Homework Help: Matrix logarithm mistake?

1. May 26, 2009

farleyknight

1. The problem statement, all variables and given/known data

Let I be the identity matrix, and let N be a nilpotent matrix order k. Then the matrix exponential function is defined as:

exp(N) := I + N + (1/2!)*N^2 + (1/3!)*N^3 + ... + (1/k!)*N^k

Similarly, we may define the matrix logarithm as the function

-log(N) := I - N + (I - N) + (I - N)^2 + (I - N)^3 ... + (I - N)^k

Show that exp(log(N)) = N.

2. Relevant equations

3. The attempt at a solution

I'm not 100% sure, but I think the prof made a mistake. I can't seem to get this into an identity. He gave us a version of the logarithm function as:

log(N) = I - N + (I - N)/1! + (I - N)^2/2! + (I - N)^3/3! ... + (I - N)^k/k!

And I had trouble getting an identity out of it. I was going to point out this mistake, but he corrected himself in class and gave us the updated version (without the factorial) and it still doesn't come out to an identity for me. When k = 2, I get

-log(exp(N)) = -2N

and when it's order 3, I get

-log(exp(N)) = -2N - (1/3)*N^3

So I'm feeling certain it's still not right. If I'm not mistaken, it supposed to be a generalization of the logarithmic function for real numbers:

http://en.wikipedia.org/wiki/Natural_logarithm#Derivative.2C_Taylor_series

So I might try out this function and hand that in, telling him the corrected version just didn't work for me. My hunch is that the first term shouldn't be I - N but instead just I, since you'd want (I - N)^0/0! = I.

Can anyone else figure out what he really meant?

2. May 26, 2009

tiny-tim

Hi farleyknight!

(try using the X2 tag just above the Reply box )

Not something I'm familiary with,

but it looks from your wikipedia link that it should be

(I - N)/1 + (I - N)2/2 + (I - N)3/3 + ...

3. May 26, 2009

Mute

The first term in the corrected version is exactly the same as the second term, and is hence mistakenly included twice, with the mistake slightly disguised by including a "/1!" on the second term. Just remove one of those, and you'll get what tiny-tim has written down, which should hopefully work out.

You don't want a (I-N)^0/0, because, well... that would just be disasterous with the zero in the denominator and all. (Really it's because when using plain old numbers the first term in the taylor series of the logarithm is about 1 is ln(1) = 0, with all the rest of the coefficients being (-1)^(n+1)(n-1)!/x^n, which give (-1)^(n+1)/n when evaluated at x=1 and dividing by the factorial included in the series definition). The -1 is eliminated by reordering (x-1)^n to -(1-x)^n for odd n and multiplying by an overall minus sign to get a series with postive coefficients.)

Last edited: May 26, 2009
4. May 26, 2009

farleyknight

An initial attempt seems to work. Thanks for your help!