btw each of the non zero eigenvalues have a corresponding eigenvector because geometric multiplicity is always less than or equal to algebraic multiplicity right?
Hmm, this is interesting, since T has 2 distinct eigenvalues in which I'll presume, furthermore since the other eigenvalues are all zero, is this enough to say that T is diagonalizable? Would that help in me in analyzing how T maps R?
is it fair to say that only the first 2 rows of T will be considered in the general case of finding the 2x2 principal submatricies since the rank of T is 2 meaning there are a lot of zero rows?
so when I do it this way I get the characteristic polynomial, p(λ)=-λ(λ2-15λ-18)
and that gives me the correct eigenvalues
the principal minors are
[5 6;8 9]
[1 3;7 9]
[1 2;4 5]
the sum of the principal minors give me -18, furthermore I see that the trace is b. but when I do...
I'm having problems when I'm calculating 4x4 matricies, my principal 3x3 minors are all 0
which makes sense since the rank of A is 2 so in this case rref(A) will have 2 rows of zeros which implies the determinant of A is 0.
Any further suggestions?
Robert:
So when I row reduced this matrix it looks something like this
[1 0 -1 -2 -3 ... 1-n;0 1 2 3 4 5 ... n-1;(remaining rows are 0)
which means that this matrix has rank 2
from here what should I do?
Also since the matrix T has rank 2, this means that there are non zero principal (2x2)...
Hi, I Like Serena!
Thanks for welcoming me!
Anyway, so far I think I was able to show that the rank of the matrix which is 2 I used a really rough method
so I think what this means for the characteristic polynomial, it's going to look something like this
p(t)=tn-2(t2-bt+c)
I'm kinda stuck...
Homework Statement
Let T be a square nxn matrix and let each entry tij=n(i-1)+j.
Calculate the eigenvalues of T of T when T is a 3x3 matrix and a 4x4 matrix.
Homework Equations
The Attempt at a Solution
For the homework I'm supposed to calculate the 3x3 case and the 4x4 case. Then it says...