Given
What are the representation of w.r.t. the basis , and the basis respectively?
Answer
We are given the matrix and we want to find its repesentation under new basis, say
Let the new represntation Matrix be called .
The ith column of is the reprsentation of the base under
Now, the vector itself is independent of its representation under different coordinates systems. Hence we can write
Hence let representation under Q be the column , and let representation under be column , then the above equation can be written as
| (1) |
Now we solve for in the above equation since we know and and .
This will give us one column of . We do the above again for each vector , and at the end we have
Hence the algorithm is
input: Old basis , new basis matrix , vector which is the representation of under A
output: column which is the representation of under by solving equation (1).
Collect all these to build up the matrix
Hence we apply the algorithm for each of these vectors, one at a time.
First we build the matrix (the new basis), this is done by just calculating , this reults in
Hence to find the representation of under , we solve for in equation (1). We do this for each , and as follows
Solving for the last equation above is shown here.
Hence we get these 4 equations
C-D
| (E) |
B-E
| (F) |
A-F
Sub. into F we get , sub into E we get , sub into D we get
Hence
Hence , the form that the matrix takes w.r.t the basis is
| (2) |
Now that we have found the representation of under we are given a new basis and asked again find representation of under
Clearly the problem is not asking us to do this computation all over again for the new basis, becuase the algorithm would be the same and we can just repeat all the above steps again but using the new basis. So we need to try to find a short cut solution.
Start by finding the charactertic polynomial for
We know that a matrix companion form can be written directly if we know the matrix charaterstic equation.
If then the matrix companion form can be written as (see page 55 in text)
| (4) |
Hence by comparing (4) and (3) we see that , hence the compaion form for is
| (5) |
Compare matrix (5) and (2) we see they are the same. (I do not think I needed to show that (5) and (2) are the same here, this is just a confirmation).
Using Cayley-Hamilton, then
Hence
| (6) |
If we multiply the above by we get
If we multiply (6) by any other vector, say we also get
But we found that the representation of under is
Hence the representation of under is also
But with the help of the cayley-hamilton theorm, we see that the coeff. of these polynomials (i.e. the values in the last column of ) are the same if the basis were or
Hence the representation of under is the same as the representation of under
Find jordan form of the following matrices
For ,
Hence 3 distinct roots. Hence Jordan form is the standard diagoonal form
For
Try factor out, then we have , hence hence . But roots of the quadratic are
Hence 3 distinct roots, hence Jordan form is
For
Hence . The eigenvalues are not all distinct.
For , we have the matrix which has rank of 1, hence nullity is , hence we can find 2 independent vectors in the null space of the matrix , so we get 2 jordan blocks for of order 1. Hence the Jordan form is or
One can also solve the above as follows:
The null space of is 2, hence we can find 2 L.I. vectors in this null space as follows. Let be the first eigenvector associated with eigenvalue .
Hence , and so and can take any values. Hence
Now is the second eigenvector associated with . It will have similar solution to the above, but since it is L.I. to , we now pick and , hence or we could have found as follows: write hence hence , then and any value, say 1. Hence as above.
Now to find . This is the eigevector associated with . Hence
Hence and is any value, say 1 hence hence
Note that this is L.I. to both and
Hence hence
Which matches earlier solution.
Hence hence we have multipicity of 3.
Matrix since Rank of A is 2, hence null space is 1, hence we find one eigenvector
But this gives hence can not use.
Matrix
This has rank =1, hence nullity=2, so we still need one more eigevector. keep going.
Matrix this has rank=0 and nullspace =3, hence we can find the last eigenvector from this null space.
So pick vector such that the above is true and also that
Try hence we pick
So now we write
And we finally write
Hence hence
Hence jordan form is