Always-True Synthesis Drills

No new theorem in this lesson — just the wiring. Rank, determinant, eigenvalues, invertibility: you already know each of them, but they’re all the same fact in disguise for a square matrix, and that’s exactly what GATE prods at year after year with the phrase “which of the following is/are ALWAYS true?”

Lean into that word always. A statement survives only if no counterexample exists, so the real skill being tested isn’t recall — it’s hunting edge cases. The same web pays off beyond the exam: it is exactly why a singular or rank-deficient feature matrix (collinear columns, det = 0) makes a linear-regression fit non-unique — the “invertible” spoke fails, so the others fail with it.

The invertibility web

For an n-by-n matrix A, the following are all equivalent — each one is true exactly when every other is true. Internalise this ring and most MSQ options answer themselves.

In words: A is invertible iff det(A) ≠ 0 iff rank(A) = n iff the columns are linearly independent iff 0 is not an eigenvalue iff Ax = 0 has only the trivial solution iff Ax = b has a unique solution for every b. Negate any one and you negate them all: a singular matrix has det = 0, rank < n, dependent columns, and 0 as an eigenvalue.

A few more always-true anchors to pair with the web:

• Rank-nullity: rank(A) + nullity(A) = n for any m-by-n matrix — always, no exceptions.
• Symmetric matrices (real) always have real eigenvalues and are diagonalisable.
• Determinant = product of eigenvalues; trace = sum of eigenvalues (counted with multiplicity), always.

The edge cases GATE weaponises

A statement is false the moment one example breaks it. Keep this small armoury of counterexamples loaded:

• Complex eigenvalues — a rotation [[0, −1], [1, 0]] is real but its eigenvalues are ±i. Kills “every real matrix has real eigenvalues.”
• Defective (non-diagonalisable) — [[1, 1], [0, 1]] has a repeated eigenvalue 1 but only one independent eigenvector, so it is not diagonalisable. Kills “every matrix / every matrix with repeated eigenvalues is diagonalisable.”
• Singular matrix — any A with det = 0 has 0 as an eigenvalue and rank < n. Kills “every matrix is invertible” and anchors the det = 0 ⇒ 0 is an eigenvalue link.
• Repeated eigenvalues that are fine — the identity I has eigenvalue 1 repeated yet is perfectly diagonalisable. So “repeated eigenvalue ⇒ not diagonalisable” is also false. Repetition alone decides nothing.

How GATE asks this

It is the MSQ “which of the following is/are always/necessarily true?” — typically four statements blending rank, eigenvalues, determinant, and invertibility, with partial marking so every box must be judged independently. Work option by option: either cite the equivalence web / an always-true anchor to confirm it, or produce one counterexample to reject it. Never accept a statement because it “usually” holds.

Worked example

A is a real n-by-n matrix. Which statements are always true? (I) If det(A) = 0 then 0 is an eigenvalue of A. (II) Every real matrix has real eigenvalues. (III) If A is symmetric, its eigenvalues are real. (IV) rank(A) + nullity(A) = n.

• (I) TRUE. det(A) = ∏ eigenvalues. If the product is 0, at least one eigenvalue is 0. Equivalently, det(A) = 0 means A is singular, and “0 is an eigenvalue” is one spoke of the invertibility web. No counterexample exists.
• (II) FALSE. The rotation [[0, −1], [1, 0]] is real with eigenvalues ±i. One counterexample is enough to sink an “every” claim.
• (III) TRUE. Real symmetric matrices have real eigenvalues (spectral theorem) — a standing always-true fact. The rotation in (II) is not symmetric, so it is no counterexample here.
• (IV) TRUE. Rank-nullity holds for every matrix: the dimensions of the column space and null space always sum to the number of columns n.

So the always-true statements are (I), (III), (IV); only (II) fails.

Quick check