Problema de forma canònica de Jordan

Problema de forma canònica de Jordan
11 de maig de 2020 No hi ha comentaris Àlgebra, Matemàtiques 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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