Problema de forma canònica de Jordan
11 de maig de 2020 No hi ha comentaris General Oscar Alex Fernandez Mora

Determini la forma canònica de Jordan $\pmb{J}$ del endomorfisme $\pmb{f}$ d’un $K$-espai vectorial la qual té com a matriu respecte d’una base $\pmb{\mathcal{B} = {v_1, v_2, v_3, v_4}}$ és:

\begin{equation}
\pmb{ A = \left( \begin{array}{cccc}
2 & 0 & 0 & 0 \\
1 & 3 & -1 & 0 \\
1 & 1 & 1 & 0 \\
1 & -1 & 1 & 2\end{array} \right)}
\end{equation}

Primer haurem de calcular el polinomi característic de $f$, o d’$A$, al polinomi de grau $n$ en la indeterminada $\lambda$} de la matriu $A$

\begin{equation}
p_f(\lambda) = det (A-\lambda I)
\end{equation}

\begin{equation}
\left( \begin{array}{cccc}
2-\lambda & 0 & 0 & 0 \\
1 & 3-\lambda & -1 & 0 \\
1 & 1 & 1-\lambda & 0 \\
1 & -1 & 1 & 2-\lambda\end{array} \right) = (2-\lambda)\left| \begin{array}{ccc}
3-\lambda & -1 & 0 \\
1 & 1-\lambda & 0 \\
-1 & 1 & 2-\lambda \end{array} \right| =
\end{equation}

\begin{equation}
= (2-\lambda)\cdot(2-\lambda)\cdot\left| \begin{array}{cc}
3-\lambda & -1 \\
1 & 1-\lambda \\
\end{array} \right| =
\end{equation}

\begin{equation}
= (2-\lambda)\cdot(2-\lambda)\cdot[(3-\lambda)\cdot(1-\lambda)-1] =
\end{equation}

\begin{equation}
= (2-\lambda)\cdot(2-\lambda)[3-3\lambda-\lambda+\lambda^2-1] =
\end{equation}

\begin{equation}
= (2-\lambda)\cdot(2-\lambda)\cdot(\lambda^2-4\lambda+2) =
\end{equation}

\begin{equation}
= (2-\lambda)\cdot(2-\lambda)\cdot(\lambda-2)^2\longrightarrow(\lambda-2)^4
\end{equation}

Ara igualem el polinomi característic a zero per tal de trobar els valors propis

Si $ p_f(\lambda) = 0\longrightarrow\lambda_1 = 2$ amb multiplicitat algebraica $a_1 = 4$

Ara trobarem els subespais propis generalitzats:

$K^1(2) = \ker(f-\lambda I)$

\begin{equation}
\left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
1 & 1 & -1 & 0 \\
1 & 1 & -1 & 0 \\
1 & -1 & 1 & 0\end{array} \right)\cdot\left(\begin{array}{c}
x_1 \\
x_2 \\
x_3 \\
x_4 \end{array}\right) = \left(\begin{array}{c}
0 \\
0 \\
0 \\
0 \end{array}\right)
\end{equation}

\begin{equation}
\left\{ \begin{array}{cc}
x_1+x_2-x_3 & = & 0 \\
x_1+x_2-x_3 & = & 0 \\
x_1-x_2+x_3 & = & 0 \\
\end{array} \right.\longrightarrow\left\{\begin{array}{cc}
x_1+x_2-x_3 & = & 0 \\
2x_1 &= & 0 \\
\end{array}\right. \sim \left\{\begin{array}{cc}
x_2 & = & x_3 \\
x_1 & = & 0 \\
\end{array}\right.
\end{equation}

Llavors tenim $\boxed{\dim K^1(2)} = \dim\ker(f-2I)=n-rang(f-2I) = 4-2 = \boxed{2}$ això és la multiplicitat geomètrica $g_1 = 2$.

Una base de $K^1$ podria ser $\mathcal{B}_{K^1} = \{u_1=(0, 1, 1, 1), u_2=(0, 1, 1, 0)\}$

$K^2(2) = \ker(f-\lambda I)^2$

\begin{equation}
\left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
1 & 1 & -1 & 0 \\
1 & 1 & -1 & 0 \\
1 & -1 & 1 & 0\end{array} \right)^2\cdot\left(\begin{array}{c}
x_1 \\
x_2 \\
x_3 \\
x_4 \end{array}\right) = \left(\begin{array}{c}
0 \\
0 \\
0 \\
0 \end{array}\right)
\end{equation}

\begin{equation} \left( \begin{array}{cccc}0 & 0 & 0 & 0 \\
1 & 1 & -1 & 0 \\
1 & 1 & -1 & 0 \\
1 & -1 & 1 & 0\end{array} \right)\cdot\left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
1 & 1 & -1 & 0 \\
1 & 1 & -1 & 0 \\
1 & -1 & 1 & 0\end{array} \right) = \left(\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0\end{array} \right)\longrightarrow \boxed{rg = 0}
\end{equation}

Per tant tenim $rg(A-2I)^2 = 0$

Amb conseqüència:

\begin{equation}
\boxed{\dim K^2(2)} = \dim\ker(f-2I)^2 = n-rg(A-2I)^2 = 4-0 = \boxed{4}\longrightarrow K^2={\RR}^4
\end{equation}

Per tant el quadre ens queda com:

\begin{array}{cccc}
&{}_2& &{}_4\\
{0} &K^1(2) &\subseteq& K^2(2)={\RR}^4 = M(2)\\
&w_2 &\longleftarrow& w_1\\
&w_4 &\longleftarrow& w_3\\
\end{array}

Sabem que:

\begin{equation}
w_1, w_3\in K^2(2)-K^1(2)\longrightarrow K^1(2)\oplus
L(w_1, w_3) = K^2(2)
\end{equation}

Tenim:

\begin{equation}
w_2 = (f-2I)\cdot(w_1)
\end{equation}

\begin{equation}
w_4 = (f-2I)\cdot(w_3)
\end{equation}

La matriu de Jordan ens quedaria com:

\begin{equation}
J = \mathfrak{M}_\mathcal{B’}(f) = \left( \begin{array}{cc|cc}
2 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 \\
\hline
0 & 0 & 2 & 0 \\
0 & 0 & 1 & 2\end{array} \right)
\end{equation}

Ara trobarem les imatges del vector $u_1$ i $u_1$:

\begin{equation}
f(u_1) = \left( \begin{array}{cccc}
2 & 0 & 0 & 0 \\
1 & 3 & -1 & 0 \\
1 & 1 & 1 & 0 \\
1 & -1 & 1 & 2\end{array} \right)\cdot\left(\begin{array}{c}
0 \\
1 \\
1 \\
1 \end{array}\right) = \left(\begin{array}{c}
0 \\
2 \\
2 \\
2 \end{array}\right) = \left(\begin{array}{c}
0 \\
1 \\
1 \\
1 \end{array}\right)
\end{equation}

\begin{equation}
f(u_2) = \left( \begin{array}{cccc}
2 & 0 & 0 & 0 \\
1 & 3 & -1 & 0 \\
1 & 1 & 1 & 0 \\
1 & -1 & 1 & 2\end{array} \right)\cdot\left(\begin{array}{c}
0 \\
1 \\
1 \\
0 \end{array}\right) = \left(\begin{array}{c}
0 \\
2 \\
2 \\
0 \end{array}\right) = \left(\begin{array}{c}
0 \\
1 \\
1 \\
0 \end{array}\right)
\end{equation}

Llavors:

\begin{equation}
f(u_1) = \alpha u_1+\beta u_2
\end{equation}

\begin{equation}
(0, 1, 1, 1) = \alpha(0, 1, 1, 1)+\beta(0, 1, 1, 0)
\end{equation}

\begin{equation}\begin{array}{l}
1 = \alpha+\beta \\
1 = \alpha+\beta \\
1 = \alpha\longrightarrow\alpha = 2; \beta = 0\longrightarrow w_1 = <(1, 0, 0, 0)>\end{array}
\end{equation}

\begin{equation}
f(u_2) = \alpha u_1+\beta u_2
\end{equation}

\begin{equation}
(0, 1, 1, 0) = \alpha(0, 1, 1, 1)+\beta(0, 1, 1, 0)
\end{equation}

\begin{equation}\begin{array}{l}
1 = \alpha+\beta \\
1 = \alpha+\beta \\
0 = \alpha\longrightarrow\alpha = 0; \beta = 1\longrightarrow w_3 = <(0, 1, 0, 0)>\end{array}
\end{equation}

On $K^1(2) = L(w_1, w_3) = L((1, 0, 0, 0), (0, 1, 0, 0))$, per tant $w_2$ i $w_4$ els calcularem:

\begin{equation}
w_2 = (f-2I)\cdot(w_1) = \left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
1 & 1 & -1 & 0 \\
1 & 1 & -1 & 0 \\
1 & -1 & 1 & 0\end{array} \right)\cdot\left(\begin{array}{c}
1 \\
0 \\
0 \\
0 \end{array}\right) = \left(\begin{array}{c}
0 \\
1 \\
1 \\
1 \end{array}\right)
\end{equation}

\begin{equation}
w_4 = (f-2I)\cdot(w_3) = \left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
1 & 1 & -1 & 0 \\
1 & 1 & -1 & 0 \\
1 & -1 & 1 & 0\end{array} \right)\cdot\left(\begin{array}{c}
0 \\
1 \\
0 \\
0 \end{array}\right) = \left(\begin{array}{c}
0 \\
1 \\
1 \\
-1\end{array}\right)
\end{equation}

Per tant la matriu $P$ ens queda formada tal com (ficant els vectors $w_1, w_2, w_3, w_4$ en columnes):

\begin{equation}
P = \left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 \\
0 & 1 & 0 & -1\end{array} \right)
\end{equation}

Sempre es complirà:

\begin{equation}
J = P^{-1}\cdot A\cdot P = \left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 \\
0 & 1 & 0 & -1\end{array} \right)^{-1}\cdot \left( \begin{array}{cccc}
2 & 0 & 0 & 0 \\
1 & 3 & -1 & 0 \\
1 & 1 & 1 & 0 \\
1 & -1 & 1 & 2\end{array} \right)\cdot\left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 \\
0 & 1 & 0 & 1 \\
0 & 1 & 0 & -1\end{array} \right)
\end{equation}

\begin{equation}
J = \left( \begin{array}{cccc}
2 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 1 & 2\end{array} \right)
\end{equation}

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Sobre l'autor
Oscar Alex Fernandez Mora Etern estudiant de la Rússia tsarista. Gran aficionat als destil·lats i als fermentats. Malaltís de llibres de la MIR i entusiasta del #LaTeX. Soci de l’ACBC. Important actiu del projecte Campana de Gauss

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