Always-True Synthesis Drills

Picture an old house with a single master switch by the front door. Flip it on, and every light in every room comes up together; flip it off, and the whole house goes dark at once. You never walk room to room checking bulbs — one switch already tells you the state of all of them. A square matrix has a master switch exactly like that. The previous lesson left you with a question: faced with a cluttered list of claims about one matrix, how do you sort the watertight facts from the traps without grinding through each in turn? The answer is that switch.

That switch is invertibility, and almost everything you learned this chapter is wired to it. Rank, determinant, eigenvalues, independent columns — they look like separate facts, but for a square matrix they all turn on and off together, like the bulbs in that house. Knowing they move as one is the whole game, because GATE asks about it every single year with the same five words: “which of the following is/are always true?”

Lean on that word always. A claim survives only if no counterexample exists, so the skill being tested is not recall — it is the hunt for a single edge case that breaks the claim. And the same wiring matters past the exam: it is precisely why a feature matrix with collinear columns (det = 0, rank short of full) makes a linear-regression fit non-unique. The “invertible” bulb is off, so every bulb wired to it is off too.

The invertibility web

Let us lay the wiring out. For an n-by-n matrix A, the statements below are all equivalent — each one is true exactly when every other is true. Internalise this single ring and most MSQ options answer themselves.

Read the ring out loud once: 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. That is the master switch in full. Negate any one spoke and you negate them all at once: a singular matrix has det = 0, rank < n, dependent columns, and 0 sitting in its spectrum — the whole house goes dark together.

The web covers square matrices that are or are not invertible. A few more facts hold for every matrix, square or not, and these are the anchors you pair with the ring:

• Rank-nullity: rank(A) + nullity(A) = n for any m-by-n matrix — always, with no exceptions, invertible or not.
• Real symmetric matrices always have real eigenvalues and are always diagonalisable (the spectral theorem).
• Determinant = product of the eigenvalues; trace = sum of the eigenvalues (each counted with multiplicity) — always.

The edge cases GATE weaponises

Now the other half of the skill. A claim is false the instant one example breaks it, so GATE keeps a small set of awkward matrices on hand to ambush over-confident “always” answers. Load the same armoury yourself:

• Complex eigenvalues. The rotation [[0, −1], [1, 0]] is a perfectly real matrix, yet its eigenvalues are ±i. That single example kills “every real matrix has real eigenvalues.”
• Defective (non-diagonalisable). The shear [[1, 1], [0, 1]] has the eigenvalue 1 repeated but only one independent eigenvector, so it cannot be diagonalised. That kills “every matrix is diagonalisable” and “a repeated eigenvalue means non-diagonalisable” in one stroke — read on for the second half.
• A singular matrix. Any A with det = 0 carries 0 as an eigenvalue and has rank < n. It kills “every matrix is invertible” and anchors the det = 0 ⇒ 0 is an eigenvalue spoke of the web.
• A repeated eigenvalue that is perfectly fine. The identity I has the eigenvalue 1 repeated n times, yet it is as diagonalisable as a matrix can be. So “repeated eigenvalue ⇒ not diagonalisable” is also false. Repetition, on its own, decides nothing.

Notice how the shear and the identity pull in opposite directions: both have a repeated eigenvalue, one is defective and one is flawless. That is the trap GATE loves — a property that looks decisive but is not.

How GATE asks this

The format is the MSQ: “which of the following is/are always (or necessarily) true?” — usually four statements blending rank, eigenvalues, determinant, and invertibility, with partial marking so each box is scored on its own. Treat it the way it is built. Walk option by option, and for each one do exactly one of two things: cite a spoke of the web (or an always-true anchor) to confirm it, or produce one concrete counterexample to reject it. Never tick a box because the claim “usually” holds — “usually” is not “always,” and the partial marking will find the gap.

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.

Take them one at a time, confirming with the web or rejecting with a counterexample.

• (I) TRUE. det(A) equals the product of the eigenvalues. A product that comes to 0 forces at least one factor to be 0, so 0 is an eigenvalue. Said the other way, det(A) = 0 means A is singular, and “0 is an eigenvalue” is one spoke of the web — no counterexample can exist.
• (II) FALSE. Reach for the rotation: [[0, −1], [1, 0]] is real with eigenvalues ±i. One counterexample sinks any “every” claim, and this is the standard one.
• (III) TRUE. Real symmetric matrices have real eigenvalues by the spectral theorem — a standing always-true anchor. The rotation from (II) is not symmetric, so it is no threat here; the symmetry hypothesis is exactly what rules it out.
• (IV) TRUE. Rank-nullity holds for every matrix: the dimension of the column space and the dimension of the null space always add to the number of columns n. No exception exists to look for.

So the always-true statements are (I), (III), (IV); only (II) fails. Three confirmed by the web and its anchors, one rejected by a single rotation — that is the entire method.

A question to carry forward

That closes the linear-algebra chapter. Every tool in it — rank, determinant, eigenvalues, the whole invertibility web — lives in a world of perfectly straight objects: lines through the origin, flat planes, transformations that stretch space by a constant factor in each direction. The slope never changes; double the input and you double the output. But most relationships worth modelling are not straight. A cost that dips to a minimum and rises again, a probability that curves from 0 to 1, a loss surface with a valley floor — these all bend. Here is the thread into the next chapter: once the rule is a curve rather than a line, how do you pin down its slope at a single point, and how do you find where the curve bottoms out?

In one breath

• For a square A, the invertibility web is one master switch: A invertible ⇔ det ≠ 0 ⇔ rank = n ⇔ columns independent ⇔ 0 not an eigenvalue ⇔ Ax = 0 only trivially ⇔ Ax = b unique. Negate one spoke, you negate them all.
• Always-true anchors beyond the web: rank-nullity (rank + nullity = n, any matrix); real symmetric ⇒ real eigenvalues + diagonalisable; det = product and trace = sum of eigenvalues.
• “Always” means zero counterexamples — confirm a statement with the web/an anchor, or reject it with one concrete edge case.
• Keep the armoury loaded: rotation [[0,−1],[1,0]] (complex eigenvalues), shear [[1,1],[0,1]] (repeated eigenvalue, defective), identity I (repeated eigenvalue, diagonalisable), any singular matrix (0 eigenvalue, rank < n).
• In the MSQ, judge every option independently — partial marking rewards exactly the boxes you can confirm or reject, and “usually true” earns nothing.

Practice