Show that matrix $A+I_{3}$ is invertible if $A$ is orthogonal with $\operatorname{trace}(A) > 1$ We have $A$ $(3×3)$ matrix with real entries. We know that A is orthogonal and $\operatorname{trace}(A)>1$. Show that matrix $A+I_{3}$ is invertible.
We can see that $\det(A)=1$ or $\det(A)=-1$. We can easily find $\operatorname{trace}(A^{*})=\det(A)\operatorname{trace}(A)$. Suppose $\det(A+I_{3})=0$. If we take the characteristic polynomial of A
$$
P(x)=-\det(A-xI_{3})=-x^{3}+\operatorname{trace}(A)x^{2}-\operatorname{trace}(A^*)x+\det(A)
$$
we can find that $P(-1)=0$ so $1+\operatorname{trace}(A)+\det(A) \operatorname{trace}(A)+\det(A)=0$. If $\det(A)=1$ we get easily a contradiction, but in the case where $\det(A)=-1$ we get something right. I tried using eigen values to get in a contradiction with the fact that $ \operatorname{trace}(A)>1$, but nothing.
 A: One quick approach using eigenvalues: suppose that $A + I$ is not invertible. It follows that $A$ has $-1$ as an eigenvalue. On the other hand, because $A$ is orthogonal, all eigenvalues of $A$ have absolute value $1$. If $A$ has eigenvalues $\lambda_1,\lambda_2,\lambda_3 = -1$, then
\begin{align}
\operatorname{trace}(A) &= \lambda_1 + \lambda_2 - 1
\leq |\lambda_1 + \lambda_2| - 1 
\\ & \leq |\lambda_1 | + |\lambda_2| - 1  = 1 + 1 - 1 = 1,
\end{align}
contradicting the premise that $\operatorname{trace}(A) > 1$.
A: Without using eigenvalues: for any unit vector $u$, complete it to an orthonormal basis $\{u,v,w\}$ of $\mathbb R^3$. Then
$$
\begin{aligned}
1&<\operatorname{tr}(A)\\
&=\langle u,Au\rangle+\langle v,Av\rangle+\langle w,Aw\rangle\\
&\le\langle u,Au\rangle+\|v\|\|Av\|+\|w\|\|Aw\|\\
&\le\langle u,Au\rangle+2\\
&=\langle u,(A+I)u\rangle+1.\\
\end{aligned}
$$
Therefore $\langle u,(A+I)u\rangle>0$ and in turn, $(A+I)u\ne0$. Since $u$ is an arbitrary unit vector, $A+I$ must be nonsingular.
A: All thw eigenvalues of $A$ are on the unit circle. There exists a real invertible $3 \times 3$ matrix $P$ such that $P^{-1}AP$ is $I$ or has the form $$\begin{bmatrix}a&-b&0\\b&a&0\\0&0&1\end{bmatrix}$$ where $0<a<1\text{ and }a^2+b^2=1$ Then $$P^{-1}(A+I)P=2I\text{ or }P^{-1}AP=\begin{bmatrix}a+1&-b&0\\b&a+1&0\\0&0&2\end{bmatrix}$$ The determinant of the last matrix is $4(a+1)>0$ so in all cases $A+I$ is invertible.
