Friday, September 22, 2017

linear algebra characteristic polynomial, matrix rank, Matrix similarity


I'm having a problem solving the following assignment, can someone please help me?


I'm given 2 n×n matrices, n>1.


A=[1.....1..........1.....1]


B=[n0..0000.........0.....0]


1) I need to find the characteristic polynomial of A using A's Rank.


2) I need to prove that the Coefficient of tn1 in the characteristic polynomial of A is equal -(trA).


3) I need to prove that A and B are similar matrices and find P so that B=P1AP


*All of A's entries = 1.


Answer




A is symmetric, so the algebraic multiplicity of an eigenvalue is equal to the geometric multiplicity.


It is not hard to see that, for any x, Ax=c(1,1,,1)T for some constant c. Thus, its rank is 1 (corresponding to eigenvalue λ=...?) and the other n1 eigenvalues are 0. Such a matrix has characteristic polynomial


(tλ)(t0)n1=tn1(tλ)


For question 2, it is easy to directly calculate the trace, and you should now have the characteristic polynomial, so just verify.


To find a similarity transform, you can find all the eigenvectors (meaning, find n linearly independent eigenvectors) of A, or of B. One will be much easier than the other.


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