Eigenvalues & Eigenvectors

The last lesson ended on a hint: most directions a matrix shoves off to a new angle, but a few it leaves sitting on their own line. Those special directions are eigenvectors, and the factor by which each one is stretched is its eigenvalue λ. Push a random arrow through a matrix and it comes out rotated and stretched; push an eigenvector through and it comes out pointing the very same way, only longer or shorter. That single idea powers covariance, PCA, PageRank, and the stability of dynamical systems — and it shows up almost every year in GATE DA.

The defining equation

A non-zero vector v is an eigenvector of A when 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, a small λ shrinks it.

Two zeros live in that equation and only one of them is banned, which is the part most people have to read twice. The eigenvector may never be zero; the eigenvalue is perfectly allowed to be. We exclude v = 0 because A·0 = λ·0 holds for every λ at once — it would make every number an eigenvalue and the definition would say nothing. But λ = 0 is a genuine, informative answer: it says A flattens that entire direction onto the origin. Zero eigenvector, meaningless; zero eigenvalue, meaningful.

Play with it below. The fan of faint arrows shows where the 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 and the eigenlines vanish — those eigenvalues are complex, more on that shortly.

Finding them — the characteristic polynomial

Rearrange A v = λ v into (A − λI) v = 0. A non-zero v can solve this only if A − λI is singular — has no inverse, because it crushes some non-zero vector to zero — and singular, from the determinant lesson, means

det(A − λI) = 0

So the search for eigenvalues is really a search for the values of λ that make that determinant collapse. Expand the left-hand side and you get a polynomial in λ, called the characteristic polynomial; its roots are exactly the eigenvalues. For a 2×2 matrix it is always a quadratic, and it has a form worth memorising on sight:

λ² − trace(A)·λ + det(A) = 0

The trace here is simply the sum of the diagonal entries — a + d for a 2×2.

Once you have the eigenvalues, plug each back into (A − λI)v = 0 and solve the linear system for the eigenvector direction v. That last step is worth watching once. For A = [[2, 2], [1, 3]] the quadratic is λ² − 5λ + 4 = 0, so λ = 1 or λ = 4. At λ = 4, A − 4I = [[−2, 2], [1, −1]], and both rows say the same thing, y = x, so v = (1, 1). Check: A(1,1) = (2+2, 1+3) = (4, 4) = 4·(1,1) ✓.

Two shortcuts ride along, and they earn a large share of the marks on this topic — for any n×n matrix with eigenvalues λ₁, …, λₙ:

So trace(A) = Σλᵢ and det(A) = Πλᵢ. They let you check an answer instantly, and often find the second eigenvalue the moment you know the first.

A worked example — a real GATE DA 2024 question

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

Read off the trace and determinant, build the quadratic, check the discriminant:

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

λ² − 3λ + 5 = 0      discriminant = (−3)² − 4·5 = 9 − 20 = −11 < 0

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

The negative discriminant means no real eigenvalues — a complex conjugate pair (this is GATE DA 2024). Geometrically M rotates as well as scales, so no real direction survives, exactly the rotation preset in the widget. For contrast, a clean real case — a triangular matrix gives its eigenvalues for free:

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

For any triangular matrix (zeros below or above the diagonal) the eigenvalues are just the diagonal entries — no algebra. Check: trace = 5 = λ₁+λ₂ ✓, det = 6 = λ₁·λ₂ ✓.

A question to carry forward

Eigenvalues are not just a number to find — they are a handle on the matrix itself. Here is the thread onward: if you know the eigenvalues of A, what are the eigenvalues of A², of A⁻¹, of A + 5I? And what makes a symmetric matrix so much better behaved than a general one that its eigenvectors come out perpendicular?

In one breath

• Eigenvector v ≠ 0: a direction the matrix only stretches, A v = λ v; the eigenvalue λ is the stretch factor.
• Find via det(A − λI) = 0; for a 2×2, λ² − trace·λ + det = 0. Then solve (A−λI)v = 0 for the direction.
• Shortcuts: trace = Σλ (sum), det = Πλ (product) — check answers, or get the second eigenvalue from the first.
• Triangular matrix → eigenvalues are the diagonal entries.
• Traps: a real matrix can have a complex conjugate pair (trace²−4det < 0); eigenvectors are non-unique (any non-zero scalar multiple).

Practice