How do I find the eigenvalues of a 2xx22×2 matrix?

1 Answer
Dec 30, 2014

You can consider an Eigenvalue as a scaling factor when you apply a matrix AA to a vector vv (which is a "special" vector called Eigenvector).
In mathematical terms: A*v=lambda*vAv=λv (=the action of AA on vv is to SCALE of an amount lambdaλ the length of vv).
You can write:
A*v-lambda*v=0Avλv=0
(A-lambdaI)v=0(AλI)v=0

Where the identity matrix II was introduced to allow vv to be factorized out and make possible the subtraction between matrices: (A-lambdaI)(AλI).

If you have a look at the expression (A-lambdaI)v=0(AλI)v=0 you'll notice that it is simply the matrix equivalent of a system of homogeneous equations of the type:
3x+2y=03x+2y=0
-2x-y=02xy=0
This kind of systems may have a trivial solution (the origin, where all your lines cross) made up of all zeros or a non-trivial solution (an oo number of solutions, basically, all your lines are coincident). You want this last one (or at least one of the oo ones).

This non-trivial solution is obtained when:
det(A-lambdaI)=0det(AλI)=0 (have a look at Cramer's Rule).
If you solve this equation for lambdaλ you get all the possible Eigenvalues of your matrix AA!!!

Let us see an example:

Consider a 2xx22×2 matrix AA:
enter image source here

(hope it is not too confusing!!!)