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EigenvectorsIntroduction.One of the mysteries of eigenvectors is "Why on Earth do we need them."I suppose that the same goes for vectors and matrices in general. Vectors and matrices provide an excellent shorthand for dealing with several variables at a time. In the 'Differential Equations' section you should see a mention of 'state variables' and 'state equations'. For example, instead of regarding ẍ=a
as a second order equation relating position to
acceleration, we can break it down into two
first-order equations with position x and velocity v
as our 'state variables' and writeẋ=v
and dv/dt = a
as two equations. Now as soon as we look at a system that has springs and viscous dampers added to it, we find that acceleration a depends on x and v. When we substitute for a we get an equation that can be expressed in matrix form as ẋ = A
x
where x is not just the position, but a vector (x, v)' And now we get to the subject of eigenvectors. If x can be expressed as an eigenvector of A, with eigenvalue λ, we can instead write ẋ = A
x
as ẋ = λ
x
and we know how to solve that one. What is an eigenvector?If we multiply a vector and a matrix, what do we get? We get another vector. For example
From the vector (1, 0)' we get (2, 1)' This new vector is not only a different 'size', it represents a different direction. Look at the demonstration here. Now
so from the vector (0, 1)' we get (1, 2)' - again in a new direction. Now we ask, "Are there any vectors that can be multiplied by the matrix
to give another vector in the same direction?" If we start with (x, y)', any other vector in the same direction will be (λx, λy)', where λ is some constant. We are looking for a vector x for which A x = λ
x
So for the special vector (x, y)' we are looking for,
Now if we multiply the unit vector by λ we get
and if we multiply this by (x, y)' we will get (λx, λy). So we have
We could write this as A x = λ
I x
and then rearrange it to give A x - λ
I x
= 0
or (A - λ I) x = 0
Now the vector 0 is not just a common-or-garden zero, it is (0, 0)' and you will recall that we could consider the matrix-vector product as a mixture of the columns of the matrix. So here, if the vector x is not 0, we have a combination of the columns of (A - λ I) that will give (0, 0)' Remember also, that to evaluate a determinant of a matrix you can first add multiples of columns to other columns of the matrix without changing the determinant's value. So we have a way to reduce a column of (A - λ I) to all zeros, and so its determinant must be zero. Now to add two matrices we just add the corresponding elements, one by one, to get
which we can expand as (2 - λ)(2 - λ)
- 1 = 0
or λ2 - 4 λ
+ 3 = 0
So we have not just one value for λ, but two, 1 and 3. If we substitute the value 3, we get
which is satisfied if x = (1, 1)' Let's try it:
so A x = 3 x, just as we hoped to find, and x is an eigenvector of A The value of λ is called an eigenvalue. Exercise 1Find the other eigenvector, corresponding to eigenvalue λ = 1.Does it match what you see on the demonstration? Exercise 2Now edit the matrix values in the demonstration to
What are the new eigenvalues? Calculate them by finding the 'characteristic equation' as described below. You will see from the demonstration that the eigenvectors are not far from each other. What are they? In generalIf the matrix A is n x n, the equation for λ will be nth order and there will be n roots. But the method is just the same:
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