Question

Hello,

I would like to discuss with someone the work that i've done on my own regarding part d).

So we have d unique eigenvalues and d < n. if d=n, then we only have a trivial solution (by the rank nullity theorem), but this is a contradiction because v is a non-zero eigen vector.

hence the determinant (A- \lambda*I) =0. where this determinant is equal to the characteristic polynomial equation.

The polynomial equation p(A)= \prod (A- \lambda_i * I) for each distinct eigenvalue lambda_i.

Am I constructing part d) correctly?

2. (4 pts. each) In this problem, you will show that if A is an nxn symmetric matrix, then its minimal polynomial is HA(t) = 1/(t - li), where li.....d are the distinct eigenvalues of A. You may use any results stated in Lecture 0 without proving them. (a) Show that for any n x n matrix M + 0 there is a v ER such that My + 0. (Hint: Consider Mel,..., Men.) (b) Show that every v ER" can be written as a linear combination v=cvi+...+ ChVn, where V1,...,Vn are eigenvectors of A. (c) Show that for any square matrix M and constants ki, k2 € R, (M – kI)(M – k I) = (M - k21)(M-kI). v=0 for all v ER (d) Use parts (b) and (c) to conclude that I Li=1 (e) Conclude that (A-\;I) = 0, and therefore pa(t) = 1/(t-li) is the minimal polynomial i=1 for A.

Answer 1

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