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The Determinant of an Operator: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

A nonidentity projection of the plane has determinant zero and is not invertible

Example

Over $R$ , the projection $T (x, y) = (x, 0)$ has determinant $0$ and is not invertible.

Facts & Assumptions

Given: $T : R^{2} \to R^{2}$ , $T (x, y) = (x, 0)$ .

[F1]

$R$ is a field (The reals form a field).

[F2]

An operator determinant is the determinant of any representing matrix (The determinant of an endomorphism of a finite-dimensional vector space: its matrix determinant in an ordered basis in positive dimension, and $1$ on the zero space).

[L1]

A finite-dimensional operator over a field is invertible exactly when its determinant is nonzero (A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero).

Verification

technique · direct

1.1

In the standard ordered basis, $[T] = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ , so [F2] gives $\det (T) = 1 \cdot 0 - 0 \cdot 0 = 0$ .

F1F2algebra

1.2

Directly, $(0, 1)$ is a nonzero vector in the kernel and the image is ${(x, 0) : x \in R}$ , so $T$ is neither injective nor surjective.

givenalgebra

2.1

The invertibility criterion [L1] agrees with step 1.2 because the determinant computed in step 1.1 is zero.

step 1.1step 1.2L1∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

One operator has matrices $diag (2, 3)$ and $(\begin{array}{cc} 2 & 0 \\ 1 & 3 \end{array})$ in two bases, both with determinant $6$

Example

Let $T (x, y) = (2 x, 3 y)$ on $R^{2}$ . In the standard ordered basis $E$ , its matrix is $diag (2, 3)$ . In the ordered basis $B = ((1, 1), (0, 1))$ , its matrix is $(\begin{array}{cc} 2 & 0 \\ 1 & 3 \end{array})$ . Both determinants are $6$ .

Facts & Assumptions

Given: $T, E, B$ as in the example.

[F1]

$R$ is a field (The reals form a field).

[L1]

If $P = P_{E \leftarrow B}$ , then $[T]_{B} = P^{- 1} [T]_{E} P$ ( $[T]_{B^{'}}^{C^{'}} = P_{C^{'} \leftarrow C} [T]_{B}^{C} P_{B \leftarrow B^{'}}$ ).

[F2]

The $2 \times 2$ determinant is the two-term Leibniz sum (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L2]

The operator determinant is independent of the ordered basis (The determinant of a linear operator is independent of the chosen ordered basis).

Verification

technique · direct

1.1

Direct evaluation gives $[T]_{E} = A = (\begin{array}{cc} 2 & 0 \\ 0 & 3 \end{array})$ and $P = (\begin{array}{cc} 1 & 0 \\ 1 & 1 \end{array})$ , whose inverse is $P^{- 1} = (\begin{array}{cc} 1 & 0 \\ - 1 & 1 \end{array})$ .

F1givenalgebra

2.1

Matrix multiplication in [L1] gives $[T]_{B} = P^{- 1} A P = (\begin{array}{cc} 2 & 0 \\ 1 & 3 \end{array}) .$

step 1.1L1algebra

3.1

By the two-term determinant formula, $\det (A) = 2 \cdot 3 = 6$ and $\det ([T]_{B}) = 2 \cdot 3 - 0 \cdot 1 = 6$ .

step 2.1F2algebra

4.1

The explicit computations in steps 3.1 and 2.1 illustrate the equality asserted abstractly by [L2].

step 3.1step 2.1L2∎

ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

Multiplication by $2$ on $Z$ is injective but not surjective: its determinant is the non-unit $2$ , its adjugate is integral, and its inverse exists after extending scalars to $Q$

Example

The coordinate endomorphism $T_{[2]} : Z \to Z$ is injective but not surjective. Its determinant is the non-unit $2$ , its adjugate is $[1]$ , and after extending scalars to $Q$ its inverse is $[1 / 2]$ .

Facts & Assumptions

Given: The $1 \times 1$ matrix $A = [2]$ .

[F1]

$Z$ is a commutative ring (The integers form a commutative ring) and its only units are $1$ and $- 1$ ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

[L1]

Integer multiplication has cancellation (The integers have no zero divisors; multiplicative cancellation).

[F2]

$T_{A} (x) = A x$ , $\det (T_{A}) = \det (A)$ , and $adj (T_{A}) = T_{adj (A)}$ (For $A \in M_{n} (R)$ , the coordinate endomorphism $T_{A} : R^{n} \to R^{n}$ , with $\det (T_{A}) : = \det (A)$ and $adj (T_{A}) : = T_{adj (A)}$ ).

[L2]

A positive-sized square matrix over a commutative ring is invertible exactly when its determinant is a unit (A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit).

[L3]

If the determinant is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ (If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ).

[F3]

$Q$ is a field (The rationals form a field).

Verification

technique · direct

1.1

From the $1 \times 1$ definitions, $T_{A} (n) = 2 n$ , $\det (A) = 2$ , and $adj (A) = [1]$ , because the unique empty minor has determinant $1$ .

F1F2algebra

1.2

If $2 m = 2 n$ , integer cancellation gives $m = n$ , so $T_{A}$ is injective. It is not surjective because $2 n = 1$ has no integer solution.

L1algebra

2.1

The element $2$ is not a unit of $Z$ by [F1], so [L2] agrees that $A$ is not invertible over $Z$ .

step 1.1F1L2

2.2

Over the field $Q$ , $2$ is a unit. Formula [L3] and step 1.1 give $A^{- 1} = 2^{- 1} [1] = [1 / 2]$ .

F3L3step 1.1

3.1

Steps 1.1 through 2.2 establish every claim.

step 1.1step 1.2step 2.1step 2.2∎

False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

FALSE: a square matrix over a commutative ring is invertible if and only if its determinant is nonzero

Statement

False claim: for every commutative ring $R$ and every positive-sized square matrix $A$ over $R$ , the matrix $A$ is invertible if and only if $\det (A) \neq 0$ .

Facts & Assumptions

Given: The claimed equivalence over arbitrary commutative rings.

[F1]

A ring in the published convention may be the zero ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[F2]

[F3]

For a positive-sized square matrix $A = (a_{i j})$ over a commutative ring, $\det (A) = \sum_{σ \in S_{n}} sgn (σ) \prod_{i < n} a_{σ (i), i}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L1]

The correct general criterion is: a positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit (A positive-sized square matrix over a commutative ring is invertible if and only if its determinant is a unit).

Refutation

technique · direct

1.1

The reverse implication fails in $Z$ . By [F3], the unique term in the $1 \times 1$ determinant gives $\det ([2]) = 2$ . This determinant is nonzero but is not a unit by [F2], so [L1] shows that $[2]$ is not invertible.

F2F3L1algebra

1.2

The forward implication also fails under the stated ring convention. In the zero ring allowed by [F1], the $1 \times 1$ identity matrix is $[0]$ ; it is its own two-sided inverse, and [F3] gives its determinant as $0$ .

F1F3algebra

2.1

Thus both directions of the claimed equivalence can fail. Replacing "nonzero" by "a unit" gives the valid equivalence [L1], including for the zero ring.

step 1.1step 1.2L1∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

A sparse $4 \times 4$ determinant is $30$ by both cofactor expansion and row reduction

Example

Over $Z$ ,

$A = (\begin{array}{cccc} 1 & 2 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 4 & 8 & 2 & 1 \\ 0 & 0 & 0 & 5 \end{array})$

has determinant $30$ , both by cofactor expansion and by row reduction.

Facts & Assumptions

Given: The displayed matrix $A$ .

[F1]

$Z$ is a commutative ring (The integers form a commutative ring).

[L1]

Laplace expansion computes a determinant along every row and column (Laplace expansion computes the determinant along every row and every column over a commutative ring).

[L2]

Adding a multiple of one row to another leaves the determinant unchanged (For every square matrix, including singular ones, a row swap negates the determinant, scaling a row by any scalar scales it, and row addition leaves it unchanged).

[L3]

A triangular determinant is the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries).

Verification

technique · direct

1.1

Expand along the last row. Its only nonzero entry is $5$ in position $(3, 3)$ , with positive cofactor sign, so $\det (A) = 5 \det (\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 3 & 0 \\ 4 & 8 & 2 \end{array}) .$

F1L1algebra

1.2

Alternatively, replace row $2$ by row $2 - 4$ times row $0$ (the third row by the third row minus four times the first row). The result is upper triangular with diagonal $1, 3, 2, 5$ , and the operation does not change the determinant.

L2algebra

2.1

Expand the $3 \times 3$ determinant along its last column. The only nonzero entry there is $2$ , again with positive sign, giving $\det (A) = 5 \cdot 2 \det (\begin{array}{cc} 1 & 2 \\ 0 & 3 \end{array}) = 5 \cdot 2 \cdot 3 = 30$ .

step 1.1L1algebra

2.2

By [L3], the determinant in step 1.2 is $1 \cdot 3 \cdot 2 \cdot 5 = 30$ .

step 1.2L3algebra

3.1

Steps 2.1 and 2.2 give the same value by the two requested methods.

step 2.1step 2.2∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

The adjugate gives the inverse of a $3 \times 3$ rational matrix with determinant $3$

Example

For

$A = (\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{array}) \in M_{3} (Q),$

one has $\det (A) = 3$ and

$A^{- 1} = \frac{1}{3} (\begin{array}{ccc} 1 & - 2 & 2 \\ 1 & 1 & - 1 \\ - 1 & 2 & 1 \end{array}) .$

Facts & Assumptions

Given: The displayed matrix $A$ .

[F1]

$Q$ is a field (The rationals form a field).

[F2]

The cofactor matrix is $(C_{i j})$ and the adjugate is its transpose (Deleted-row-and-column minors, cofactors, the cofactor matrix and the adjugate over a commutative ring).

[L1]

$A adj (A) = \det (A) I$ (For every positive-sized square matrix over a commutative ring, $A adj (A) = adj (A) A = \det (A) I$ ).

[L2]

If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ (If $\det (A)$ is a unit, then $A^{- 1} = \det (A)^{- 1} adj (A)$ ).

Verification

technique · direct

1.1

Computing the nine signed $2 \times 2$ minors gives $cof (A) = (\begin{array}{ccc} 1 & 1 & - 1 \\ - 2 & 1 & 2 \\ 2 & - 1 & 1 \end{array}), adj (A) = (\begin{array}{ccc} 1 & - 2 & 2 \\ 1 & 1 & - 1 \\ - 1 & 2 & 1 \end{array}) .$

F2algebra

2.1

Direct multiplication gives $A adj (A) = 3 I_{3}$ . By [L1], this also confirms $\det (A) = 3$ .

step 1.1L1algebra

3.1

The scalar $3$ is a unit of $Q$ by [F1], so [L2] gives the displayed inverse. Multiplying it by $A$ on either side gives $I_{3}$ .

step 2.1F1L2algebra∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

Cramer's rule solves $2 x + y = 5$ and $x - y = 1$ as $(x, y) = (2, 1)$

Example

Cramer's rule solves $2 x + y = 5$ , $x - y = 1$ over $Q$ as $(x, y) = (2, 1)$ .

Facts & Assumptions

Given: $A = (\begin{array}{cc} 2 & 1 \\ 1 & - 1 \end{array})$ and $b = (\begin{array}{c} 5 \\ 1 \end{array})$ .

[F1]

$Q$ is a field (The rationals form a field).

[F2]

A $2 \times 2$ determinant is its two-term Leibniz sum (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[L1]

Over a field, a nonzero determinant gives the unique solution $x_{j} = \det (A_{j} (b)) / \det (A)$ (Over a field, for each right-hand side $b$ , $A x = b$ has a unique solution exactly when $\det (A)$ is nonzero, and then Cramer's quotient formula holds).

Verification

technique · direct

1.1

Direct calculation gives $\det (A) = 2 (- 1) - 1 = - 3 \neq 0$ .

F1F2algebra

1.2

Replacing the first and second columns gives $\det (\begin{array}{cc} 5 & 1 \\ 1 & - 1 \end{array}) = - 6, \det (\begin{array}{cc} 2 & 5 \\ 1 & 1 \end{array}) = - 3.$

F2algebra

2.1

By [L1], $x = (- 6) / (- 3) = 2$ and $y = (- 3) / (- 3) = 1$ .

step 1.1step 1.2L1

3.1

Substitution gives $2 (2) + 1 = 5$ and $2 - 1 = 1$ , verifying both equations and the claimed solution.

step 2.1algebra∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-13Open item page →

Determinant is additive in one selected column but not under simultaneous whole-matrix addition

Example

Over $Z$ , determinant is additive when one column varies and the other is fixed, but it is not additive as a function of the whole matrix.

Facts & Assumptions

Given: $u = (1, 0)^{T}$ , $v = (0, 1)^{T}$ , $w = (0, 1)^{T}$ , and $A = B = I_{2}$ .

[F1]

$Z$ is a commutative ring (The integers form a commutative ring) and the displayed columns and matrices belong to the corresponding matrix sets (Finite rectangular matrices over a commutative ring, their entries, rows and columns).

[L1]

Determinant is multilinear in its columns (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring).

[F2]

The $2 \times 2$ determinant is the two-term Leibniz sum (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

Verification

technique · direct

1.1

Direct calculation gives $\det [u + v, w] = 1, \det [u, w] + \det [v, w] = 1 + 0 = 1.$

F1F2algebra

1.2

For simultaneous whole-matrix addition, $A + B = 2 I_{2}$ , so $\det (A + B) = 4$ , whereas $\det (A) + \det (B) = 1 + 1 = 2$ .

F1F2algebra

2.1

This is the instance $\det [u + v, w] = \det [u, w] + \det [v, w]$ of columnwise multilinearity [L1], with the second column fixed.

step 1.1L1

3.1

Steps 2.1 and 1.2 isolate the distinction: additivity holds in a single selected column, but fails for the whole matrix argument.

step 2.1step 1.2∎

Sources

Standard references

Recommended treatments; not extraction sources.

The Determinant of an Operator: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A nonidentity projection of the plane has determinant zero and is not invertible

Example

Facts & Assumptions

Verification

One operator has matrices diag⁡(2,3)diag(2,3) and (2013)(21​03​) in two bases, both with determinant 66

Example

Facts & Assumptions

Verification

Multiplication by 22 on ZZ is injective but not surjective: its determinant is the non-unit 22, its adjugate is integral, and its inverse exists after extending scalars to QQ

Example

Facts & Assumptions

Verification

FALSE: a square matrix over a commutative ring is invertible if and only if its determinant is nonzero

Statement

Facts & Assumptions

Refutation

A sparse 4×44×4 determinant is 3030 by both cofactor expansion and row reduction

Example

Facts & Assumptions

Verification

The adjugate gives the inverse of a 3×33×3 rational matrix with determinant 33

Example

Facts & Assumptions

Verification

Cramer's rule solves 2x+y=52x+y=5 and x−y=1x−y=1 as (x,y)=(2,1)(x,y)=(2,1)

Example

Facts & Assumptions

Verification

Determinant is additive in one selected column but not under simultaneous whole-matrix addition

Example

Facts & Assumptions

Verification

Sources

Standard references

One operator has matrices $diag (2, 3)$ and $(\begin{array}{cc} 2 & 0 \\ 1 & 3 \end{array})$ in two bases, both with determinant $6$

Multiplication by $2$ on $Z$ is injective but not surjective: its determinant is the non-unit $2$ , its adjugate is integral, and its inverse exists after extending scalars to $Q$

A sparse $4 \times 4$ determinant is $30$ by both cofactor expansion and row reduction

The adjugate gives the inverse of a $3 \times 3$ rational matrix with determinant $3$

Cramer's rule solves $2 x + y = 5$ and $x - y = 1$ as $(x, y) = (2, 1)$