hardddd linear algebra problem. Helppp?
thanks so much to anyone who is able to help. i will choose best answer.
Suppose λ is an eigenvalue of an n x n matrix A with eigenvector x is not equaled to 0. Show that λ² is an eigenvalue of A² with the same eigenvector x.
Comments
You can do this! A^2 x = A(Ax). What does Ax equal?
lambda x, right? So A^2 x=A(lambda x)=...?
the 1st step, somewhat, is to declare what your variables recommend! If A = fee of a sophisticated fee tag, and S = fee of a similar-day fee tag, then the equations are: a) A + S = sixty 5 (no longer 40) and b) 40*A + 35*S = 2425. you pick for a and S, so remedy the equipment. From a), we've A = sixty 5-S, word signs and indicators, subs this into b) b) 40*(sixty 5-S) + 35*S = 2425, improve 40*sixty 5 - 40S + 35S = 2425, collect -40 S + 35S = 2425 - 40*sixty 5 or -5S = -a hundred seventy five, div by making use of -5 or S = 35, so a similar-day fees $ 35, and A = sixty 5 - 35 = 30 $ for stepped forward tickets.
The answer is always Pi.