✨Eigenvalues & Eigenvectors

Lens-keeper Ash of the Eigen Spire has one job, and she describes it the same way every time: find the directions the lens leaves alone.

Push an arrow through a matrix and it normally comes out pointing somewhere else. Take A = [[2, 1], [1, 2]] and the arrow (1, 0). Out comes (2, 1) — longer, and tilted. Now try (1, 1):

A(1, 1) = (2*1 + 1*1, 1*1 + 2*1) = (3, 3) = 3 * (1, 1)

Same line. Same direction. Three times as long. An arrow that behaves like that is an eigenvector of A, and the factor it is multiplied by — here 3 — is its eigenvalue. Written out, the whole idea is one short equation:

Av = (lambda)v, with v not the zero vector

The word "lambda" is just the name every book gives that factor; there is no Greek on the isle, so it is written out in words in every question here.

Three things are worth saying straight away. First, an eigenvalue can be negative: the arrow comes back along the same line but pointing the other way. Second, it can be zero: Av is the zero vector, which happens exactly when A is not invertible. Third, an eigenvector is a direction, not a length — if v works, so does 2v, and so does -v. That is why questions ask for the eigenvector "scaled to the smallest whole entries with a positive first entry": one arrow per direction, so there is one answer to type.

The zero vector is deliberately not allowed. A(0, 0) = (0, 0) = anything times (0, 0), so it would be an eigenvector for every number at once and would tell you nothing.

Testing arrows one at a time is no way to find them. Rearrange instead:

Av = (lambda)v => Av - (lambda)v = 0 => (A - (lambda)I)v = 0

The I is the identity matrix, and it is there because you cannot subtract a plain number from a matrix — lambda has to be spread down the diagonal first.

Now the key move. That equation says the matrix A - (lambda)I sends a non-zero arrow to zero. A matrix that squashes something non-zero to nothing cannot be invertible, and a 2 by 2 matrix is not invertible exactly when its determinant is 0. So:

det(A - (lambda)I) = 0

For A = [[a, b], [c, d]] this expands to

(a - lambda)(d - lambda) - bc = lambda^2 - (a + d)lambda + (ad - bc) = 0

and both coefficients are things you can read off the matrix without any work at all:

lambda^2 - (trace A)lambda + (det A) = 0

The trace is the two diagonal entries added; the determinant is the usual ad - bc. That quadratic is the characteristic equation, and its roots are the eigenvalues.

The minus sign in front of the trace is where most slips live. Subtracting lambda from each diagonal entry is what puts it there, so if your lambda term comes out with the trace's own sign, that subtraction went the wrong way.

Try it on A = [[4, 1], [2, 3]]: trace 7, determinant 12 - 2 = 10, so lambda^2 - 7lambda + 10 = 0, which factors as (lambda - 2)(lambda - 5) = 0. The eigenvalues are 2 and 5.

Each eigenvalue comes with its own line of arrows, and you find it by going back to (A - (lambda)I)v = 0 with a number in place of lambda.

For A = [[4, 1], [2, 3]] and lambda = 5:

A - 5I = [[4 - 5, 1], [2, 3 - 5]] = [[-1, 1], [2, -2]]

Look at what happened: the second row is -2 times the first. That is not luck. The determinant of A - (lambda)I is 0 by construction, so the two rows always say the same thing, and one row is all you need. Ignore the other one.

The first row reads -x + y = 0, so y = x, and every arrow on the line through (1, 1) works. In general a row reading

px + qy = 0

is solved by (q, -p): swap the two numbers over and change the sign of one of them. Forgetting that sign change is the single most common slip here, and it is worth checking your answer by pushing it back through A — that takes ten seconds and settles it completely.

If the two rows ever come out saying different things, the eigenvalue was wrong; go back to the characteristic equation.

Four facts save a great deal of arithmetic, and questions on this craft lean on all of them.

The trace is the sum of the eigenvalues, and the determinant is their product. Multiply the factored form out and compare:

(lambda - r)(lambda - s) = lambda^2 - (r + s)lambda + rs

so r + s = trace and rs = determinant. That is a free check on every answer, and it is often a faster route than the quadratic: knowing det A = 12 and that one eigenvalue is 3, the other must be 4, so the trace is 7.

A triangular matrix wears its eigenvalues on the diagonal. If everything below the diagonal is 0, then A - (lambda)I is triangular too, and the determinant of a triangular matrix is the product straight down it. So det(A - (lambda)I) = (a - lambda)(d - lambda), which is 0 exactly at the diagonal entries. No working at all — and it is true for 3 by 3 and larger just the same.

Powers, inverses and shifts keep the eigenvectors and move the eigenvalues. Start from Av = (lambda)v and apply the same v to each:

• ◆A² v = A((lambda)v) = lambda(Av) = (lambda^2)v — the eigenvalues get squared
• ◆A^-1 v = (1/lambda)v — the eigenvalues get turned upside down (which needs lambda not to be 0)
• ◆(A + kI)v = (lambda + k)v — every eigenvalue shifts by k, and none of them is scaled

Every eigenvalue in this kingdom is a whole number, because the matrices are built that way. If a surd appears, an arithmetic slip appeared first — recompute the trace and the determinant before you reach for the quadratic formula.