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I am trying to demonstrate next assert about matrices:

$A$ is a matrix of $n$ order, with $n$ odd, that obeys $A A^T =I$ and $\det\, A=1$. Then $\det\,(A-I)=0$.

I have tried a number of things but none of them work. That $n$ is odd seems to indicate to the trace of the matrix an its properties but I have also tried to find a product of matrix useful but it did not work out.

All ideas well be apreciated.

Diglett
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user210390
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3 Answers3

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We have $\left(-1\right)^n = -1$ (since $n$ is odd). But \begin{equation} \left(A-I\right)A^T = \underbrace{AA^T}_{=I} - A^T = I-A^T = \left(I-A\right)^T . \end{equation} Taking determinants of both sides of this equality, we obtain \begin{align} \det\left(\left(A-I\right)A^T\right) &= \det\left(\left(I-A\right)^T\right) = \det\left(\underbrace{I-A}_{=-\left(A-I\right)}\right) \\ & = \det\left(-\left(A-I\right)\right) = \left(-1\right)^n \det\left(A-I\right) = - \det\left(A-I\right) \end{align} (since $\left(-1\right)^n = -1$). Thus, \begin{align} -\det\left(A-I\right) = \det\left(\left(A-I\right) A^T\right) = \det\left(A-I\right) \cdot \underbrace{\det \left(A^T\right)}_{= \det A = 1} = \det\left(A-I\right) , \end{align} so that $0 = 2 \cdot \det\left(A-I\right)$ and thus $\det\left(A-I\right) = 0$, assuming that your matrices are over a field of characteristic $\neq 2$.

darij grinberg
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We know that $\|Ax\|^2=\langle Ax,Ax\rangle=\langle AA^Tx,x\rangle=\|x\|^2$. Since $n$ is odd, we know that there’s at least one zero of the characteristic polynomial of $A$, name it $\lambda$ and let $v$ be a correspondent eigenvector. Then $$\|v\|=\|Av\|=\|\lambda v\|=|\lambda|\|v\|,$$ hence $\lambda=\pm1$, that is, any real eigenvalue is either $1$ or $-1$.

Since $\det(A)$ is the product of all eigenvalues of $A$ and that determinant equals $1$, the multiplicity of eigenvalue $-1$ must be even, say $2k$. Then the characteristic polynomial factorized over the real numbers is $(\lambda+1)^{2k}p(\lambda)$ where $p$ has an odd degree and therefore a zero, which must be $1$. That means $\det(A-1\cdot I)=0$.

Michael Hoppe
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An orthogonal $n\times n$ matrix is certainly diagonalizable over the complex numbers, being normal. However, its eigenvalues may be complex.

Your task is to prove that in the case of odd $n$, $1$ is an eigenvalue.

Since $A$ is a real matrix, its nonreal eigenvalues come in pairs $\lambda$ and $\bar{\lambda}$. So, assume the eigenvalues (possibly repeated) are $$ \lambda_1,\dots,\lambda_k, \lambda_{k+1},\lambda_{k+2}=\bar{\lambda}_{k+1},\dots, \lambda_{n-1},\lambda_n=\bar{\lambda}_{n-1} $$ where the first $k$ are real. Note that $k$ is odd. Also the module of every eigenvalue is $1$, so we have $$ \det A=\lambda_1\dots\lambda_k \lambda_{k+1}\lambda_{k+2}\dots\lambda_{n-1}\lambda_n= \lambda_1\dots\lambda_k $$ If all the real $\lambda_i$ were $-1$, for $i=1,2,\dots,k$, we'd have $\det A=(-1)^k=-1$.

egreg
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  • I think the problem doesn't seem to assume that $A$ is a real matrix. – Tengu Oct 23 '18 at 02:02
  • @Tengu Then it's not necessarily true that $1$ is an eigenvalue. When “orthogonal” is mentioned, it's usually understood that the framework is real matrices. – egreg Oct 23 '18 at 06:48
  • It seems to me that @darijgrinberg proof is true for matrices over any field of characteristic $\ne 2$ so I was just wondering about that. You mentioned that it is not necessarily true for $1$ to be an eigenvalue. Could you please give an example for such matrix? Thank you. – Tengu Oct 23 '18 at 07:56
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    @Tengu I hadn't seen that proof, so I withdraw my conjecture. But the title seems quite clear about “orthogonal” and “linear algebra first course”. – egreg Oct 23 '18 at 08:36