Invariants don't determine

Two objects sharing an invariant were treated as the same object.

Would you have fallen for these?

Each is a real option from a previous-year paper. Read it cold and decide before opening it.

• Option A

  A is diagonalisable

  Why this is wrong, and why it looks right

  False: the minimal polynomial $(x-1)(x-2)^{2}$ has a repeated factor $(x-2)^{2}$, so A has a genuine 2×2 Jordan block and cannot be diagonalized — matching eigenvalues to the characteristic polynomial isn't enough.

  From this question

  Let A be a 4×4 complex matrix with characteristic polynomial $(x - 1)^{2}(x - 2)^{2}$ and minimal polynomial $(x - 1)(x - 2)^{2}$. Which of the following are true?

  Part C · free

• Option D

  A is similar to diag(1, 1, 2, 2)

  Why this is wrong, and why it looks right

  False: from options (1)–(2), the Jordan form is diag(1,1) plus a single 2×2 block for eigenvalue 2, not four separate 1×1 blocks — so A is similar to diag(1,1)⊕$J_{2}(2)$, never to diag(1,1,2,2).

  From this question

  Let A be a 4×4 complex matrix with characteristic polynomial $(x - 1)^{2}(x - 2)^{2}$ and minimal polynomial $(x - 1)(x - 2)^{2}$. Which of the following are true?

  Part C · free

• Option D

  T is surjective.

  Why this is wrong, and why it looks right

  T = 0 is a linear map $\mathbb{R}^{5}\to\mathbb{R}^{3}$ that is about as far from surjective as possible. Having a larger domain than codomain permits surjectivity but never forces it.

  From this question

  Let $T : \mathbb{R}^{5} \to \mathbb{R}^{3}$ be a linear transformation. Which of the following must be true?

  Part C · free

• Option A

  The eigenspace of A for the eigenvalue 0 is two-dimensional.

  Why this is wrong, and why it looks right

  False — this confuses algebraic with geometric multiplicity. $T^{2}$ dividing p means 0 has algebraic multiplicity at least 2, but the eigenspace can still be one-dimensional if the Jordan block has size 2. SymPy takes A = [[0,1,0],[0,0,0],[0,0,1]], whose characteristic polynomial is $\lambda^{3} - \lambda^{2} = \lambda^{2}(\lambda - 1)$, and finds rank 2 — so the 0-eigenspace has dimension 3 − 2 = 1, not 2.

  From this question

  Let A be a 3×3 real matrix whose characteristic polynomial p(T) is divisible by $T^{2}$. Which of the following statements is true?

  See the full question

  Part B · june 2023 · free

• Option D

  A is diagonalizable.

  Why this is wrong, and why it looks right

  False, refuted by the same matrix: it has a size-2 Jordan block at 0, so it is not diagonalizable. Diagonalizability would need the geometric multiplicity of 0 to reach 2, which is exactly what (A) wrongly assumes.

  From this question

  Let A be a 3×3 real matrix whose characteristic polynomial p(T) is divisible by $T^{2}$. Which of the following statements is true?

  See the full question

  Part B · june 2023 · free

• Option B

  Suppose that for every vector $v \in \mathbb{R}^{5}$ there exists n such that $T^{n}v = 0$. Then $T^{2}$ must be zero.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option D

  rank(AB) = rank(BA)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option A

  Quadratic forms (a) and (b) are equivalent.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2023

• Option A

  If $G_{1}, G_{2}$ are two groups such that Aut$(G_{1})$ is isomorphic to Aut$(G_{2})$, then $G_{1}$ is isomorphic to $G_{2}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  X = {$e^{i\theta}$ | $\theta \in [-\pi/2, \pi]$}.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option B

  n = 4 and A has two distinct eigenvalues.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option C

  n = 5 and A has some eigenvalue for which the dimensions of the eigenspaces of A and B are the same.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  S is in bijection with the power set of S.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2023

• Option A

  There exists an injective $\mathbb{Q}-$linear transformation $T : \mathbb{R} \to \mathbb{Q}[x]$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option C

  The $\mathbb{Q}-$vector spaces $\mathbb{Q}[x]$ and $\mathbb{R}$ are isomorphic

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option A

  1

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option C

  The characteristic polynomial of A is $x^{3} - 1$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option C

  Two 4×4 matrices with the same characteristic and minimal polynomials

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option D

  The matrix with rows (0, 1, 0, 0), (1, 0, 0, 0), (0, 0, 0, 1), (0, 0, 1, 0)

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option B

  dim W ≥ 2.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option A

  (A) is true but (B) is false.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  Both (A) and (B) are false.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option A

  Both V and W are infinite-dimensional

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  V is infinite-dimensional and dim W = 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2024

• Option D

  𝔽$_{8}$ is isomorphic to $\mathbb{Z}/8\mathbb{Z}$ as a ring.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C

• Option C

  not diagonalisable

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  singular

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option B

  The minimal polynomial of A is $x^{2}(x - 1)(x - 2\pi{}i)^{3}$.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option C

  The dimensions of the eigenspaces for $0, 1, 2\pi{}i$ are 2, 1, 3 respectively.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2023

• Option A

  B(w, w) = 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2025

• Option D

  $\infty$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option D

  countably infinite

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · december 2023

• Option D

  The minimal polynomial of T is equal to its characteristic polynomial

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option A

  $\mu(A) = \mu(B), \sigma(A) = \sigma(B)$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part A · june 2024

• Option C

  W and W^⊥ have the same dimension over $\mathbb{R}$

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · june 2024

• Option A

  The set {$I, A, \dots, A^{n}$} is linearly independent but the set {$I, A, \dots, A^(n^{2})$} is not linearly independent.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option B

  If A is a singular matrix, then the set {I, A, …, A^k} spans a (k + 1)-dimensional subspace of V for all k ≤ rank(A).

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part C · december 2024

• Option D

  trace(A) = 0

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B

• Option B

  $\ker \varphi$ is finite dimensional.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

• Option D

  X is finite dimensional.

  Why this is wrong, and why it looks right

  The option is above and the answer is on the question page, both free. This is the part that is not: the reasoning that makes this particular wrong option feel like the right one.

  Part B · june 2025

The counterexamples that settle these

One example is enough to kill a false statement. These are the standard ones for the claims above.

• A subset of ℝ with measure zero is countable
• Same characteristic polynomial ⇒ similar
• Diagonalisable ⇒ invertible
• Real matrix with real eigenvalues is diagonalisable
• AB = I ⇒ BA = I for all matrices
• Every real symmetric matrix is positive definite if det > 0

Other mistakes that cost marks