# If $Q$ is a proper orthogonal transformation matrix, deduce that $\det(1-Q)=0$.

Show that if $$Q$$ is orthogonal transformation matrix, then $$Q^t(Q-1)=(1-Q)^t$$. Deduce that if $$Q$$ is also proper, then $$\det(1-Q)=0$$. Hence show that transformation has nonzero vector that has the same components in both coordinate system.

I tried to solve this problem.I think I got the first part right,

$$Q^t(Q -1)= Q^t Q- Q^t=1- Q^t=(1- Q)^t$$

The second part,

$$-Q ^t(1-Q)=(1-Q)^t$$

$$\det(-Q^t)\det(1-Q)=\det((1-Q)^t$$

$$(-1)^n\det(Q)\det(1-Q)=\det((1-Q)^t)$$

since the orthogonal matrix is proper which means $$\det(Q)=1$$ and for any matrix, its determinant equals the determinant of its transpose.

$$(-1)^n\det(1-Q)=\det(1-Q)$$

So, it's always true for $$\det(1-Q)=0$$

But that's not what the question asks. I haven't done linear algebra for a while and I am not sure from the concepts I used, so I would be glad if you clarify any mistake I made.

• Unrelated: which context did this come from? I've almost never seen $\lambda$ used to denote a matrix instead of an eigenvalue before. – YiFan Feb 15 at 23:44
• Me neither. In linear algebra, I always see λ denotes an eigenvalue, but this is how my professor wrote it. I will edit it to avoid any misunderstanding. – K.ali Feb 15 at 23:56
• not mentioned in the question. – K.ali Feb 16 at 0:15

Since $$Q$$ satisfies $$Q^t(Q-1)=(1-Q)^t$$, then $$\det(Q)\det(Q-1)=\det(1-Q)=(-1)^n\det(Q-1)$$, as you observed. Suppose $$\det(Q-1)$$ is nonzero, then we can divide both sides by it to get $$\det Q=(-1)^n$$. If $$n$$ is odd, we get a contradiction with the fact that $$Q$$ is proper.
The statement is false when $$n$$ is even. Consider $$\mathbb R^2$$ and $$Q=-1=\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$$. We have $$\det Q=1$$, but $$\det(1-Q)=4\neq0$$.
For the last part, we need to show the existence of $$x$$ in the vector space so that $$Qx=x$$, i.e. one of the eigenvalues of $$Q$$ is $$1$$. But we've just done that!