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aula12.html
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<!doctype html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0, maximum-scale=1.0, user-scalable=no">
<title>Aula 12 - Forma de Jordan </title>
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<div class="reveal">
<div class="slides">
<section>
<div style="background-color: rgb(126, 15, 34);
border: slategray;
border-width: 10px;
border-style: groove;">
<h2 style="color: rgb(250, 250, 250);">Tópico 12 <br> Forma Canônica de Jordan
<h4 style="color: white;">Pedro Aladar Tonelli</h4></div>
</section>
<section>
<h4 style="color: green;"> A Forma de Jordan </h4>
<p> Uma matriz $A\in \mathbb{C}^{n\times n}$ com $s$ autovetores
linearmente independentes, é semelhante a uma matriz em blocos:
</p>
$$ J = \begin{bmatrix} J_1 & 0 & \cdots & 0 \\
0 & J_2 & \cdots & 0 \\
\vdots & & \ddots & \vdots \\
0 & 0 & \cdots & J_s\end{bmatrix} $$
</section>
<section>
<p> Cada bloco de Jordan tem a forma.</p>
$$ J_i = \begin{bmatrix}
\lambda_i & 1 & 0 & \cdots & 0 \\
0 & \lambda_i & 1 & \cdots & 0 \\
\vdots & & \ddots & & \vdots \\
0 & 0 & \cdots & 0 & \lambda_i \end{bmatrix}$$
</section>
<section>
<h4>Autovetores generalizados</h4>
<p>Para achar a forma de Jordan, precisamos dos autovetores generalizados</p>
Se $\color{green}{x_1}$ é um autovetor de $A$ associado a $\lambda$, então o conjunto de vetores
$\{ \color{green}{x_1}, x_2, \dots, x_k\}$ que satisfaz
$$ \left( A - \lambda I\right)x_{i+1} = x_i $$ são os autovetores generalizados de ordem $i$.
</section>
<section>
<p>Para achar os autovetores generalizados, estudamos as matrizes nilpotentes </p>
<p> $A$ é nilpotente de ordem $k$ se $A^k=0$ e $A^{k-1}\neq 0$</p>
ou
<p>$A$ é nilpotente se e somente se $0$ é um autovalor de multiplicidade $n$</p>
</section>
<section>
<p> vamos construir uma sequência de autovetores generalizados de $A$ </p>
<p>Como $A^{k-1}\neq 0$ então existe um vetor $x \neq 0$ tal que $y=A^{k-1}x$ também
é diferente de zero.</p>
</section>
<section>
<p>Note $Ay = A^{k}x = 0$ então vamos renomear $\color{green}{x_1}=y$.</p>
<p>Note que neste caso todos os autovalores são nulos, então $(A-\lambda I)=A$</p>
<p>Como $x_1 = A^{k-1}x = A(A^{k-2}x)$ renomeamos $x_2 = A^{k-2}x$ ou ... </p>
$$ x_i = A^{k-i}x \text{ e } x_k =x $$
</section>
<section>
<p>$\left\{ \color{green}{x_1},x_2,\dots,x_k\right\}$ é linearmente independente.</p>
</section>
<section>
<p>$\left\{ \color{green}{x_1},x_2\right\}$ é LI pois se $x_2 = a x_1$ então $x_1=A
x_2 = a Ax_1 =0$ contra a hipótese.
</p>
<p>Agora suponha que $\left\{ \color{green}{x_1},x_2,\dots,x_{l-1}\right\}$ é LI e que </p>
\begin{gather}\alpha_1 x_1 + \cdots + \alpha_{l-1}x_{l-1} + \alpha_lx_l = 0 \implies \\
\alpha_1 Ax_1 + \cdots + \alpha_{l-1}Ax_{l-1} + \alpha_lAx_l = 0 \\
\alpha_2 x_1 + \cdots + \alpha_{l}x_{l-1} = 0 \implies \alpha_i =0
\end{gather}
</section>
</div>
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