Eigenvalues and Eigenvectors

Most columns are turned

Take A = (4 −1; 2 1). It sends the column (3; 1) to (4 × 3 + (−1) × 1; 2 × 3 + 1 × 1) = (11; 7). The arrow to (3, 1) climbs 1 for every 3 across, and the arrow to (11, 7) climbs 7 for every 11 across, so A has turned it to point in a different direction as well as making it longer. Most columns are turned like this.

The column v = (3, 1) and its image Av = (11, 7): the image points in a different direction.

Some columns are only stretched

Now try (1; 1). A sends it to (4 × 1 + (−1) × 1; 2 × 1 + 1 × 1) = (3; 3), which is 3 times (1; 1). The image points in exactly the same direction, three times as long.

Try (1; 2) as well. A sends it to (4 − 2; 2 + 2) = (2; 4), which is 2 times (1; 2). This one is doubled, and again its direction does not change.

A sends (1, 1) to (3, 3), along the same line, and (1, 2) to (2, 4), along the same line: each short arrow lies on its own image.

Eigenvectors and eigenvalues

A column that a matrix only stretches is called an eigenvector of the matrix, and the number it is stretched by is its eigenvalue. Written as an equation, v is an eigenvector with eigenvalue $\lambda$λ when $Av=\lambda v$A v = λ v and v is not the zero column. Along v, the matrix acts like multiplication by the single number $\lambda$λ.

So (1, 1) is an eigenvector of A with eigenvalue 3, and (1, 2) is an eigenvector with eigenvalue 2. The column (3, 1) is not an eigenvector, because (11, 7) is not a multiple of it: 11 is $\frac{11}{3}$11/3 times 3, but 7 is 7 times 1.

Here the matrix is (2 1; 1 2). Turn the column v and watch its image Av: it is turned away from v until v points along a line at 45° to the axes, where Av lines up with it. Along y = x the stretch is 3, and along y = −x it is 1.

An eigenvector names a direction

Double the eigenvector (1, 1): A sends (2; 2) to (8 − 2; 4 + 2) = (6; 6), which is 3 times (2; 2). It is an eigenvector too, with the same eigenvalue. This always happens. If $Av=\lambda v$A v = λ v and k is any number, then $A\left(kv\right)=k\left(Av\right)=k\left(\lambda v\right)=\lambda \left(kv\right)$A(k v) = k(A v) = k(λ v) = λ(k v), so every non-zero multiple of an eigenvector is an eigenvector with the same eigenvalue.

So an eigenvector names a whole direction, the line through the origin along it, and A sends that line onto itself. The eigen directions of a matrix are its invariant lines through the origin.

The zero column is left out on purpose. A times (0; 0) is (0; 0), which is $\lambda$λ times (0; 0) for every number $\lambda$λ, so it would make every number an eigenvalue and tell us nothing.

Negative and zero eigenvalues

The reflection in the x-axis, (1 0; 0 −1), leaves (1, 0) where it is, so (1, 0) is an eigenvector with eigenvalue 1. It sends (0, 1) to (0, −1), which is −1 times (0, 1), so (0, 1) is an eigenvector with eigenvalue −1. A negative eigenvalue keeps the line and reverses the direction along it.

The matrix C = (2 4; 1 2) has determinant 2 × 2 − 4 × 1 = 0. It sends (2; −1) to (4 − 4; 2 − 2) = (0; 0), which is 0 times (2; −1), so (2, −1) is an eigenvector with eigenvalue 0: that whole direction collapses to the origin. Its other eigenvector is (2, 1), which goes to (4 + 4; 2 + 2) = (8; 4), 4 times itself.

An eigenvalue of 0 always comes with det = 0, because a non-zero column sent to zero means the matrix cannot be undone. Some matrices have no real eigenvectors at all: the quarter turn (0 −1; 1 0) turns every direction through a right angle, so it keeps none of them.

Why eigenvectors are useful

Along an eigenvector, applying A again and again only multiplies by $\lambda$λ again and again. Applying A five times to (1, 1) gives $3^5=243$3⁵ = 243 times (1, 1), which is (243, 243).

Other columns can be built from the eigenvectors. The column (3, 1) is 5 times (1, 1) minus 2 times (1, 2), since 5 − 2 = 3 and 5 − 4 = 1. A stretches the first part by 3 and the second by 2, so A sends (3, 1) to 15 times (1, 1) minus 4 times (1, 2), which is (15 − 4, 15 − 8) = (11, 7), as before. Applying A again stretches the parts by 3 and 2 once more: 45 times (1, 1) minus 8 times (1, 2) is (37, 29), and multiplying A by (11; 7) directly gives (44 − 7; 22 + 7) = (37; 29).

The usual mistakes

Calling a column an eigenvector because its image is longer. (3, 1) is lengthened to (11, 7), but it is turned as well, so it is not an eigenvector.

Checking only one entry. (11, 7) against (3, 1) gives $\frac{11}{3}$11/3 in the first entry and 7 in the second; both entries must be multiplied by the same number.

Giving det A as the eigenvalue. det A = 4 × 1 − (−1) × 2 = 6, which is the product of the two eigenvalues, 3 × 2, not one of them.

Calling the zero column an eigenvector. It satisfies $Av=\lambda v$A v = λ v for every $\lambda$λ, which is why eigenvectors must be non-zero.

A town map and a logo

In the first application below, three roads run out from a town hall and a map is redrawn by a matrix. Each road is multiplied by the matrix, and a road that stays on its own line is an eigenvector, with its stretch factor as the eigenvalue. In the second, one entry of a matrix is chosen so that an arrow in a logo stays on its own line, and the second eigenvalue turns out to be negative.