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where P is a permutation matrix and each Bi is a square matrix that is either irreducible or zero. Now if A is non-negative then so too is each block of PAP-1, moreover the spectrum of A is just the union of the spectra of the Bi.

[ The inverse of PAP-1 (if it exists) must have diagonal blocks of the form Bi-1 so if any Bi isn't invertible then neither is PAP-1 or A. Conversely let D be the "diagonal" component of PAP-1, in other words PAP-1 with the asterisks zeroised. If each Bi is invertible then so is D and D-1(PAP-1) is equal to the identity plus a nilpotent matrix. But such a matrix is always invertible (if Nk = 0 the inverse of 1 - N is 1 + N + N2 + ... + Nk-1) so PAP-1 and A are both invertible. ]

Therefore many of the spectral properties of A may be deduced by applying the theorem to the irreducible Bi. For example the Perron root is the maximum of the spectral radii of the Bi. Whilst there will still be eigenvectors with non-negative components it is quite possible that none of these will be positive.

[ To see this let D be the "diagonal" component of PAP-1, in other words PAP-1 with the asterisks zeroised. Then D is invertible iff each Bi is invertible and D-1(PAP-1) is equal to the identity plus a nilpotent matrix. But such a matrix is always invertible (if Nk = 0 the inverse of 1 - N is 1 + N + N2 + ... + Nk-1) so PAP-1 and A are both invertible. XXXX ]

Formula
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Uninteresting centers
Assume a,b,c are real variables and let α,β,γ be any three real constants.



f(a,b,c) = \begin{cases} \alpha & \text{if } a < b \text{ and } a < c \quad \text{ (equivalently the first variable is the smallest)}, \\ \gamma & \text{if } a > b \text{ and } a > c \quad \text{ (equivalently the first variable is the largest)}, \\ \beta & \text{otherwise} \quad \qquad \qquad \text{ (equivalently the first variable is in the middle)}. \end{cases} $$

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Let &Delta; be the area of the triangle with sides a, b, c. Whenever no angle of &Delta;ABC exceeds 120&deg; the sum L of the distances from the Fermat point to the vertices is given by the formula L = √ ( (a2 + b2 + c2 + 4&Delta;&radic;3) /2 This formula also occurs in the following two seemingly unrelated situations.


 * a, b, c are the distances from an internal point to the vertices of an equilateral triangle of side L.


 * L is the height of the largest equilateral triangle that can be circumscribed about a triangle with sides a, b, c.

Despite the square roots there are infinitely many solutions (a,b,c,L) in integers. Some like (3,5,7,8) and (7,8,13,15) are simple in the sense that the triangle has an angle of exactly 120&deg; but there are plenty of more interesting ones such as (57,65,73,112) and (73,88,95,147).

Napoleon's Theorem is said to be one of the most-often rediscovered results in mathematics. C&G reference - whilst the theorem has been attributed to Napoleon the possibility of his knowing enough geometry for this feat is as questionable as the possibility of his knowing enough English to compose the famous palindrome : able was i ere i saw elba.

see the web page on the square root symbol in html and css - ugh! best way is probably to resize and drop the superscripts slightly.

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$$L= \sqrt{\mbox{if}~a\ \mbox{is even}^2+b^2+c^2+4\Delta\surd3)/2}$$&emsp;. A(&radic; 3 ,-30°) and B(&radic; 3 ,30°).



Here is the original diagram from Bell's Theorem that was removed on 11/3/22.

The image downloaded as a png file but the text is separate. If it is to be used in LaTeX then the png will need converted to pdf.