Linear algebra - Matrix similarity and diagonalization?

Question

(A) If A is similar to B, show that A^n is similar to B^n for n greater than or equal to 1

(B) If A is diagonalizable and p(x) is a polynomial such that p(v)=0 for all eigenvalues v of A, show that p(A)=0. In particular, show the characterististic polynomial of A (Ca(A)) =0. It also has a note saying Ca(A)=0 for all square matrices A.

If you can help with either of these questions that would be great!

Answer 1

(A) If A is similar to B, show that A^n is similar to B^n for n greater than or equal to 1

A= PBP^{-1}
A^2 = PBP^{-1}PBP^{-1}=PB^2 P^{-1}
A^3=PBP^{-1} PBP^{-1} PBP^{-1} = PB^3 P^{-1}
....
A^n = PBP^{-1}... PBP^{-1}
= PB^n P^{-1}

(B) If A is diagonalizable and p(x) is a polynomial such that p(v)=0 for all eigenvalues v of A, show that p(A)=0. In particular, show the characterististic polynomial of A (Ca(A)) =0. It also has a note saying Ca(A)=0 for all square matrices A.

here if A is diagonalizable, then
A= P D P^{-1} ie A is similar to a diagonal matrix.
assume that p(x) = an x^n + ...+ a0
then p(A) = an A^n + ... + a0 I
= an P D^n P^{-1} + ...+ a0 I (by part (a))
= P ( an D^n + ... + a0 I)P^{-1}

the elements of D^n are the eigenvalues of A,
and since p(v)=0 for all eigenvalues of A, then
an D^n + ... + a0 I = 0 (matrix)
therefore p(A)= 0(matrix)

(the first answer is kind of wrong... )

Answer 2

(B) If A is diagonalizable A=XDY, then the eigenvectors are the columns of D.

If p(v) = 0 , for all eigenvalues then, show that p(A) = 0.

D has as diagonal elements the eigenvalues of A, so if p(v) = 0 for every eigenvalue, yes then p(A) = p(XDY) = p(Y)p(D)p(X) = p(Y)0p(X) = 0.
----^ the matric with only zeros.since p(v)=0.

the rest i dont understand the notation.
for (A) i dont know what you mean with similair.