$a_{11}^2+a_{22}^2+a_{12}^2+a_{21}^2\leq 1/10 $. Prove that $I+A$ is invertible . Let $A=(a_{ij}) $ be a $2\times 2$ real matrix such that $a_{11}^2+a_{22}^2+a_{21}^2+a_{12}^2 < 1/10 $. 
Prove that $I+A$ is invertible .
My approach : Suppose $I+A$ were singular . Then it has a non-zero $2\times 1$ column vector such that $(I+A)v=0 $ i.e $Av=-v $. Hence $-1$ is an eigenvalue of $A$. Let $a$ be the other eigenvalue $A$. Then $A=P \mbox{diag}(-1,a)P^{-1} $ for some invertible matrix and diagonal matix with entries $-1$ and $a$ . 
Hence $1/10 > a_{11}^2+a_{22}^2+a_{33}^2+a_{44}^2\geq a_{11}^2+a_{22}^2+2a_{12}a_{21}=\mbox{trace} (A^2)=1+a^2>1   $ which is contradiction because $1>1/10$. Hence the assumption on the existence  on $v$ is wrong. So $I+A$ must be invertible .
Is my solution correct? I feel like I'm missing out something because the proof  heavily relied on the fact that $1>1/10 $ . So if the proof was correct then the statement is true when $1/10$ is replaced by any constant less $1$ .  
 A: I think we can get a simpler proof. Note that the inequality
$$a_{11}^2+a_{22}^2+a_{21}^2+a_{12}^2 < \frac{1}{10}$$
implies that $|a_{ij}|< \frac{1}{\sqrt{10}}$. Hence
\begin{align*}|\det(A+I)|&=|1+a_{11}+a_{22}+a_{11}a_{22}-a_{21}a_{12}|\\
&\geq 1-|a_{11}|-|a_{22}|-|a_{11}||a_{22}|-|a_{21}||a_{12}|\\
&> 1-\frac{2}{\sqrt{10}}-\frac{2}{10}>0.\end{align*}
Therefore $\det(A+I)\not=0$ and $A+I$ is invertible.
A: If $X = (x,y)^T$ is any column vector, you have $AX = (a_{11} x + a_{12} y, a_{21}x + a_{22}y)^T$. So 
$$
\begin{align*}
|AX|^2 &= [(a_{11},a_{12}) \cdot (x,y)]^2 + [(a_{21},a_{22}) \cdot (x,y)]^2 \\ &\leq |(a_{11},a_{12})|^2|(x,y)|^2 + |(a_{21},a_{22})|^2|(x,y)|^2 \\
&=(a_{11}^2 + a_{12}^2 + a_{21}^2 + a_{22}^2)|X|^2
\end{align*}
$$
by the Cauchy-Schwarz inequality.
Using the hypothesis, this shows that $|AX| < \frac{1}{\sqrt{10}} |X|$ for any $X \ne 0$.
Therefore, if $AX = - X$ for any vector, we must have $X = 0$. This shows that the kernel of $I + A$ is trivial, hence that $I+A$ is invertible.
