Linear-Algebra-Done-Right · Unit 29 · Video 2 · Interactive Practice

Similarity Invariance: One Determinant in Every Basis

IKey Formulas

Formula	Name	Hypotheses
$\det(S^{-1}TS) = \det T$	Similarity invariance	$T \in \mathcal{L}(V)$ , $S \colon W \to V$ invertible linear (here $W$ may differ from $V$ )
$\det T = \det \mathcal{M}\bigl(T, (v_1, \dots, v_n)\bigr)$	Determinant of an operator via a matrix	Any basis $v_1, \dots, v_n$ of $V$
$\det T = \lambda_1 \cdots \lambda_n$	Product of eigenvalues	$\mathbf{F} = \mathbf{C}$ ; each eigenvalue repeated as often as its multiplicity

Key Insight: When $W \neq V$ , $\det S$ is not defined, so $\det(S^{-1})\det T\det S$ cannot be cancelled; instead an alternating form $\tau$ on $W$ is carried to $V$ by $\alpha(v_1, \dots, v_n) = \tau(S^{-1}v_1, \dots, S^{-1}v_n)$ .

IIChange the Basis, Keep the Determinant

For $T$ with standard matrix $\begin{pmatrix} 0 & 1 \\ -3 & 4 \end{pmatrix}$ , every basis gives a different matrix with the same determinant $3$ .

IIIThe Determinant Is the Product of the Eigenvalues

Over $\mathbf{C}$ , the two eigenvalues of a $2$ -by- $2$ matrix multiply to $ad - bc$ , whether they are real or a conjugate pair.

a 1 b −2 c 1 d 3

IVQuiz Questions

Problem 1 · Counting Multiplicity

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ has eigenvalues $2$ , with multiplicity $2$ , and $-3$ , with multiplicity $1$ . Find $\det T$ .

✅ Correct!

\det T = 2 \cdot 2 \cdot (-3) = -12

❌ Not quite.

2 + 2 - 3 = 1

is the trace; the determinant is the product.

❌ Count multiplicity. The eigenvalue

2

appears twice on the diagonal of an upper-triangular matrix of

T

❌ Check the sign. One factor,

-3

, is negative.

Show solution

Over $\mathbf{C}$ there is a basis in which $\mathcal{M}(T)$ is upper triangular with diagonal $2, 2, -3$ (each eigenvalue as often as its multiplicity). The determinant of $T$ is the determinant of this matrix, the product of the diagonal entries:

\det T = 2 \cdot 2 \cdot (-3) = -12.

Problem 2 · Conjugation Cancels

Given: $3$ -by- $3$ matrices $A$ and $P$ with $\det A = 5$ and $\det P = 2$ . Find $\det(P^{-1}AP)$ .

✅ Correct!

\det(P^{-1})\det A\det P = \tfrac{1}{2} \cdot 5 \cdot 2 = 5

❌ Not quite. That is

\det(PAP)

; the inverse contributes

\det(P^{-1}) = \tfrac{1}{2}

, not

2

❌ Not quite. Only one factor is

P^{-1}

; the other is

P

, contributing

2

❌ Not quite. Both

P^{-1}

and

P

contribute; their factors

\tfrac{1}{2}

and

2

cancel.

Show solution

By the product rule and $\det(P^{-1}) = 1/\det P$ ,

\det(P^{-1}AP) = \det(P^{-1})\,\det A\,\det P = \tfrac{1}{2} \cdot 5 \cdot 2 = 5.

Similar matrices always have the same determinant: they are matrices of one operator in two bases.

Problem 3 · An Operator on Polynomials

Given: $T \in \mathcal{L}(\mathcal{P}_2(\mathbf{R}))$ defined by $(Tp)(x) = p(2x + 1)$ . Find $\det T$ .

✅ Correct! In the basis

1, x, x^2

the matrix is upper triangular with diagonal

1, 2, 4

❌ Not quite.

1 + 2 + 4 = 7

is the trace; multiply the diagonal entries.

❌ Not quite.

4

is only the

x^2

diagonal entry; the entries for

1

and

x

also count.

❌ Not quite.

T1 = 1

gives the first diagonal entry

1

, but

Tx = 2x + 1

and

Tx^2 = 4x^2 + 4x + 1

give

2

and

4

Show solution

Use the basis $1, x, x^2$ : $T1 = 1$ , $Tx = 2x + 1$ , $Tx^2 = (2x + 1)^2 = 4x^2 + 4x + 1$ . So

\mathcal{M}(T) = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 2 & 4 \\ 0 & 0 & 4 \end{pmatrix}, \qquad \det T = \det \mathcal{M}(T) = 1 \cdot 2 \cdot 4 = 8.

Any other basis gives a different matrix with the same determinant $8$ .

Problem 4 · A Conjugate Pair

Given: a $3$ -by- $3$ real matrix $A$ , viewed as an operator on $\mathbf{C}^3$ , has eigenvalues $3$ , $1 + i$ and $1 - i$ . Find $\det A$ .

✅ Correct!

3(1 + i)(1 - i) = 3 \cdot 2 = 6

❌ Not quite. The non-real eigenvalues count too; their product

(1 + i)(1 - i) = 2

is real.

❌ Not quite.

3 + (1 + i) + (1 - i) = 5

is the trace.

❌ Not quite.

2

is the product of the conjugate pair alone; include the eigenvalue

3

Show solution

Over $\mathbf{C}$ , the determinant is the product of the eigenvalues with multiplicity:

\det A = 3\,(1 + i)(1 - i) = 3\,(1 - i^2) = 3 \cdot 2 = 6.

Non-real eigenvalues of a real matrix come in conjugate pairs, and each pair contributes the positive real number $|1 + i|^2 = 2$ .

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