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 tn−1 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=P−1AP
*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 n−1 eigenvalues are 0. Such a matrix has characteristic polynomial
(t−λ)(t−0)n−1=tn−1(t−λ)
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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