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!

  • $\begingroup$ $I-A^{m+1}$ tends to $I$. $\endgroup$ Jan 14 '16 at 15:50
  • 2
    $\begingroup$ 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

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 $A.$


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 $B = A, B= -A, B= A^2$.

  • $\begingroup$ 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

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 $\lim_k A^k = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.