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Ok so I have this question I have had some problem solving (I've done part 1.):

We have two $n \times n$ matrices $A$ and $B$ and it says that $A=I-AB$.

  1. Prove that $A$ is regular and $AB=BA$. (done this one is pretty easy)

  2. Prove that if $B$ is symmetrical, so is $A$.

  3. Prove that $B^3=0$ if and only if $A=I-B+B^2$.

Thanks in advance...

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1 Answer 1

up vote 2 down vote accepted

For part 2, factor $A$ and obtain $A(I+B) = I$. Check the transpose and notice the symmetric parts

For part 3, the necessity direction is shown by $$ \begin{align} AB &= I-A\\ AB^2 &= B-AB = B-I+A\\ AB^3 &= B^2 - B + AB = B^2 - B + I - A \end{align} $$ if $B^3=0$ then, $B^2 - B + I - A = 0$.

For sufficiency, $$\begin{align} A &= I-B+B^2\\ I - AB &= I - B + B^2\\ -AB &= -B + B^2\\ -AB^2 &= -B^2 + B^3\\ (I-A)B^2 &= B^3\\ AB^3 &= B^3 \end{align} $$ from this, we see that $(I-A)B^3=0$

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