An eigenvector is a nonzero vector that, when multiplied against a given square matrix, yields back itself times a multiple. This multiple is a scalar called an "eigenvalue." Finding eigenvalues and eigenvectors is necessary in solving problems in differential equations, such as quantum mechanics and thermodynamics, for example. You will need to understand matrix algebra and determinants to calculate them.

## Step 1

Find the eigenvalues for a square matrix A. An eigenvalue is a scalar and is symbolized by the Greek letter lambda, but for simplification, it is abbreviated to L. Then, for a nonzero vector x where Ax = Lx, x is called an eigenvalue of A. Eigenvalues are found by using the characteristic equation det (A -- LI) = 0. Det stands for the determinant, and I is the identity matrix.

## Step 2

Calculate the eigenvector for each eigenvalue. To do this, find an eigenspace E(L), which is the null space of the characteristic equation. The nonzero vectors of E(L) are the eigenvectors of A. These are found by plugging the eigenvectors back into the characteristic matrix and by finding a basis for A -- LI = 0.

## Step 3

Practice Steps 1 and 2 by studying the matrix in the image. Shown is a square 2 x 2 matrix.

## Step 4

Calculate the eigenvalues with the use of the characteristic equation. Det (A -- LI) = (1 -- L)(--4 -- L) -- 3*2 = L^2 + 3L -- 10 = 0, the characteristic polynomial. Factoring yields (L + 5)(L -- 2) = 0, or L1 = --5 and L2 = 2. These are the eigenvalues of the matrix.

## Step 5

Find the eigenvector for the eigenvalue L1 = --5 by calculating the null space. Do this by placing L = --5 in the characteristic matrix and finding the basis for A -- (--5)I = A + 5I = 0. The two equations are 6x + 3y = 0 and 2x + y = 0. Choosing the second since they are equivalent gives the solution 2x = --y. If x = 1, then y = --2, so v1 = (1,--2) is an eigenvector that spans the eigenspace of L1 = --5.

## Step 6

Find the eigenvector for the eigenvalue L = 2. The equation is x -- 3y = 0, or x = --3y. If x = 3, then y = --1, so v2 = (3,--1) is an eigenvector that spans the eigenspace of L = 2.

#### References

- Linear Algebra: A Modern Introduction; David Poole; 2006
- 3000 Solved Problems in Linear Algebra; Seymour Lipschutz; 1989

#### Photo Credits

- Lyudmil Antonov Lantonov