Prove that $\det(M-I)=0$ if $\det(M)=1$ and $MM^T=I$

Question

$M$ 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.

Answer 1

$MM^T=I\implies(M-I)M^T=I-M^T\implies\det(M-I)\det(M^T)=\det(I-M^T)$

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 $n$.

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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.

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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 $\det(M-I)=0$.

Answer 2

Lauds to Mr. T for a very elegant solution which nicely avoids the mundane approach based on eigenvalues, which I in all humility submit in the following:

Assuming $M$ is a real matrix, we have the following terse and perhaps overpacked verbiage:

Since $MM^T = I$, $M$ is orthogonal; since $M$ is $3 \times 3$, the characteristic polynomial $p_M(\lambda) = \det (M - \lambda I)$ is of odd degree $3$; since $p_M(\lambda)$ is a real polynomial of odd degree, it has a real root $\lambda_0$; since $\lambda_0$ is a root of $p_M(\lambda)$, it is an eigenvalue of $M$; since $\lambda_0$ is an eigenvalue of $M$, there is a nonzero vector $\mathbf x \in \Bbb R^3$ with $M\mathbf x = \lambda_0 \mathbf x$; since $M\mathbf x = \lambda_0 \mathbf x$ we have $\langle \mathbf x, \mathbf x \rangle = \langle \mathbf x, I \mathbf x \rangle = \langle \mathbf x, M^TM \mathbf x \rangle = \langle M \mathbf x, M\mathbf x \rangle = \langle \lambda_0 \mathbf x, \lambda_0 \mathbf x \rangle = \lambda_0^2 \langle \mathbf x, \mathbf x \rangle$ where $\langle \cdot, \cdot \rangle$ is the Euclidean inner product on $\Bbb R^3$; since $\mathbf x \ne 0$ we have $\lambda_0^2 = 1$; since $\lambda_0^2 = 1$, we have $\lambda_0 = \pm 1$; since the preceeding argument applies to any real root of $p_M(\lambda)$, we must have that the real eigenvalues of $M$ lie in the set $\{1, -1\}$; since $p_M(\lambda)$ is a real polynomial of degree $3$ with at least one real root, the remaining two roots if complex must be a conjugate pair $\mu, \bar \mu$, whence $\lambda_0 \mu \bar \mu = \det M = 1$ whence $\mu \bar \mu > 0$ whence $\lambda_0 = 1$; since the roots of $p_M(\lambda)$ if real must lie in the set $\{1, -1\}$ and $\det M = 1$ is the product of these eigenvalues, we must have at least one eigenvalue of $M$ equal to $1$ in this case; since in either case $\lambda_0 = 1$ is an eigenvlaue of $M$, we must have a non-zero vector $\mathbf x \in \Bbb R^3$ with $M \mathbf x = \mathbf x$; since $M \mathbf x = \mathbf x$ we have $(M - I)\mathbf x = 0$, i.e. $M - I$ is singular; since $M - I$ is singular we must have $\det(M - I) = 0$. QED.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!