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 September 20th, 2012, 06:49 AM #1 Member   Joined: Aug 2012 Posts: 30 Thanks: 0 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...
September 21st, 2012, 08:47 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: $I_n(I_n+ A+ A^2+ \cdot\cdot\cdot+ A^k)= I_n+ A+ A^2+ \cdot\cdot\cdot+ A^k$ and $-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 $A^j= 0$, 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}$.

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, $I_4- A= \begin{bmatrix}-1 &-2=&1=&1 \\ 1=&1=&0 \\ 1=&1=&0 \\ 0=&-1=&1=&0\end{bmatrix}=$ and you can calculate the powers directly. What do you get?

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