# Show all matrices very similar to the identity matrix are invertible

Please don't post completed solutions for 24 hours, as this is a homework problem, and I just need a bit of guidance.

Given a $N\times N$ matrix A, let $|\mathbf{*}|_{2}$ be the sum of the squares of each element of such a matrix. If $|I-A|_2 \leq \frac{1}{2}$, then $A^{-1}$ must exist.

Following is an overview of my attempts, not the question itself.

From the assumption, it follows that the absolute value of every element of $|I-A|_2$ is less than $\frac{1}{2}$, or one is exactly $\frac{1}{2}$ and all others are $0$. I've attempted to show that such a matrix can be changed to row echelon form, by using the bounds of each element of $A$, showing that if I subtract some multiple of the first row from all following rows, $A_{2,2}$ must be positive, but it doesn't hold for the general case, specifically not for $A_{3,3}$.

If the function $L_A$ (the linear map taking in vectors $v$ from $R^n$, and outputting $A\cdot v$) could be shown to map only $0_n$ to $0_n$, that would be sufficient, but my attempts to prove that have reduced to attempting to show that A can be reduced to ref form, and if I could show that directly, that would be sufficient proof.

Similarly, if I could show $L_A$ to be injective or surjective, or that the kernel of $L_A$ was $\{0_n\}$, I could show that $A^{-1}$ must exist, but I have been unable to do so.

Let $x$ in $\mathbb{R}^n=E$. You want to show that if $Ax=0$, then $x=0$. Suppose not, put $x=(x_1,\cdots,x_n)$, $B=I-A=(b_{i,j})$ and note that you have for $i=1,\cdots,n$ $x_i=\sum_{j=1}^n b_{i,j}x_j$. By what can you bound $|x_i|^2$, and then $S=\sum_{i=1}^n (x_i)^2$ ?
Hint: $$\frac{1}{1-x} = 1 + x + x^2 + \cdots \quad \text{when |x|<1}$$
Let $B=I-A$. When $\|B\|_2<1$ we can apply the above to get the inverse of $I-B=A$.