Let A and B be n x n matrices. Show that: (a) If "w" is a nonzero eigenvalue of AB, then it is also an eigenva

If w is an eigenvalue of AB, then there is a non-zero eigenvector x such that ABx = wx.

If w is not zero then y = Bx is a non-zero vector, because otherwise Ay = ABx = 0.

The vector y satisfies:

BAy = (BA)Bx = B(ABx) = B(wx) = wBx = wy, so w is an eigenvalue of BA. That proves (a).

For (b), the above proof doesn't work because Bx could be the zero vector, so you argue with determinants. If 0 is an eigenvalue of AB, then det(AB) = 0 but det(AB) = det(A)det(B) = det(BA), so det(BA) = 0 and so BA has a zero as an eigenvalue.

(I am using the fact that 0 is an eigenvalue of an nxn matrix if and only if the determinant of that matrix is 0!)

section a is ordinary. this is the final concept. If A in an m-with the aid of-n matrix and B is an n-with the aid of-m matrix the place m ? n then the function polynomial of BA is on the subject remember of the function polynomial of AB contained in here way: p_BA (?)=?^(n-m) p_AB (?) with the aid of fact they share exceptionally lots a similar function polynomial, they'd have a similar eigenvalues. with the aid of fact AB and BA would be sq. matrices of distinctive dimensions the greater desirable product could have some greater zeros as eigenvalues. on your concern, you have 2 sq. matrices, so it is even much less complicated. they have precisely a similar function polynomial and hence a similar eigenvalues. in case you should tutor section a, try this: Take here identity the place the matrices are partitioned into here blocks, the sizes of that are set above, { {AB, 0}, {B, 0} } * { {I, A}, {0, I} } = { {AB, ABA}, {B, BA} } = { {I, A}, {0, I} } * { {0, 0}, {B, BA} } the 2d matrix, enable’s call it S, is invertible with the aid of fact its determinant is a million (i.e. it’s nonzero). S = { {I, A}, {0, I} } After multiplying the 1st and third blocks with the aid of S^(-a million), one gets { {AB, 0}, {B, 0} } = S * { {0, 0}, {B, BA} } * S^(-a million) revealing that those are comparable matrices and hence ought to share a similar eigenvalues. those are the eigenvalues of AB plus a team of zeroes which must be a similar with the aid of fact the eigenvalues of BA plus a team of zeroes.