Let A and B be two similar matrices Thus there is a matrix P

Let A and B be two similar matrices. Thus, there is a matrix P such that A = P^-l BP. Show that A and B have the same eigenvalues with the same multiplicities. If W is the eigenspace of A corresponding to A, what is the eigenspace of B corresponding to A?

Solution

a)

Similar matrices have same determinant and hence same characteristic Polynomial. Having same chacteristic polynomial means having same eigenvalues with same multiplicity .

Proof :--

As A and B are similar matrices , an invertible matrix P such that B = P^-1AP .

Now B-x.I = P^-1AP -x.I = P^-1AP - P^-1.xI.P ........................ ( as P^-1.xI.P = xP^-1IP = xI )

= P^-1( A - x.I )P

Determinant(B-x.I) = Det(P^-1)Det(A-x.I)Det(P) = Det(P^-1)Det(P)Det(A-x.I) ........Det denotes determinant .

= Det(P^-1P)Det(A-x.I)

= 1.Det(A-x.I)

= Det(A-x.I)

Therefore matrices A and B have same characteristic polynomial so they give same eigenvalues with same multiplicities .

b.Eigen Subspace of a matrix corresponding to its eigenvalue --

Every non zero vector X satisfying the equation |A- I| .X= 0 i)

is an eigenvector of A corresponding to the eigenvalue .

If rank of matrix | A-I is r <n , then , (n-r) linearly independent solutions to eqn i) .i.e. n-r eigenvectors for a given which are linearly independent among themselves.

Therefore all linear combinations of these ( means 1.X1 + 1.X1 +1.X1....+ 1.X1, 1,2... etc. being scalars and where at least one i 0) eigenvectors ,with one Zero vector of dimension n   taken together , form a subspace of V_n . This subspace is called Eigenspace of matrix A .

This subspace is called Eigenspace of matrix A corresponding to its particular eigenvalue .

The corresponding eigenspace of B ---- For an eigenvalue of matrix A if X is a corresponding eigenvector then , P^-1X is the corresponding eigenvector of B for same eigenvalue , as --

B(P^-1X) =( P^-1AP)(P^-1X) =  P^-1A.X = P^-1.X ... because A.X = X . being eigenvalue of A and X being a corresponding eigenvector of A

= (P^-1X)

Therefore P-1X is an eigenvector of B corresponding to its eigenvalue . Similarly n-r independent solutions to

|B- I| .Y= 0  give n-r linearly independent eigen vectors of type P^-1X , and by the same arguments as given above they too form a subspace of Vn .

For matrix B , its eigenspace is spanned by linearly independent vectors of type P^-1X with a zero vector provided , where X is eigenvector of matrix A .

Any vector from eigenspace of B shall have the form P^-1.Z , where Z would belong to eigenspace of matrix A ..