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I have a linear algebra question I need help with.

Let $A$ be an $m\times m$ matrix with $\|A\|_2 < 1$ where $\|A\|_2$ is the $2$-norm of $A$. Show that $I - A$ is invertible where $I$ is the identity matrix.

I know that $\|Ax\|_2 \leq C\|x\|_2$ for some constant $C$ and a vector $x$. However I don't know the definition of $\|x\|_2$. I also don't see how this definition can help solve this problem.

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Since $\|A\|_2 < 1$, the sequence of matrices $S_n = \sum_{i=0}^n A^n$ is Cauchy (as a sequence in $L^2$) and so converges in the $L^2$-norm to some matrix $S$. Now, $S_n(I - A) = I - A^n$, so taking the limit in $L^2$ gives $S(I-A) = I$. Something similar works for the other side.

There may be some other details you need to fill in, but this is the basic idea I would use to solve the problem.

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Thanks! I'm still a little fuzzy on the details but this definitely put me in a good place =) – user1855952 Jan 31 '13 at 5:56

A related problem. Note this,

$$ (I-A)^{-1} = \sum_{n=0}^{\infty}A^n. $$

Now, for the above series to converge, you need to impose the condition $||A||_2<1$. Just compare with the geometric series

$$ \sum_{n=0}^{\infty}x^n, $$

where you need the condition $|x|<1$ for convergence. Note that

$$ ||A^n||\leq ||A||^n. $$

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