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I need to prove $I - A$, $I + A$ and $I - A^2$ are nonsingular where $||A|| < 1$.

Also need to show $(I - A)^{-1} = \sum_{k=0}^{\infty}A^k$.

So far I've got that because $||A||< 1$ then $p(A) < 1$. Therefore $p(I-A) < 1$ (as $p(I) = 1$). I want this lead to me stating that this means none of the eigenvalues of $I-A$ are $0$, therefore $A$ is nonsingular.

For $(I - A)^{-1} = \sum_{k=0}^{\infty}A^k$:

If I say $(I-A)\sum_{k=0}^{m}A^k = I - A^{m+1}$

Then $\sum_{k=0}^{m}A^k = (I - A)^{-1}(I - A^{m+1})$ and because $||A|| < 1$ then $I-A^{m+1}$ tends to $0$ as $m$ tends to inf.

I appreciate my answers are far from complete and appreciate any help!

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  • $\begingroup$ $I-A^{m+1}$ tends to $I$. $\endgroup$ Jan 14 '16 at 15:50
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    $\begingroup$ I think you're not that far from complete. What you need to do is to proof read your proof and remove mistakes like $I-A^{m+1}$ tends to $0$ and $\sum_0^\infty A^k = I-A^{m+1}$ etc. $\endgroup$
    – skyking
    Jan 14 '16 at 15:53
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Computing an explicit inverse by a norm convergent geometric power series is legitimate but you need to adapt your argument a little: at one point you are confusing the $m$-th partial sum of a series with the infinite sum.

The following argument rather follows the "eigenvalue" line that you indicated.

If one of these three matrices is singular then that matrix has a nontrivial kernel, i.e., at least one eigenvector with eigenvalue $0.$ By linearity this means that $A,$ or $-A,$ or $A^2$ has an eigenvector with eigenvalue $\pm1$ and this means that the norm of that matrix cannot be less than $1.$ Also remember that the norm of $A^2$ is bounded by the square of the norm of $A.$

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Show the second point first. For all $n \in \mathbb{N}$ and all matrices with $\|B\| < 1$, we have

$$(I-B)\sum_{k=0}^{n} B^{k} = \sum_{k=0}^{n} B^{k} (I-B) = I - B^{n+1}$$ Because $\|B\| < 1$, taking the limit yields $$(I-B)^{-1} = \sum_{k=0}^{\infty}B^{k}$$

which shows that $(I-B)$ is invertible. Now apply this reasoning to $B = A, B= -A, B= A^2$.

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  • $\begingroup$ Minor note: At second occurrence of $\|B\|$, it should be $\|B\|<1$, I guess (i.e. strict inequality). $\endgroup$
    – mickep
    Jan 14 '16 at 15:54
  • $\begingroup$ @mickep thanks, fixed it. $\endgroup$
    – user159517
    Jan 14 '16 at 15:55
  • $\begingroup$ This only works if the matrix norm is submultiplicative. $\endgroup$
    – copper.hat
    Jan 14 '16 at 16:05
  • $\begingroup$ @copper.hat true, I assumed that we were talking about an operator norm here. $\endgroup$
    – user159517
    Jan 14 '16 at 16:21
  • $\begingroup$ @user159517: It is true in general see Gelfand's formula. $\endgroup$
    – copper.hat
    Jan 14 '16 at 19:40
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Suppose $(A-I)x = 0$. Without loss of generality we can take $\|x\|=1$. Then $Ax = x$, which implies $\|A\| \ge 1$ a contradiction. Hence $\ker (A-I) $ is trivial and so $A-I$ is invertible.

Exactly the same argument applies to $A+I$.

Since $I-A^2 = (I-A)(I+A)$ we see that $I-A^2$ is also invertible since it is the product of two invertible matrices.

For the second part we can use Gelfand's formula which says that $\lim_{k} \sqrt[k]{\|A^k\|} = \rho(A)$. In particular, if $\rho(A)<\lambda <1$, then there is some $K$ such that for all $k \ge K$ we have $\|A^k\| \le \lambda^k$, and so $\lim_k A^k = 0$.

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