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September 20th, 2012, 07:49 AM   #1
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invertible matrix

I have some difficulty proving the bolded part of the exercise:

If A is an nxn matrix establish the identity
Isub(n)-A^(k+1)=(Isub(n)-A)(Isub(n)+A+A^2+...+A^k).
Deduce that if some power of A is the zero matrix then Isub(n)-A is invertible.
Suppose now that
A=2 2 -1 -1
-1 0 0 0
-1 -1 1 0
0 1 -1 1
Compute the powers (Isub(n)-A)^i for i=1,2,3,4 and, by considering
A=Isub(4)-(Isub(4)-A), prove that A is invertible and determine A-1.

I would be grateful for any help you are able to provide...
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September 21st, 2012, 09:47 AM   #2
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Re: invertible matrix

Quote:
Originally Posted by shine123
I have some difficulty proving the bolded part of the exercise:

If A is an nxn matrix establish the identity
Isub(n)-A^(k+1)=(Isub(n)-A)(Isub(n)+A+A^2+...+A^k).
That's pretty much direct multiplication: and [itex]-A(I_n+ A+ A^2+ \cdot\cdot\cdot+ A^k)= -A- A^2- A^3-\cdot\cdot\cdot- A^{k-1}[/tex]. Now add those.

Quote:
Deduce that if some power of A is the zero matrix then Isub(n)-A is invertible.
If , the taking k= j- 1 in the formula above [latex]I_n- A^j= I_n= (I_n-A)(I_n+ A+ A^2+ \cdot\cdot\cdot+ A^{j-1}[/itex].

Quote:
Suppose now that
A=2 2 -1 -1
-1 0 0 0
-1 -1 1 0
0 1 -1 1
Compute the powers (Isub(n)-A)^i for i=1,2,3,4 and, by considering
A=Isub(4)-(Isub(4)-A), prove that A is invertible and determine A-1.

I would be grateful for any help you are able to provide...
With that A, and you can calculate the powers directly. What do you get?
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