Relationships between x and y
2 answers:
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Answer:
Only d) is false.
Step-by-step explanation:
Let be the characteristic polynomial of B.
a) We use the rank-nullity theorem. First, note that 0 is an eigenvalue of algebraic multiplicity 1. The null space of B is equal to the eigenspace generated by 0. The dimension of this space is the geometric multiplicity of 0, which can't exceed the algebraic multiplicity. Then Nul(B)≤1. It can't happen that Nul(B)=0, because eigenspaces have positive dimension, therfore Nul(B)=1 and by the rank-nullity theorem, rank(B)=7-nul(B)=6 (B has size 7, see part e)
b) Remember that . 0 is a root of p, so we have that
.
c) The matrix T must be a nxn matrix so that the product BTB is well defined. Therefore det(T) is defined and by part c) we have that det(BTB)=det(B)det(T)det(B)=0.
d) det(B)=0 by part c) so B is not invertible.
e) The degree of the characteristic polynomial p is equal to the size of the matrix B. Summing the multiplicities of each root, p has degree 7, therefore the size of B is n=7.
We can write the sequence out more fully, as we can see each time it is divided by 6.
60, 60/6, 60/6^2, 60/6^3, and so on.
Therefore we know the sequence can be written as
You can think of this as a graph, i.e. y=60/6^(x-1)
As a result, as x tends to infinity, y tends to 0 (since it effectively becomes 60/infinity). Therefore the sequence converges toward zero.
60, 60/6, 60/6^2, 60/6^3, and so on.
Therefore we know the sequence can be written as
You can think of this as a graph, i.e. y=60/6^(x-1)
As a result, as x tends to infinity, y tends to 0 (since it effectively becomes 60/infinity). Therefore the sequence converges toward zero.