Skip to content
Part BCSIR NET June 2023algebraic-vs-geometric-multiplicity

Algebraic vs geometric multiplicity

Let A be a 3×3 real matrix whose characteristic polynomial p(T) is divisible by T2T^{2}. Which of the following statements is true?

  1. A.The eigenspace of A for the eigenvalue 0 is two-dimensional.
  2. B.All the eigenvalues of A are real.✓ Correct
  3. C.A3=0A^{3} = 0.
  4. D.A is diagonalizable.

The idea, in plain words

A real cubic that is divisible by a square factor has a repeated root at zero and one remaining root, and that remaining one must be real because complex roots come in pairs. What stays open is whether the repeated root brings one independent direction or two.

Solution

p(T)=T2(T−c)p(T) = T^{2}(T - c) with c real (degree-3 real polynomial).

So all eigenvalues are real.

The geometric multiplicity of 0 may be 1 (Jordan block)

so (1), (4) fail; A need not be nilpotent (c ≠ 0).

Why each option is right or wrong

Checked against 3 symbolic computations
  • A.WrongInvariants don't determine

    False — this confuses algebraic with geometric multiplicity.

    T2T^{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 λ3−λ2=λ2(λ−1)\lambda^{3} - \lambda^{2} = \lambda^{2}(\lambda - 1), and finds rank 2 — so the 0-eigenspace has dimension 3 − 2 = 1, not 2.

  • B.Correct

    True. p is a real cubic, and T2T^{2} divides it

    so p(T)=T2(T−c)p(T) = T^{2}(T - c) for some real c — the quotient of a real polynomial by T2T^{2} is real.

    Hence the eigenvalues are 0, 0 and c, all real.

    The trigger: once two roots are pinned to 0, the third is forced real because complex roots of a real polynomial come in conjugate pairs, and there is only one root left.

  • C.WrongHypothesis dropped

    False unless c = 0

    which is not given.

    With the same example A = [[0,1,0],[0,0,0],[0,0,1]], SymPy shows A3A^{3} has rank 1 — non-zero.

    A is nilpotent only when the whole characteristic polynomial is T3T^{3}, and T2T^{2} dividing p permits a non-zero third eigenvalue.

  • D.WrongInvariants don't determine

    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.

3 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Invariants don't determine

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

Drill statements like this

Related counterexample: Same characteristic polynomial ⇒ similar

More on this topic

The chapter behind this: Characteristic vs minimal polynomial — what each tells you — free to read

From Eigenvalues and Canonical Forms › Eigenvalues, characteristic & minimal polynomials, Cayley–Hamilton

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram