A = λI + N, N is a nilpotent matrix

[MaxManus]

Homework Statement

A is a quadratic matrix

A = λI + N

where λ is a real number, I is the identity matrix and N is a nilpotent matrix

Compute exp(tA) as a function of t, λ and N^i for i ? 1,2..,n-1

My attempt

exp(At) = exp((λI + N)t) = exp(λIt)exp(Nt) =
$$\sum_{n=0}^{\infty} \frac{(Iλt)^n}{n!} \sum_{n=0}^{\infty} \frac{(Nt)^n}{n!}$$

Its supposed to be Iλt and not I955;t

 

[CompuChip]

Then try making a proper lambda in LaTeX: $I \lambda t$ ;)

What is exp(c I), where c is a constant and I is the identity matrix?

What happens when you start writing out the sum
$$\sum_{n=0}^{\infty} \frac{(Nt)^n}{n!} = \sum_{n = 0}^\infty \frac{t^n}{n!} N^n$$?

 

[MaxManus]

Thanks

exp(cI) = ones + exp(c)I - I
where one is a sqare matrix with only ones,
and exp(ctI) = ones + exp(ct)I - I

and e^N=
$$I + N \frac{1}{2}N^2 + \frac{1}{6}N^3 + \cdots + \frac{1}{(n-1)!}N^{n-1}.$$

so
$$e^{At} = ( ones + e^{ \lambda t} I - I )(I + N \frac{1}{2}N^2 + \frac{1}{6}N^3 + \cdots + \frac{1}{(n-1)!}N^{n-1} )$$

correct?

 

[Mark44]

Thanks

exp(cI) = ones + exp(c)I - I

No, this doesn't make sense at all. exp(A) = I + A + A²/2! + ... + Aⁿ/n! + ... Replace A by cI and see what you get.

where one is a sqare matrix with only ones,
and exp(ctI) = ones + exp(ct)I - I

and e^N=
$$I + N \frac{1}{2}N^2 + \frac{1}{6}N^3 + \cdots + \frac{1}{(n-1)!}N^{n-1}.$$

so
$$e^{At} = ( ones + e^{ \lambda t} I - I )(I + N \frac{1}{2}N^2 + \frac{1}{6}N^3 + \cdots + \frac{1}{(n-1)!}N^{n-1} )$$

correct?

 

[MaxManus]

No, this doesn't make sense at all. exp(A) = I + A + A²/2! + ... + Aⁿ/n! + ... Replace A by cI and see what you get.

Thanks, it should be exp(c)*I

 

[Mark44]

exp(c)*I is a diagonal matrix with exp(c) in each position along the diagonal.

 

[MaxManus]

Am I wrong again?

 

[Mark44]

exp(At) = exp((λI + N)t) = exp(λIt)exp(Nt)

I think that what you need to do now is to replace each factor on the right by its corresponding series (not in closed form), and use the fact that N is nilpotent, which will simplify exp(Nt) considerably.

 

[MaxManus]

Thanks

 

[CompuChip]

So what you got is
$$e^c \left( I + \sum_{i = 1}^{n - 1} \frac{t^i}{i!} N^i \right)$$
right?

 

[MaxManus]

Yes, thanks to both of you for all the help
$$e^{At} = (e^{\lambda t})(I + \sum_{i = 1}^{n-1} \frac{t^i}{i!}N^i)$$