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Eigenvalues & Eigenvectors

Some directions a matrix only stretches: those are eigenvectors, and the stretch factors are eigenvalues. The backbone of GATE Linear Algebra.

9 min read Intermediate GATE DA Lesson 25 of 122

What you'll learn

  • The defining equation `A v = λ v` and why v must be non-zero
  • Eigenvalues solve the characteristic polynomial `det(A − λI) = 0`
  • Two shortcuts: trace(A) = sum of eigenvalues, det(A) = product of eigenvalues
  • Eigenvectors are non-unique — any non-zero scalar multiple is the same eigenvector

Before you start

Push a random arrow through a matrix and it usually comes out somewhere new — rotated and stretched. But hidden inside every square matrix are a few special directions that only get stretched, not turned. Those directions are eigenvectors, and the stretch factor along each is its eigenvalue λ.

It’s the same idea that powers covariance, PCA, PageRank and the stability of dynamical systems — and it shows up almost every year in GATE DA. Find the eigenvalues of a small matrix and you’ve nailed one of the most reliable marks in the paper.

The defining equation

A non-zero vector v is an eigenvector of A if applying A only scales it:

A v = λ v        (v ≠ 0)

The scalar λ says by how much: positive keeps the direction, negative flips it, zero collapses it. We insist v ≠ 0 because A·0 = λ·0 holds for every λ and tells us nothing.

A v = λ vrearrange →(A − λI) v = 0non-zero v exists only if det(A − λI) = 0
Moving everything to one side turns the eigenvalue search into a determinant equation.

Play with it. The fan of faint arrows shows where a matrix sends every input direction — most swing to a new angle, but the highlighted lines stay put and only scale by their eigenvalue. Drag the test vector onto one to feel a direction survive; switch to the rotation preset to watch the eigenlines vanish (those have complex eigenvalues — more on that below).

Finding eigenvalues: the characteristic polynomial

Rearrange A v = λ v into (A − λI) v = 0. A non-zero v can solve this only if the matrix A − λI is singular, i.e.

det(A − λI) = 0

This is the characteristic polynomial. For a 2×2 matrix it is always a quadratic, and it has a memorable form:

λ² − trace(A)·λ + det(A) = 0

Once you have the eigenvalues, plug each back into (A − λI)v = 0 and solve the linear system for the eigenvector direction v.

Two shortcuts you must memorise

For any n×n matrix with eigenvalues λ₁, λ₂, …, λₙ:

  • trace(A) = Σλᵢ — the sum of the diagonal equals the sum of the eigenvalues.
  • det(A) = Πλᵢ — the determinant equals the product of the eigenvalues.

These let you check answers instantly and often find the second eigenvalue once you know the first. They are the source of a large fraction of GATE marks on this topic.

2×2 matrix Aabcdtrace = a + d = λ₁ + λ₂det = ad − bc = λ₁ · λ₂
Sum of the diagonal = sum of eigenvalues; determinant = product of eigenvalues.

How GATE asks this

Nearly every year there is a NAT or MCQ that hands you a small matrix (2×2 or 3×3) and asks for an eigenvalue, the sum/product of eigenvalues, or which value is not an eigenvalue. The drill is fixed: write λ² − trace·λ + det = 0 for a 2×2, solve, and sanity-check with the trace/det shortcuts. GATE DA 2024 posed exactly this on the matrix [[2, −1], [3, 1]] — and the twist was that the eigenvalues turn out to be complex (worked below).

Worked example — a real GATE DA 2024 question

Find the eigenvalues of M = [[2, −1], [3, 1]].

Read off the trace and determinant, then build the characteristic quadratic:

trace(M) = 2 + 1 = 3
det(M)   = (2)(1) − (−1)(3) = 2 + 3 = 5

characteristic polynomial:  λ² − 3λ + 5 = 0

discriminant = (−3)² − 4·1·5 = 9 − 20 = −11  < 0

Because the discriminant is negative, the roots are not real:

λ = ( 3 ± √(−11) ) / 2 = (3 ± i√11) / 2     →  a complex conjugate pair

So this real matrix has no real eigenvalues — the answer is a complex conjugate pair. (This is GATE DA 2024.) Geometrically M is a rotate-and-scale, so no real direction survives — exactly the “rotation preset” state in the widget above.

Now a clean real case for contrast — a triangular matrix:

A = [[2, 1],            det(A − λI) = (2−λ)(3−λ) − (1)(0)
     [0, 3]]                        = (2−λ)(3−λ) = 0

                        →  λ = 2  or  λ = 3

For a triangular matrix (zeros below or above the diagonal) the eigenvalues are just the diagonal entries — read them straight off, no algebra needed. Check: trace = 2 + 3 = 5 = λ₁ + λ₂ ✓ and det = 6 = λ₁·λ₂ ✓.

Quick check

Quick check

0/5
Q1A 2×2 matrix has trace 7 and determinant 12. One eigenvalue is 3. What is the other eigenvalue?numerical answer — type a number
Q2Find the larger eigenvalue of A = [[4, 2], [1, 3]].numerical answer — type a number
Q3The matrix M = [[2, −1], [3, 1]] (the 2024 question). How many REAL eigenvalues does it have?numerical answer — type a number
Q4Which statements about eigenvalues are ALWAYS true? (select all that apply)select all that apply
Q5v is an eigenvector of A with eigenvalue 5. Which of the following are ALSO eigenvectors of A with eigenvalue 5? (select all that apply)select all that apply

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