Invertible matrix

[Lily@pie]

Homework Statement

(a) Show matrix A+I_n is invertible and find (A+I_n)^-1 in terms of A.

(b) exp(A) =I_n +A +(1/2!)A²+ (1/3!)A³+..+(1/2010!)A²⁰¹⁰

Show exp(A) is invertible and find (exp(A))^-1 in terms of A

Homework Equations

A is square matrix with size n*n such that A²⁰¹¹=0

The Attempt at a Solution

From A²⁰¹¹=0
I made it to A*(A²⁰¹⁰)=0
it's in the same form as Ax=0, so can I say that solution, x=A²⁰¹⁰
(well, I don't know whether this will help me so I just wrote it down)

(a) To show matrix (A+I_n) is invertible, I tried to let (A+I_n)x=0
so
Ax+x=0
and I don't know how to continue anymore T^T
I just know if I can find out that solution for (A+I_n)x=0 is trival then it is invertible.
But, am I approaching it in a correct path?

(b) I guess we use the same approach as a?

 

[lanedance]

call the inverse B, then
I=B(A+I)
not zero

for b, the series will terminate as further terms are zero

 

[Lily@pie]

what do you mean by the series will terminate? A=0?

 

[lanedance]

for (b) teh srioes for e(A) will terminate at the A^2010 term, as all follwing terms are zero

 

[Dick]

What lanedance is getting at is that I/(I+A) has a power series expansion. Just like 1/(1+r) where r is a real number with |r|<1. Do you know it? It might not converge, but if it terminates it will converge.

 

[LPerrott]

A quick an easy way to show that the inverse of A+I exists is to use the determinant. Find the determinant of A using the equation A^2011=0 and then find the determinant of A+I ... what does this tell you?

 

[Lily@pie]

A quick an easy way to show that the inverse of A+I exists is to use the determinant. Find the determinant of A using the equation A^2011=0 and then find the determinant of A+I ... what does this tell you?

This may sound silly,
does
det (A²⁰¹¹) = 0
det (A) * det(A²⁰¹⁰)=0
so, det(A) = 0?

but how do I use this in for det (A+I)?

The way I though of is as
A*A²⁰¹⁰=0
is the same form as Ax=0
so, by letting (A+I)x=0
Ax+x=0
x=0
this shows that columns of (A+I) is linearly independent, hence invertible.

Can I do it this way?

 

[Lily@pie]

call the inverse B, then
I=B(A+I)
not zero

For this part, by calling the inverse B such that
B(A+I)=I
so now we need to find what is B right?

do we expand it
BA+B=I

I'm so lost...

 

[Lily@pie]

What lanedance is getting at is that I/(I+A) has a power series expansion. Just like 1/(1+r) where r is a real number with |r|<1. Do you know it? It might not converge, but if it terminates it will converge.

Erm, not quite...

I googled it and I know that 1/(1+r) is the sum of an infinite geometric progression when |r|<1. But what do you mean by converge? The values gets smaller and smaller? And how does it relates to the I/(I+A)? Does this means the sum of an infinite list of matrices?

 

[Dick]

Erm, not quite...

I googled it and I know that 1/(1+r) is the sum of an infinite geometric progression when |r|<1. But what do you mean by converge? The values gets smaller and smaller? And how does it relates to the I/(I+A)? Does this means the sum of an infinite list of matrices?

Yes, 1/(1+r)=1-r+r^2-r^3+... Sort of. It doesn't work, for example, if r=2. It does work if r=1/2. That's what I mean by 'converge'. And it will work with your matrix, because the sum is finite. Why?

 

[Lily@pie]

Yes, 1/(1+r)=1-r+r^2-r^3+... Sort of. It doesn't work, for example, if r=2. It does work if r=1/2. That's what I mean by 'converge'. And it will work with your matrix, because the sum is finite. Why?

Is it because as the A²⁰¹¹=0, so the terms after it will also be zero as they can be written as multiple of A²⁰¹¹?

 

[Dick]

Is it because as the A²⁰¹¹=0, so the terms after it will also be zero as they can be written as multiple of A²⁰¹¹?

Sure it is. So what is the inverse of I+A? Check by multiplying (I+A) by the inverse and show you get I.