91影视

In linear algebra, a symmetric matrix is a square matrix that stays unchanged when its transpose is calculated. In that instance, a symmetric matrix is one whose transpose equals the matrix itself.

Consider n+1 distinct unit vectors in ${\mathrm{鈩�}}^{\mathrm{n}}$such that the angle between any two of them is $胃$. If ${\stackrel{鈬赌}{\mathrm{v}}}_0,{\stackrel{鈬赌}{\mathrm{v}}}_1,{\stackrel{鈬赌}{\mathrm{v}}}_2...{\stackrel{鈬赌}{\mathrm{v}}}_{\mathrm{n}}$are n+1 distinct unit vectors in ${\mathrm{鈩�}}^{\mathrm{n}}$such that the angle between any two of them is $\mathrm{胃}$, let

$\mathrm{A}=\left[{\stackrel{鈬赌}{\mathrm{v}}}_0{\stackrel{鈬赌}{\mathrm{v}}}_1{\stackrel{鈬赌}{\mathrm{v}}}_2...{\stackrel{鈬赌}{\mathrm{v}}}_{\mathrm{n}}\right]$

Then the non-invertible matrix ${\mathrm{A}}^{\mathrm{T}}\mathrm{A}$has 1's on the diagonal and cosA everywhere else as represented below:

${\mathrm{A}}^{\mathrm{T}}\mathrm{A}=\left[\begin{array}{cccc}1& \mathrm{肠辞蝉胃}& 鈥�& \mathrm{肠辞蝉胃}\\ \mathrm{肠辞蝉胃}& 1& 鈥�& \mathrm{肠辞蝉胃}\\ 鈰�& 鈰�& 鈰�& 鈰�\\ \mathrm{肠辞蝉胃}& \mathrm{肠辞蝉胃}& 鈥�& 1\end{array}\right]$

Therefore to find $\mathrm{胃}$we must take the inverse of the cosine and as we have n dimensions we take the inverse of the cosine of $-\frac{1}{\mathrm{n}}$as represented below:

$\mathrm{胃}={\cos }^{-1}\left(-\frac{1}{\mathrm{n}}\right)$