Prove that e^A is similar to I_n+A, where I_n is the identity matrix of order n.

sorrowandsongto

sorrowandsongto

Answered question

2022-10-23

If A is a complex matrix of order n and A is nilpotent, that is, there exists a positive integer s such that A s = 0. Let's say that e A = k = 0 A k k ! . Prove that e A is similar to I n + A, where I n is the identity matrix of order n.

Answer & Explanation

wespee0

wespee0

Beginner2022-10-24Added 15 answers

It suffices to show that this holds in the case that A is a nilpotent Jordan block (of size n) with eigenvalue 0.
Note that it is equivalent to show that e A I is similar to A, and that that a matrix B C n × n is similar to a nilpotent Jordan block (of size n) if and only if it is nilpotent with B n 1 0. So, let B = e A I. By considering the series expansion B = A k = 0 A k ( k + 1 ) ! , note that for all exponents m, we have
B m = A m ( I + a 1 A 1 + + a n 1 A n 1 )
for some coefficients a 1 , , a n 1 C . Conclude therefore that B n 1 = A n 1 0 and that B n = 0. Thus, B must be similar to a Jordan block of size n, which is to say that it is similar to A.
Dylan Nixon

Dylan Nixon

Beginner2022-10-25Added 2 answers

This is equivalent to show that A is similar to e A I = A + A 2 ! + . We show a more general statement that A is similar to B = A + a 2 A 2 + a 3 A 3 + for arbitrary coefficients a 2 , a 3 ,
It's enough to show this when A has a single Jordan block, i.e. there is a vector v 0 such that { v 0 , v 1 = A v 0 , v 2 = A v 1 , } form a basis of C n . Now it's enough to show that v 0 , B v 0 , B 2 v 0 , B 3 v 0 are linearly independent. Note that the coefficients of v i in B j v 0 is δ i j for i j. The matrix of v 0 , B v 0 , under the basis v 0 , v 1 , is triangular with only 1 on the diagonal, hence they are linearly independent.

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