If you find nothing out of this rapid review of linear algebra you should get this section. Without this section you will not be capable to do any of the differential equations work which is into this section.
Therefore let's start with the subsequent. If we multiply an n x n matrix with an n x 1 vector we will find a new n x 1 vector back. Conversely,
A?h= y?
What we want to know is if it is possible for the following to happen. Instead of just getting a brand new vector out of the multiplication is it possible instead to find the following,
A?h=l?h
Conversely is it possible, at least for specific λ and ?h, to contain matrix multiplication be similar as just multiplying the vector by a constant? Obviously, we probably wouldn't be talking regarding to this if the answer was no. Thus, it is possible for this to occur, however, it won't occur for just any value of λ or?h. If we do occur to have a λ and‾?h for that this works (and they will λ always come in pairs) so we call λ an eigen-value of A and ?h an eigenvector of A.
Thus, how do we go about find the eigen-values and eigenvectors for a matrix? Well firstly see that if ?h= 0 then (1) is intended for be true for any value of λ and therefore we are going to make the assumption that
‾?h≠ 0?,
With such out of the way let's rewrite (1) a little.
A?h - l?h = 0?
A?h - In l?h = 0?
(A - In l) ?h= 0?
Remember that before we factored out the ?hwe added in the appropriately sized identity matrix. It is equivalent to multiplying things with a one and so doesn't modify the value of anything.
Therefore, with this rewrite we notice that
(A - lIn) ?h = 0?
It s equivalent to eq.(1). So as to find the eigenvectors for a matrix we will require solving a homogeneous system. Recall the fact from the earlier section which we know that we will either contain exactly one solution (?h = 0?) or we will have infinitely several nonzero solutions. As we've already said as don't needs this means that we want the second case.
Knowing it will permit us to find the eigenvalues for a matrix. Recall from such fact which we will get the second case only if the matrix in the system is particular. Thus we will require determining the values of λ for that we get,
det (A - l I ) = 0
Once we have the eigen-values we can after that go back and find out the eigenvectors for each eigen-value.
To determine eigenvalues of a matrix all we require to do is solve a polynomial. It is generally not too bad provided we maintain n small. Similarly in above also we see that for an nxn matrix, matrix A, we will contain n eigenvalues if we comprise all repeated eigenvalues.
The usefulness of these facts will become apparent when we get back into differential equations as wherein work we will want linearly independent solutions.
Let's work a couple of illustrations now to check how we in fact go about finding eigen-values and eigenvectors.