Proving determinant value zero: $\det(M-I)=0$

Question

If $M$ is a $3\times3$ matrix, where M'M=I and $\det(M)=1$, then prove that $\det(M-I)=0$

By the information given I know that given matrix is orthogonal. How can prove the above determinant zero?

Answer 1

Since $M^\prime M = I$, it follows that $$ M - I = (I - M^\prime) M. $$ Also, since $M$ is a $3 \times 3$ matrix, $$ \begin{aligned} \det(I - M^\prime) &= \det\left(\left(I - M\right)^\prime\right) = \det(I - M) \\ &= \det(-(M - I)) = \left(-1\right)^3 \det(M - I) = -\det(M - I). \end{aligned} $$ Therefore since $\det(M) = 1$, $$ \det(M - I) = \det(I - M^\prime) \det(M) = \det(I - M^\prime) = -\det(M - I). $$ Thus, $\det(M - I) = 0$.

Answer 2

An orthogonal transformation of $\mathbb{R}^3$ has to be a rotation about some axis. So $M$ is such a transformation. A vector on the axis of rotation is an eigenvector for $M$ of eigenvalue $1$. Therefore $(M-I)$ annihilates such vectors, and $(M-I)$ has $0$ for one of its eigenvalues. From this, it follows that $(M-I)$ has determinant $0$.