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

The idea in words first: if A stretches some direction by λ and leaves it pointing the same way, then anything you build out of A — squaring it, inverting it, adding a multiple of the identity — still leaves that direction alone.

All that changes is the stretch factor, and it changes in the obvious way. Start from A v = λ v and watch that same eigenvector v survive each operation:

• 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.

Now the over-generalisation that costs marks. These rules are so easy that people quietly invent a fourth one — that the eigenvalues of A + B are the eigenvalues of A plus those of B. They are not.

Look at why the shift rule worked: cI scales every vector by c, so it shares all of A’s eigenvectors, and two stretches along the same line simply add. A general B has its own, different eigenvectors — hand it A’s eigenvector and it comes back pointing somewhere else, so there is nothing to add.

The same caution covers products: AB does not get λ_A · λ_B. Aᵏ is safe only because a matrix always shares eigenvectors with itself.

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.

The symmetric claim is worth watching happen once. Take A = [[2, 1], [1, 2]]: trace 4, determinant 3, so λ² − 4λ + 3 = 0 gives λ = 3 and λ = 1. At λ = 3 the surviving equation is −x + y = 0, so v₁ = (1, 1); at λ = 1 it is x + y = 0, so v₂ = (1, −1).

Their dot product is 1·1 + 1·(−1) = 0 — perpendicular, as promised. Symmetric matrices are also the one shape guaranteed to be diagonalisable: they always supply a full set of n independent eigenvectors, enough to write any direction in the space as a combination of them.

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 keeps 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