$M$ is a $3 \times 3$ matrix such that $\det(M)=1$ and $MM^T=I$, where $I$ is the identity matrix. Prove that $\det(M-I)=0$
I tried to take $M$ $=$ $$ \begin{pmatrix}
a &b & c \\
d & e & f \\
g & h & i
\end{pmatrix} $$
but its expansion is too long.
Best Answer
$MM^T=I\implies(M-I)M^T=I-M^T\implies\det(M-I)\det(M^T)=\det(I-M^T)$
Since $\det(M^T)=\det(M)=1$ we have:
\begin{eqnarray}\det(M-I)&=&\frac{\det(I-M^T)}{\det M^T}\\ &=& \frac{\det((I-M^T)^T)}{\det M}\\&=&\det(I-M)\\ &=&\det((-1)(M-I))\\&=&(-1)^3\det(M-I).\end{eqnarray}
and therefore $\det(M-I)=0$. This proof also works for any $n\times n$ matrix $M$ satisfying the conditions with odd $n$.
In case $n$ is even, a counter example is provided by $\pmatrix{\cos\theta&\sin\theta\\-\sin\theta&\cos\theta}$, hence the property is no longer true.
Another proof might be the following. $MM^T=I$ then $M$ is orthogonal. Since $n$ is odd, by the canonical form of orthogonal matrix, $M$ has an eigenvalue $1$. Therefore $\det(M-I)=0$.