Friday, 27 June 2014

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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