Linear Algebra/Eigenvalues and eigenvectors

•One mathematical tool, which has applications not only for Linear Algebra but for differential equations, calculus, and many other areas, is the concept of eigenvalues and eigenvectors.

•Eigenvalues and eigenvectors are based upon a common behavior in linear systems.

An example.Let

and

What happens with x and y if they are transformed by A? Well,

But what is remarkable is that

• •So when we operate on the vector x with the matrix A, instead of getting a different vector (as we would normally do), we get the same vector x multiplied by some constant. And the same goes for vector y.

•We call the values 1 and -2 the eigenvalues of the matrix A, and the vectors x and y are called eigenvectors for the matrix A.

Definition

We now generalize this concept of when a matrix/vector product is the same as a product by a scalar as above:

essentially if we have a n×n matrix A, we seek solutions in v to find the eigenvectors, and solutions in λ to find the eigenvalues for the equation

Av=λv

Let us rearrange the equation

Av-λv=0

(A-λI)v=0 But (A-λI) is a matrix, so we are trying to solve Bv=0 where B=(A-λI), and this solution is merely the kernel of B, ker B. So the eigenvectors are in ker (A-λI), where λ is an eigenvalue

•Bv=0 has nonzero solution if |B| = det(B) is zero. So to find the eigenvalues, we let |A-λI|=0 and then solve for λ. •We will thus obtain a polynomial equation over the complex numbers (eigenvalues can be complex), known as the characteristic equation.

•The roots of the characteristic equation are the eigenvalues.

• we exclude 0 as an eigenvector, because it is trivially a solution to Av=λv and is not really interesting to consider.

• Additionally, if the zero vector were to be included, it would allow for an infinite number of eigenvalues, since any value of λ satisfies A0=λ0.

•we have an eigenvalue λ of a matrix A, together with a corresponding eigenvector, x, then any multiple of x is also an eigenvector for the same eigenvalue.

• To see that kx is also an eigenvector, follow this argument: If Ax=λx, then A(kx)=kAx=kλx=λ(kx). (Here k may be any scalar.) Thus, every multiple of an eigenvector is also an eigenvector.

ExampleLet's work through an example to show these ideas.

So what do we do if we want to find A14? Let's use the method we've just described.Find the eigenvalues:|A-λI|=0(3-λ)(-λ)-4=0λ2-3λ-4=0λ=-1, 4

Find the eigenvectors:

for λ=-1 

for λ=4 

The eigenvectors are then

so put the eigenvectors together to form the matrix P

Now -1 generated the eigenvector in the first column, and 4 generated the eigenvector in the second column, so form D in this way:

We can easily calculate (-1)14=1, so we get

and we have the fast method for creating inverses of 2×2 matrices:

So now we can now directly multiply out

Simplifying we get