Eigenvalues

IKey Formulas

Formula	Name	Where it comes from
$\det(ST) = (\det S)(\det T)$ , $\ \det(AB) = (\det A)(\det B)$	Product rule ( $S, T \in \mathcal{L}(V)$ ; $A, B$ square of the same size)	$\alpha_{ST} = (\det S)\,\alpha_T = (\det S)(\det T)\,\alpha$
$T$ invertible $\iff \det T \neq 0$ , $\ \det(T^{-1}) = \dfrac{1}{\det T}$	Invertibility test	$1 = \det I = \det(TT^{-1})$ ; converse from a basis $v, e_2, \dots, e_n$
$\lambda$ eigenvalue of $T \iff \det(\lambda I - T) = 0$	Eigenvalue test ( $\lambda \in \mathbf{F}$ )	$\lambda$ eigenvalue $\iff \lambda I - T$ not invertible

Key Insight: The product rule needs only the definition $\alpha_T = (\det T)\,\alpha$ , used once for $S$ and once for $T$ ; both tests then follow from it.

IISigned Areas Multiply

On $\mathbf{R}^2$ , $T$ scales the unit square's signed area by $\det T$ , and $S$ then scales it again by $\det S$ .

Operator S

IIIEigenvalues Are the Zeros of det(λI − T)

For $T$ with matrix $\begin{pmatrix} 3 & 1 \\ 2 & 2 \end{pmatrix}$ , $\lambda I - T$ fails to be invertible exactly where $\det(\lambda I - T)$ crosses zero.

λ 2.5

IVQuiz Questions

Problem 1 · Products and Inverses

Given: $S, T \in \mathcal{L}(V)$ with $\det S = 3$ and $\det T = -2$ . Find $\det(S^2 T^{-1})$ .

✅ Correct! The product rule gives

(\det S)^2 \cdot \dfrac{1}{\det T} = 9 \cdot \left(-\tfrac{1}{2}\right)

❌ Not quite.

T^{-1}

contributes

\det(T^{-1}) = 1/\det T

, not

\det T

❌ Check the sign.

\det(T^{-1}) = 1/(-2)

is negative, and

(\det S)^2 = 9

is positive.

❌ Not quite.

S^2 = SS

contributes

\det S

twice:

\det(S^2) = 9

Show solution

Apply the product rule to $S \cdot S \cdot T^{-1}$ and use $\det(T^{-1}) = 1/\det T$ :

\det(S^2T^{-1}) = (\det S)(\det S)\det(T^{-1}) = 3 \cdot 3 \cdot \frac{1}{-2} = -\frac{9}{2}.

Problem 2 · Reading Off Eigenvalues

Given: $T \in \mathcal{L}(\mathbf{R}^2)$ with matrix $\begin{pmatrix} 4 & -2 \\ 1 & 1 \end{pmatrix}$ in the standard basis. Find the eigenvalues of $T$ using $\det(\lambda I - T) = 0$ .

✅ Correct!

\det(\lambda I - T) = (\lambda - 4)(\lambda - 1) + 2 = (\lambda - 2)(\lambda - 3)

❌ Not quite. The diagonal entries are the eigenvalues only for a triangular matrix; here the off-diagonal entries matter.

❌ Check the sign. The zeros of

\lambda^2 - 5\lambda + 6

are positive.

❌ Check the off-diagonal product.

\lambda I - T

has off-diagonal entries

2

and

-1

, whose product

-2

is subtracted:

(\lambda - 4)(\lambda - 1) - (2)(-1)

Show solution

\lambda I - T = \begin{pmatrix} \lambda - 4 & 2 \\ -1 & \lambda - 1 \end{pmatrix}, \qquad \det(\lambda I - T) = (\lambda - 4)(\lambda - 1) - (2)(-1) = \lambda^2 - 5\lambda + 6.

This is $(\lambda - 2)(\lambda - 3)$ , which is $0$ exactly when $\lambda = 2$ or $\lambda = 3$ . Check: $T(1, 1) = (2, 2)$ and $T(2, 1) = (6, 3)$ .

Problem 3 · Undoing a Scalar

Given: $\dim V = 3$ , $T \in \mathcal{L}(V)$ , and $\det(2T) = 40$ . Find $\det(T^{-1})$ .

✅ Correct!

\det(2T) = \det(2I)\det T = 8\det T

, so

\det T = 5

and

\det(T^{-1}) = \tfrac{1}{5}

❌ Not quite.

40

\det(2T)

, not

\det T

; remove the factor coming from

2I

first.

❌ Not quite.

2T = (2I)T

and

\det(2I) = 2^3

on a

3

-dimensional space, not

2

❌ Almost.

5

\det T

; the inverse has the reciprocal determinant.

Show solution

Write $2T = (2I)T$ . The matrix of $2I$ in any basis is diagonal with entries $2, 2, 2$ , so $\det(2I) = 8$ and

40 = \det(2T) = \det(2I)\det T = 8 \det T \implies \det T = 5.

Since $\det T \neq 0$ , $T$ is invertible and $\det(T^{-1}) = 1/\det T = \tfrac{1}{5}$ .

Problem 4 · When Does a Cyclic Pattern Fail?

Given: $A_c = \begin{pmatrix} 1 & c & 0 \\ 0 & 1 & c \\ c & 0 & 1 \end{pmatrix}$ with $c \in \mathbf{R}$ , so $\det A_c = 1 + c^3$ . Find every real $c$ for which $A_c$ is not invertible.

✅ Correct!

1 + c^3 = (c + 1)(c^2 - c + 1)

, and

c^2 - c + 1 > 0

for every real

c

❌ Not quite.

A_0 = I

has determinant

1

, so it is invertible.

❌ Not quite.

1 + 1^3 = 2 \neq 0

; the cube keeps the sign of

c

❌ Not quite. A real cubic always has a real zero; try

c = -1

Show solution

A square matrix is invertible exactly when its determinant is nonzero, so we need the real zeros of $1 + c^3$ . Factor:

1 + c^3 = (c + 1)(c^2 - c + 1), \qquad c^2 - c + 1 = \left(c - \tfrac{1}{2}\right)^2 + \tfrac{3}{4} > 0.

So $A_c$ fails to be invertible only for $c = -1$ . Check: the rows of $A_{-1}$ sum to $(0, 0, 0)$ , so they are linearly dependent.

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