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Linear Algebra: Prove rank(A) <= rank(exp(A))

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Let a be all real numbers, nxn. Prove that

    a) rank(A) less than or equal to rank(exp(A))
    b) rank(exp(A)-I) less than or equal to Rank(A)

    2. Relevant equations

    I'm new to math proofs and so don't really know where to start. Could someone point me in the right direction? Would I need to prove that A is diagonalizable and somehow proceed from there?

    3. The attempt at a solution

    rank(A) + dim N(A) = n, N(A) = nullspasce of A
    This means rank(A) less than or equal to n

    exp(A) = I + A + (1/2)A^2 +...+ (1/(r-1)!)*A^(r-1) Taylor Series Expansion

    Using Sylvester's Inequality: [rank(A) + rank(exp(A)) -n ] less than or equal to rank(Aexp(A))

    Aexp(A) = A + A^2 + (1/2)A^3 + ... + (1/(r-1)!)*A^r
     
    Last edited: Dec 1, 2011
  2. jcsd
  3. Dec 1, 2011 #2

    Office_Shredder

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    I don't really have a whole lot of insight right now but thought I'd just point out real quick that you're missing a -n in Sylvester's inequality
     
  4. Dec 1, 2011 #3
    Fixed Thanks.
     
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