Eigen-properties & Transforms

Finding eigenvalues is the work; here is the payoff the last lesson promised. Once you have them for A, you get them for free for A², for A⁻¹, for A + 5I — no re-solving — because the same eigenvector carries through and only the number λ changes. GATE leans on this hard, almost always as a “which of these is always true” multi-select, where knowing the handful of rules turns the whole question into instant marks.

How eigenvalues transform

Start from A v = λ v and watch what happens to that same eigenvector v:

• Powers: A²v = A(Av) = A(λv) = λ(Av) = λ²v. In general Aᵏ has eigenvalue λᵏ.
• Inverse: from Av = λv, apply A⁻¹ and divide by λ, giving A⁻¹v = (1/λ)v. So A⁻¹ has eigenvalue 1/λ (when 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” concrete: an eigenline of A is also an eigenline of A² and of A + cI. The line stays fixed; only the scaling along it changes.

Three shapes that come with guarantees

Beyond the transforms, three matrix shapes hand you eigenvalue facts outright — prime “always true” material, and the second answers the question the last lesson left open:

• Symmetric real (A = Aᵀ): all eigenvalues are real, and eigenvectors for distinct eigenvalues come out perpendicular. This is why a covariance matrix (symmetric) always yields real, orthogonal variance directions — the well-behaved case eigenvalues hinted at.
• Triangular: the eigenvalues are exactly the diagonal entries — read them off.
• Idempotent (P² = P — applying it twice does no more than once): every eigenvalue is in {0, 1}. 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 example.

A worked example — a real GATE DA 2025 question

A real matrix A satisfies A³ = A. Is “A and A² have the same rank” always true?

Push it through the eigenvalues. From Av = λv, applying A twice more gives A³v = λ³v; but A³ = A, so λ³v = λv, hence

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

Every eigenvalue is 0, 1, or −1. Now compare ranks through 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 stays zero — so the count of non-zero eigenvalues, and hence the rank, is unchanged. “A and A² have the same rank” is always true (GATE DA 2025). A quick transform drill to close: if A has eigenvalues 2 and 3, then A² has 4, 9; A⁻¹ has 1/2, 1/3; A + 5I has 7, 8 — each keeping A’s eigenvectors, only the eigenvalues moving.

A question to carry forward

A symmetric matrix was special because its eigenvectors come out perpendicular. That hints at another kind of special matrix — not one with perpendicular eigenvectors, but one whose action preserves perpendicularity and length, turning the whole space rigidly like a rotation. Here is the thread onward: what are those length-preserving matrices, and what tidy property does Aᵀ then have?

In one breath

• Transforms (same eigenvector, λ moves): Aᵏ → λᵏ, A⁻¹ → 1/λ (if λ≠0), A + cI → λ + c. So e.g. A² + 3I on eigenvalues 2,3 → 7, 12.
• Symmetric real (A=Aᵀ): eigenvalues real, eigenvectors orthogonal.
• Triangular: eigenvalues = diagonal entries.
• Idempotent (P²=P): eigenvalues only {0,1} — so trace = number of 1’s = rank.
• Traps: eigenvectors don’t change under these transforms; not every matrix is diagonalisable (defective repeated eigenvalues). 2025: A³=A → λ∈{0,1,−1} → A, A² same rank.

Practice