Thread: problem on Trace of the sum of two matrices

The following question is in the exercise of my text book.

A is 4x4 matrix with 1,-1,2 and -2 as its four eigen values. If B = A^4 - 5A^2 + 5I, which of the following is true?
a) det(A+B)= 0 b) trace (A-B) = 0 c) det(B) = 1 d) trace of (A+B) = 4.

I know for sure that trace of A is 0 as the sum of the eigen values is the same.
But i'm not sure how to arrive at the conclusion for A-B without a little more help in defining the matrix B.

Originally Posted by MAX09

A is 4x4 matrix with 1,-1,2 and -2 as its four eigen values. If B = A^4 - 5A^2 + 5I, which of the following is true? a) det(A+B)= 0 b) trace (A-B) = 0 c) det(B) = 1 d) trace of (A+B) = 4.

is diagonalizable and there exists invertible such that with . Then,



Using that similar matrices have the same trace, you'll obtain .

Originally Posted by FernandoRevilla

is diagonalizable and there exists invertible such that

I'm a bit confused here. You have a matrix P such that A = PDP^{-1} and D = PDP^{-1}. Which means A = D and P is the identity.

Is this a typo or am I missing something?

-Dan

Originally Posted by topsquark

Is this a typo or am I missing something?

A typo, thanks. I've just changed it.

Since the only information you are given is the eigenvalues of the matrix, any matrix having those eigenvalues must give the same answer. So I would take, say, .

Then it is very easy to find B!

Since the only information you are given is the eigenvalues of the matrix, any matrix having those eigenvalues must give the same answer. So I would take, say, .

Then it is very easy to find B. (The fact that the characteristic equation of A is guarentees that!)

@hallsofIvy

I actually tried the same method, but for some stupid reason, I made a calculation error and ended up getting the wrong answer for A^4 i got a diagonal matrix with {1, 1, 64, 64} which was the most stupid mistakes one could ever make. I retried the same procedure after i got your suggestion and VOILA!! there it is!!

Thanks HallsofIvy

@FernandoRevilla

Your solution was as elegant as it can ever get!!! Thanks for opening my mind to new ideas!!! I'm working on the solution you had suggested! THanks...