Eigen-properties & Transforms

Finding eigenvalues is work. The payoff: once you have them for A, you get them for free for A², A⁻¹, A + 5I, and so on — no re-solving. GATE leans on this hard, usually as a multi-select (“which of the following are always true”). Knowing the handful of transform rules and the symmetric/triangular facts turns those questions into instant marks.

How eigenvalues transform

Start from A v = λ v. Watch what happens to that same eigenvector v under common operations:

• Powers: A² v = A(Av) = A(λv) = λ(Av) = λ²v. In general A^k has eigenvalue λ^k.
• Inverse: from Av = λv, multiply by A⁻¹ and divide by λ: A⁻¹v = (1/λ)v. So A⁻¹ has eigenvalue 1/λ (provided λ ≠ 0).
• Shift: (A + cI)v = Av + cv = λv + cv = (λ + c)v. So A + cI has eigenvalue λ + c.

The explorer makes the “direction is preserved” part concrete: an eigenline of A is also an eigenline of A² and A + cI — the line is fixed, only the scaling along it changes.

Three structural facts GATE loves

Beyond the transforms, three matrix shapes come with guaranteed eigenvalue properties — prime “always true” material:

• Symmetric real matrix (A = Aᵀ): all eigenvalues are real, and eigenvectors for distinct eigenvalues are orthogonal. (This is why covariance matrices, which are symmetric, always give real variance directions.)
• Triangular matrix: the eigenvalues are exactly the diagonal entries — read them straight off.
• Idempotent matrix (P² = P — applying it twice does nothing more than once): every eigenvalue is in {0, 1}. Proof: from Pv = λv, apply P again — P²v = λ²v, but P² = P so λ²v = λv, giving λ² = λ, i.e. λ(λ − 1) = 0. Projection matrices are the classic example.

How GATE asks this

Almost always a MSQ (“which of the following statements is/are always TRUE?”). The options mix the transform rules (A², A⁻¹, A + cI) with the symmetric/triangular/idempotent facts, often planting one false statement (e.g. “every matrix is diagonalizable”). GATE DA 2025 ran exactly this style around the condition A³ = A (worked below). Evaluate each option independently against the rules above — there is no shortcut, but no real computation either.

Worked example — a real GATE DA 2025 question

A real matrix A satisfies A³ = A. Among several statements, decide whether “A and A² have the same rank” is always true.

Work it through the eigenvalues. From Av = λv, applying A twice gives A³v = λ³v. But A³ = A, so λ³v = λv, hence:

λ³ = λ   →   λ(λ² − 1) = 0   →   λ ∈ {0, 1, −1}

So every eigenvalue is 0, 1, or −1. Now compare ranks via A², whose eigenvalues are λ²:

λ  ∈ { 0,  1, −1 }
λ² ∈ { 0,  1,  1 }

A non-zero eigenvalue of A (±1) stays non-zero in A² (becomes 1), and a zero eigenvalue stays zero. The count of non-zero eigenvalues — and hence the rank — is unchanged, so “A and A² have the same rank” is always true. (This is GATE DA 2025.)

A second quick drill on the transforms. Suppose A has eigenvalues 2 and 3:

A²        →  2² , 3²        =  4 , 9
A⁻¹       →  1/2 , 1/3
A + 5I    →  2 + 5 , 3 + 5  =  7 , 8

Each result keeps the same eigenvectors as A; only the eigenvalues move.

Quick check