Showing invertibility of matrix

Question

Suppose $A = \pmatrix{a_{11} & a_{12} \cr a_{21} & a_{22} } $ satisfies $a_{11}^2 + a_{12}^2 + a_{21}^2 + a_{22}^2 < 1/10$

Why must $I + A$ be invertible?

Answer 1

Look at the determinant of $I+A$, which is equal to $$\det(I+A)=1+a_{11}+a_{22}+a_{11}a_{22}-a_{12}a_{21}.$$ Since $$ab\geq-\frac{1}{2}(a^2+b^2)$$ for all $a,b$, this implies that $$\det(I+A)\geq 1+a_{11}+a_{22}-\frac{1}{2}(a_{11}^2+a_{22}^2+a_{12}^2+a_{21}^2)\geq 1+a_{11}+a_{22}-\frac{1}{20}.$$ Also, $a_{11}^2\leq\frac{1}{10}$, so $|a_{11}|\leq\frac{1}{\sqrt{10}}$, and similarly for $a_{22}$, therefore $$\det(I+A)\geq 1+a_{11}+a_{22}-\frac{1}{20}\geq 1-\frac{2}{\sqrt{10}}-\frac{1}{20}>0,$$ so $I+A$ is invertible.

Answer 2

The other works well, but here's one that puts the question into a more general context.

The function $$ \|A\|_F = \sqrt{\sum_{i,j = 1}^2 a_{ij}^2} $$ is the Frobenius norm, which is an example of a (submultiplicative) matrix norm. See also my own explanation here. For any matrix norm $\|\cdot\|$, it is well known that if $\|A\| < 1$, then $I + A$ is invertible with inverse $$ (I + A)^{-1} = \sum_{k=0}^\infty (-A)^k $$ notably, because of the matrix norm, we can guarantee that this infinite sum is meaningful.