Part CCSIR NET December 2023

Eigenvalues, characteristic & minimal polynomials, Cayley–Hamilton

Let A be an n × n real symmetric matrix. Which of the following statements are necessarily true?

A.A is diagonalizable.✓ Correct
B.If Aᵏ = I for some positive integer k, then $A^{2} = I$ .✓ Correct
C.If Aᵏ = 0 for some positive integer k, then $A^{2} = 0$ .✓ Correct
D.All eigenvalues of A are real.✓ Correct

The idea, in plain words

A symmetric matrix is just a stretch along a set of perpendicular directions, with a real stretch factor for each. Every claim here becomes a claim about those factors, and each one is easy once you ask what a single number would have to satisfy.

Solution

Spectral theorem: A = QDQᵀ with real D.

Aᵏ = I forces each real eigenvalue to satisfy $\lambda^{k} = 1$

so $\lambda = \pm1$ and $A^{2} = I; A^{k} = 0$ forces all $\lambda = 0$ so A = 0.

Why each option is right or wrong

Checked against 1 symbolic computation

A.Correct
True by the spectral theorem: a real symmetric matrix is orthogonally diagonalisable, A = QDQᵀ with Q orthogonal and D real diagonal.

No hypothesis about distinct eigenvalues is needed — symmetry alone suffices.
B.Correct
True.

Diagonalise as A = QDQᵀ; then Aᵏ = QDᵏQᵀ = I forces $\lambda^{k} = 1$ for every eigenvalue $\lambda$ .

The eigenvalues are real, and the only real k-th roots of unity are ±1

so $D^{2} = I$ and hence $A^{2} = I$ .

Over $\mathbb{C}$ this argument fails — complex roots of unity are available — which is why realness matters here.
C.Correct
True.

Aᵏ = QDᵏQᵀ = 0 forces $\lambda^{k} = 0$

so every eigenvalue is 0

so D = 0 and therefore A = 0 outright — not merely $A^{2} = 0$ .

A symmetric nilpotent matrix is the zero matrix; the usual nilpotent examples such as [[0,1],[0,0]] are not symmetric.
D.Correct
True.

For a real symmetric A and eigenpair Av $= \lambda{}v$ with v ≠ 0, v*Av is real

giving $\lambda = v$ Av / v $v \in \mathbb{R}$ .

SymPy confirms on [[1,2],[2,1]], whose eigenvalues are 3 and −1.

All four options here are correct — a Part C question with no false statement

which can make you doubt sound work.

Related counterexample: Same characteristic polynomial ⇒ similar

Eigenvalues, characteristic & minimal polynomials, Cayley–Hamilton

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