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

[aznkid310]

Homework Statement

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)

Homework 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?

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

 

[Office_Shredder]

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

 

[aznkid310]

Fixed Thanks.