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Invertible matrix

  • Thread starter Lily@pie
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  • #1
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Homework Statement



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

(b) exp(A) =In +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+In) is invertible, I tried to let (A+In)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+In)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?
 

Answers and Replies

  • #2
lanedance
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call the inverse B, then
I=B(A+I)
not zero

for b, the series will terminate as further terms are zero
 
  • #3
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what do you mean by the series will terminate? A=0?
 
  • #4
lanedance
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for (b) teh srioes for e(A) will terminate at the A^2010 term, as all follwing terms are zero
 
  • #5
Dick
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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.
 
  • #6
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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?
 
  • #7
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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?
 
  • #8
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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.....
 
  • #9
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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?
 
  • #10
Dick
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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?
 
  • #11
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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?
 
  • #12
Dick
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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.
 

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