Show that matrix $A+I_{3}$ is invertible if $A$ is orthogonal with $\operatorname{trace}(A) > 1$

Question

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.

Answer 1

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 $\operatorname{trace}(A) > 1$.

Answer 2

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.

Answer 3

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.