The function jordan(A) computes the Jordan form of a matrix A.

When the function is called using the form J = jordan(A), the returned value of J is the Jordan form of the matrix A.

When the function is called using the form [V,J] = jordan(A), the returned value of J is the Jordan form of the matrix A, and the returned value of V is the transformation matrix corresponding to this Jordan form.

$\mathrm{with}\left(\mathrm{MTM}\right)\:$

$A≔\mathrm{Matrix}\left(\left[\left[1\,2\,1\right]\,\left[2\,4\,2\right]\,\left[2\,8\,1\right]\right]\right)$

$A≔\left[\begin{array}{ccc}1& 2& 1\\ 2& 4& 2\\ 2& 8& 1\end{array}\right]$

$J≔\mathrm{jordan}\left(A\right)$

$J≔\left[\begin{array}{ccc}0& 0& 0\\ 0& 3-\sqrt{22}& 0\\ 0& 0& 3+\sqrt{22}\end{array}\right]$

$V,J≔\mathrm{jordan}\left(A\right)$

$V,J≔\left[\begin{array}{ccc}\frac{12}{13}& \frac{2\sqrt{22}-11}{22\left(-3+\sqrt{22}\right)}& \frac{\left(10+\sqrt{22}\right)\sqrt{22}}{572}\\ -\frac{2}{13}& \frac{2\sqrt{22}-11}{11\left(-3+\sqrt{22}\right)}& \frac{\left(10+\sqrt{22}\right)\sqrt{22}}{286}\\ -\frac{8}{13}& \frac{7\sqrt{22}-22}{22\left(-3+\sqrt{22}\right)}& \frac{\left(1+4\sqrt{22}\right)\sqrt{22}}{286}\end{array}\right],\left[\begin{array}{ccc}0& 0& 0\\ 0& 3-\sqrt{22}& 0\\ 0& 0& 3+\sqrt{22}\end{array}\right]$