Let $A$ and $B$ be two orthogonal matrices (of order $n\geqslant 2$) such that
$$\det A=1 \qquad\qquad \det B=-1$$
Can we say that:
- there is $\lambda \in [0,1]$ such that $\lambda A + (1-\lambda) B$ defines a projection operator?
- there is $\lambda \in [0,1]$ such that $\lambda A + (1-\lambda) B$ is a singular matrix?
I was wondering why can't we say that every orthogonal matrix is a projection matrix that projects onto $\mathbb{R^n}$ (so that the first question is answered by $\lambda = 0$ or $\lambda = 1$)?
As user251257 points out in the comment, the statement (2) is true because the determinant is continuous. Here is a counterexample for (1). Let's take $$A=\begin{pmatrix}\sqrt{0.5}&-\sqrt{0.5}\\\sqrt{0.5}&\sqrt{0.5}\end{pmatrix},\,\,\, B=\begin{pmatrix}0&1\\1&0\end{pmatrix},$$ then $A, B$ are orthogonal and $\det A=1$, $\det B=-1$. Their convex combination is, with $\lambda\in[0,1]$, $$C(\lambda)=\begin{pmatrix}\lambda\sqrt{0.5}&-\lambda\sqrt{0.5}+1-\lambda\\\lambda\sqrt{0.5}+1-\lambda&\lambda\sqrt{0.5}\end{pmatrix}.$$ Clearly, $C(\lambda)$ cannot be the identity matrix, so it must be singular if it is a projector. We have $$\det C(\lambda)=0.5\lambda^2-(1-\lambda+\sqrt{0.5}\lambda)(1-\lambda-\sqrt{0.5}\lambda)=\lambda^2-(1-\lambda)^2=-1+2\lambda.$$ Therefore, $C(\lambda)$ can be a projector only if $\lambda=0.5$, but then $$C(0.5)=\begin{pmatrix}0.5\sqrt{0.5}&-0.5\sqrt{0.5}+0.5\\0.5\sqrt{0.5}+0.5&0.5\sqrt{0.5}\end{pmatrix},$$ and we can check that the $(1,1)$ entry of the square of the latter matrix is $1/4$, so $C(0.5)$ is not a projector as well.