Let n be a positive integer. Let A be an element of the vector space Mat(n,n,F), which has dimension n2 over F. Show that the span of the infinite set of matrices span(In, A, A2, A3, ...) has dimension not exceeding n over F.
Defn of the linear space Mat(n,n,F):
The set of all n-by-n matrices with entries in F. Mat(n,n,F ) is a vector space. The additive identity in Mat(n,n,F ) is the n-by-n matrix all of whose entries equal zero.
Clues to solving the problem:
A is an element of Mat(n,n,F )
dim Mat(n,n,F ) = n2
In = Identity matrix. 1's on the diagonal and 0's else.
In, A, A2, ... , An^2 -- n+1 matrices are linearly dependent.
Want to show that dim span(In, A, A2, A3, ...), all infinitely many of them, are less than or equal to n.
Assume that A is invertible (that is that A-1 exists) then A-1 is an element of span (In, A, A2, A3, ...,An-1).
Reformulation: A-1 = p(A) for a polynomial P
Proof: Interpret A as M(T) for some T in L(Fn). let c be the minimum polynomial for T.
c = a0 +a1z+ ... + amzm, m ≤ n
means that c(T) = 0, assume am ≠ 0
a0 +a1T + ... + amTm = 0 If a0 = 0, then c(0)=0, and 0 is an eigenvalue for T. T is not injective and A-1 does not exist. Since it DOES exist(due to the hypothesis), c(0)≠0, therefore a0 ≠ 0
So, 1 = -(a1/a0)T - ... - (am/a0)Tm
So, In = -(a1/a0)A - ... - (am/a0)Am
So, A-1 = -(a1/a0) In - ... - (am/a0)Am-1
A vector space proof is provided. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.