Eigenpairs in 3D

The purpose of this web page is to explain finding eigenpairs of 3x3 matrices

Definitions: An eigenvector of a square matrix M is a non-zero vector v such that M*v=λ*v.
λ is called an eigenvalue of M and (λ,v) is an eigenpair of M.
M*v=λ*v means M*v=λ*I*v, where I is the identity matrix of the same dimension as $M.
That translates to (M-λ*I)*v=0, i.e. v is perpendicular to all rows of M-λ*I. That implies det(M-λ*I)=0.
det(M-λ*I)=0 is called the characteristic equation for M.
Step 1: Solve det(M-λ*I)=0.
If M is a 3x3 matrix, det(M-λ*I)=0 is a cubic equation, so solving it is not so easy. The best situation is if we already know one particular eigenvalue, say k. In this case we can divide the characteristic polynomial det(M-λ*I) by λ-k and find roots of the resulting quadratic polynomial.
For example, if det(M)=0, then λ=0 is an eigenvalue of M.
Step 2: Given an eigenvalue, find all corresponding eigenvectors.
If M is a 3x3 matrix, det(M-λ*I)=0 means that rows of M-λ*I are on the same plane, so to find the corresponding eigenvector v one chooses two non-parallel rows of M-λ*I, computes their vector product and then transposes it to get a column vector u. In that case all eigenvectors corresponding to λ are multiples k*u of u, where k is not zero scalar.
If all rows of M-λ*I are parallel, then any non-zero vector perpendicular to them is an eigenvector corresponding to λ.
Step 3: Checking if your answers are correct.
Matrix: M =
Candidate for eigenvalue of M: x =
Candidate for eigenvector of M: v = T

M*v-x*v = T



Basic Calculator for Linear Algebra

1st constant: a= 1st variable: x=
2nd constant: b= 2nd variable: y=
3rd constant: c= 3rd variable: z=



Result of the calculation: