### M Theory Lesson 164

Let's do some really basic algebra. On Friday we wondered whether or not the inverses of Riemann zeroes might be related to that damned number. Note that inverse eigenvalues naturally occur for inverse matrices, since

$A^{-1} A v = v = A^{-1} (\lambda v)$

where $\lambda$ is an eigenvalue for $A$. If $A$ (assumed complex) is both Hermitian and unitary, it satisfies $A^{\dagger} = A^{-1} = A$, from which it follows that $A^{2} = I$. What are the solutions to this equation?

In the $2 \times 2$ case one quickly finds that either $A = I$ or $A = \sigma_{x}$. For the $3 \times 3$ case, relations of the form

$a_{11}^{2} - a_{22}^{2} - a_{33}^{2} - 2 a_{23} \overline{a_{23}} = 1$

suggest setting off diagonal elements to be real, since diagonal elements are already real. So if $a_{12} = 1$, it immediately follows that $A$ is completely specified by the circulant matrix

0 1 0

1 0 0

0 0 1

The remaining possibilities are left to the reader. Now note that if $A = A^{-1}$ then the real eigenvalues come in pairs $(\lambda , \lambda^{-1})$, with $\lambda = \pm 1$ considered a double eigenvalue. As a factor of the characteristic polynomial, these 2 roots give

$x^{2} - x(\lambda + \frac{1}{\lambda}) + 1$

which in the $2 \times 2$ case is precisely the statement that $\textrm{det} (A) = 1$ (the constant term) and $\textrm{tr} (A) = \lambda + \frac{1}{\lambda}$.

$A^{-1} A v = v = A^{-1} (\lambda v)$

where $\lambda$ is an eigenvalue for $A$. If $A$ (assumed complex) is both Hermitian and unitary, it satisfies $A^{\dagger} = A^{-1} = A$, from which it follows that $A^{2} = I$. What are the solutions to this equation?

In the $2 \times 2$ case one quickly finds that either $A = I$ or $A = \sigma_{x}$. For the $3 \times 3$ case, relations of the form

$a_{11}^{2} - a_{22}^{2} - a_{33}^{2} - 2 a_{23} \overline{a_{23}} = 1$

suggest setting off diagonal elements to be real, since diagonal elements are already real. So if $a_{12} = 1$, it immediately follows that $A$ is completely specified by the circulant matrix

0 1 0

1 0 0

0 0 1

The remaining possibilities are left to the reader. Now note that if $A = A^{-1}$ then the real eigenvalues come in pairs $(\lambda , \lambda^{-1})$, with $\lambda = \pm 1$ considered a double eigenvalue. As a factor of the characteristic polynomial, these 2 roots give

$x^{2} - x(\lambda + \frac{1}{\lambda}) + 1$

which in the $2 \times 2$ case is precisely the statement that $\textrm{det} (A) = 1$ (the constant term) and $\textrm{tr} (A) = \lambda + \frac{1}{\lambda}$.

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