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$M$ 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.

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

Remember $\det(M^T)=\det(M)$ and

\begin{eqnarray}\det(I-M^T) &=& \det((I-M^T)^T)\\&=&\det(I-M)\\ &=&\det((-1)(M-I))\\&=&(-1)^3\det(M-I).\end{eqnarray}

Do you see now?

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    $\begingroup$ The last equality is true because $(-1)^3=-1$ $\endgroup$ – Fabien Jul 8 '14 at 3:59
  • $\begingroup$ This is a identity $det(I−M^T)=(-1)^ndet(M−I)$?? With $M_{n\times n}$? $\endgroup$ – AsdrubalBeltran Jul 8 '14 at 4:12
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    $\begingroup$ $det(I-M^T)=det((I-M^T)^T)=det(I-M)=(-1)^ndet(M-I)$ hope you see now $\endgroup$ – BQT Jul 8 '14 at 4:16
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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!!!

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    $\begingroup$ I think this is the natural solution though, something which strikes the very first time you see the problem. Correct me if I am wrong, but doesn't the following logic work? $MM'=I$ is positive definite since $M$ has full rank $\Rightarrow$ eigenvalues of $M$ are positive $\Rightarrow$ $1$ is the only distinct eigenvalue of $M$ (as $M$ is orthogonal) $\Rightarrow \det (M-1I)=0$. $\endgroup$ – StubbornAtom Mar 23 '17 at 15:45
  • $\begingroup$ Thanks. I was not familiar with the assertion that $M$ has full rank $\Rightarrow$ eigenvalues of $M$ are positive. Can you hit at the proof? Cheers! $\endgroup$ – Robert Lewis Apr 4 '17 at 0:45
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    $\begingroup$ You are not familiar because I was wrong to say that. Only the eigenvalues of $MM'$ are positive. $\endgroup$ – StubbornAtom Apr 5 '17 at 20:03

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