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lista12.tex
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% Filename: lista12.tex
%
% This code is part of 'Solutions for MT402, Matrizes'
%
% Description: This file corresponds to the solutions of homework sheet 12.
%
% Created: 16.06.12 07:56:48 AM
% Last Change: 29.06.12 05:37:50 PM
%
% Authors:
% - Raniere Silva (2012): initial version
%
% Copyright (c) 2012 Raniere Silva <[email protected]>
%
% This work is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/3.0/ or send a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA.
%
% This work is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
%
\begin{questions}
\question Considere $A : m \times n$, sua decomposi\c{c}\~{a}o SVD, $A = U D V^t$ e pseudoinversa $A^\dagger = V D^\dagger U^t$. Demonstre que $A^\dagger$ satisfaz as quatro condi\c{c}\~{o}es que caracterizam uma matriz como pseudoinversa de $A$: $B A B = B$, $A B A = A$, $(B A)^t = BA$ e $(A B)^t = A B$.
\begin{solution}
Pela constru\c{c}\~{a}o da decomposi\c{c}\~{a}o SVD temos que
\begin{align*}
D &= \begin{bmatrix}
\hat{D} & 0 \\
0 & 0
\end{bmatrix}
\end{align*}
e pela constru\c{c}\~{a}o da pseudoinversa temos que
\begin{align*}
D^\dagger &= \begin{bmatrix}
\hat{D}^{-1} & 0 \\
0 & 0
\end{bmatrix}.
\end{align*}
Logo,
\begin{align*}
D D^\dagger = \begin{bmatrix}
\hat{D} & 0 \\
0 & 0
\end{bmatrix} \begin{bmatrix}
\hat{D}^{-1} & 0 \\
0 & 0
\end{bmatrix} = \begin{bmatrix}
I & 0 \\
0 & 0
\end{bmatrix} = \begin{bmatrix}
\hat{D} & 0 \\
0 & 0
\end{bmatrix} \begin{bmatrix}
\hat{D}^{-1} & 0 \\
0 & 0
\end{bmatrix} = D^\dagger D.
\end{align*}
Outras propriedades que decorrem da constru\c{c}\~{a}o da decomposi\c{c}\~{a}o SVD s\~{a}o:
\begin{align*}
U U^t &= I, & U^t U &= I, \\
V V^t &= I, & V^t V &= I.
\end{align*}
Para $B A B = B$, temos
\begin{align*}
B A B &= A^\dagger A A^\dagger \\
&= \left( V D^\dagger U^t \right) \left( U D V^t \right) \left( V D^\dagger U^t \right) \\
&= V D^\dagger I D I D^\dagger U^t \\
&= V \left( D^\dagger D \right) D^\dagger U^t \\
&= V D^\dagger U^t \\
&= B.
\end{align*}
Para $A B A = A$, temos
\begin{align*}
A B A &= A A^\dagger A \\
&= \left( U D V^t \right) \left( V D^\dagger U^t \right) \left( U D V^t \right) \\
&= U D I D^\dagger I D V^t \\
&= U \left( D D^\dagger \right) D V^t \\
&= U D V^t \\
&= A.
\end{align*}
Para $\left( B A \right)^t = B A$, temos
\begin{align*}
\left( B A \right)^t &= \left( A^\dagger A \right)^t \\
&= \left( V D^\dagger U^t U D V^t \right)^t \\
&= V D^t U^t U (D^\dagger)^t V^t \\
&= V D U^t U D^\dagger V^t \\
&= V D \left( I \right) D^\dagger V^t \\
&= V D D^\dagger V^t \\
&= V D^\dagger D V^t \\
&= V D^\dagger \left( I \right) D V^t \\
&= V D^\dagger \left( U^t U \right) D V^t \\
&= \left( V D^\dagger U^t \right) \left( U D V^t \right) \\
&= A^\dagger A \\
&= B A
\end{align*}
Para $\left( A B \right)^t = A B$, temos
\begin{align*}
\left( A B \right)^t &= \left( A A^\dagger \right)^t \\
&= \left( U D V^t V D^\dagger U^t \right)^t \\
&= U \left( D^\dagger \right)^t V^t V D^t U^t \\
&= U D^\dagger V^t V D U^t \\
&= U D^\dagger \left( I \right) D U^t \\
&= U D^\dagger D U^t \\
&= U D D^\dagger U^t \\
&= U D \left( I \right) D^\dagger U^t \\
&= U D \left( V^t V \right) D^\dagger U^t \\
&= \left( U D V^t \right) \left( V D^\dagger U^t \right) \\
&= A A^\dagger \\
&= A B.
\end{align*}
\end{solution}
\question Considere a matriz $A : m \times n$, $m \geq n$ e o sistema linear $A x = b$. Usando a SVD demonstre que:
\begin{parts}
\part a solu\c{c}\~{a}o de quadrados m\'{i}nimos na norma-2 m\'{i}nima \'{e} $\overline{x} = \sum_{i = 1}^r (\sigma_i)^{-1} u_i^t b v_i$,
\begin{solution}
Temos que
\begin{align*}
\| A x - b \|_2^2 &= \| U D V^t x - b \|_2^2 \\
&= \| U^t \left( U D V^t x - b \right) \|_2^2 \\
&= \| U^t U D V^t x - U^t b \|_2^2 \\
&= \| I D V^t x - U^t b \|_2^2 \\
&= \| D V^t x - U^t b \|_2^2 \\
&= \| D \hat{x} - \hat{b} \|_2^2 \\
&= \sum_{i = 1}^n | \sigma_i \hat{x}_i - \hat{b}_i |^2 \\
&= \left( \sum_{i = 1}^r | \sigma_i \hat{x}_i - \hat{b}_i |^2 \right) + \left( \sum_{i = r + 1}^n \hat{b}_i^2 \right).
\end{align*}
Portanto
\begin{align*}
\min \| A x - b \|_2^2 &= \min \left[ \left( \sum_{i = 1}^r | \sigma_i \hat{x}_i - \hat{b}_i |^2 \right) + \left( \sum_{i = r + 1}^n \hat{b}_i^2 \right) \right]
\end{align*}
que ocorre quando
\begin{align*}
\left( \sum_{i = 1}^r | \sigma_i \hat{x}_i - \hat{b}_i |^2 \right) &= 0.
\end{align*}
Ent\~{a}o
\begin{align*}
\sigma_i \hat{x}_i - \hat{b}_i = 0, \forall i = 1, \ldots, r,
\end{align*}
que implica em
\begin{align*}
\hat{x}_i &= \hat{b}_i / \sigma_i, \forall i = 1, \ldots, r, \\
x &= V \hat{x} \\
&= \sum_{i = 1}^n \hat{x}_i V_i \\
&= \sum_{i = i}^r \left( \hat{b}_i / \sigma_i \right) V_i \\
&= \sum_{i = 1}^r \left( \sigma_i \right)^{-1} U_i^t b V_i.
\end{align*}
\end{solution}
\part o valor m\'{i}nimo da norma-2 do res\'{i}duo \'{e} $\sqrt{\sum_{i = r + 1}^m (u_i^t b)^2}$.
\begin{solution}
Pelo item acima temos que o res\'{i}duo corresponde a $\sum_{i = r + 1}^n \hat{b}_i^2$ que corresponde a $\sum_{i = r + 1}^m (u_i^t b)^2$.
\end{solution}
\end{parts}
\question Seja $A : m \times n$, $m \geq n$ e o sistema linear $A x = b$. Seja $\overline{x}$ a solu\c{c}\~{a}o de norma-2 m\'{i}nima de $\min_x r = \| b - A x \|_2$. Demonstre que: $\overline{x} = A^\dagger b$.
\begin{solution}
Pelo exerc\'{i}cio anterior, a solu\c{c}\~{a}o de norma-2 \'{e}
\begin{align*}
\overline{x} &= \sum_{i = 1}^r \left( \sigma_i \right)^{-1} U_i^t b V_i
\end{align*}
que pode ser colocada na forma
\begin{align*}
\overline{x} &= \sum_{i = 1}^r \left( \sigma_i \right)^{-1} V_i U_i^t b \\
&= \hat{V} \hat{D}^{-1} \hat{U}^t b,
\end{align*}
onde
\begin{align*}
D = \begin{bmatrix}
\hat{D} & 0 \\
0 & 0
\end{bmatrix}.
\end{align*}
Portanto,
\begin{align*}
\overline{x} &= V D^\dagger U^t b \\
&= A^\dagger b.
\end{align*}
\end{solution}
\question \'{E} correto afirma que $A A^\dagger$ \'{e} uma matriz de proje\c{c}\~{a}o ortogonal? Em que subespa\c{c}o? Justifique usando pelo menos duas argumenta\c{c}\~{o}es diferentes.
\begin{solution}
Observer que $\left( A A^\dagger \right)^t = A A^\dagger$ e $\left( A A^\dagger \right)^2 = A A^\dagger A A^\dagger = A A^\dagger$. Logo, $A A^\dagger$ \'{e} uma matriz de proje\c{c}\~{a}o ortogonal.
Observe ainda que $x = A^\dagger b$ \'{e} solu\c{c}\~{a}o de quadrados m\'{i}nimos e como $A x \in \mathcal{I}(A)$, i.e., $A x = A A^\dagger b$, conclui-se que $A A^\dagger$ \'{e} a matriz de proje\c{c}\~{a}o em $\mathcal{I}(A)$.
\end{solution}
\question \'{E} correto afirma que $A^\dagger A$ \'{e} uma matriz de proje\c{c}\~{a}o ortogonal? Em que subespa\c{c}o? Justifique usando pelo menos duas argumenta\c{c}\~{o}es diferentes.
\begin{solution}
Observer que $\left( A^\dagger A \right)^t = A^\dagger A$ e $\left( A^\dagger A \right)^2 = A^\dagger A A^\dagger A = A^\dagger A$. Logo, $A^\dagger A$ \'{e} uma matriz de proje\c{c}\~{a}o ortogonal.
Observe ainda que $y \in \mathcal{I}(A^t) \Rightarrow y = A^t b$ e $A^t x = b \Rightarrow x = \left( A^\dagger \right)^t b$. Portanto, $A^t x = y = A^t \left( A^\dagger \right)^t b \Rightarrow y = \left( A^\dagger A \right)^t b = A^\dagger A b$. Portanto, $A^\dagger A$ \'{e} a proje\c{c}\~{a}o ortogonal de $b$ na imagem de $A^\dagger$.
\end{solution}
\question Demonstre que se $A : m \times n$, $m \geq n$ e $\textrm{posto}(A) = n$, ent\~{a}o $A^\dagger = (A^t A)^{-1} A^t$.
\begin{solution}
A solu\c{c}\~{a}o de quadrados m\'{i}nimos para $A x = b$ \'{e} $\overline{x} = \left( A^t A \right)^{-1} A^t b$ pois $A^t A : n \times n$ tem posto completo. Logo, como a solu\c{c}\~{a}o \'{e} \'{u}nica ($A^t A$ invers\'{i}vel) temos que $A^\dagger b = \left( A^t A \right)^{-1} A^t b \Rightarrow A^\dagger = \left( A^t A \right)^{-1} A^t$.
\end{solution}
\question Apresente diferentes demonstra\c{c}\~{o}es para a igualdade: $\| A A^\dagger \|_2 = 1$.
\begin{solution}
\textbf{Primeira demonstra\c{c}\~{a}o:}
$A A^\dagger$ \'{e} uma matriz de proje\c{c}\~{a}o e portanto $\| A A^\dagger \|_2 = 1$.
\textbf{Segunda demonstra\c{c}\~{a}o:}
Temos que
\begin{align*}
\| A A^\dagger \|_2 = \sqrt{\lambda_{\max}(A A^\dagger A A^\dagger)}.
\end{align*}
Mas
\begin{align*}
\left( A A^\dagger \right)^t \left( A A^\dagger \right) = A A^\dagger A A^\dagger = A A^\dagger = U D D^t U^t
\end{align*}
e os autovalores de $D D^t$ s\~{a}o iguais a $1$. Ent\~{a}o $\| A A^\dagger \|_2 = 1$.
\end{solution}
\question Demonstre que a pseudoinversa de $A$, $A^\dagger = U D^\dagger V^t$, \'{e} a matriz de norma de Frobenius m\'{i}nima que resolve o problema $\min \| A B - I_m \|_F$, $B : n \times m$.
A matriz $A^\dagger$ \'{e} a \'{u}nica solu\c{c}\~{a}o deste problema? Depende dos fatores $U$ e $V$?
\begin{solution}
% TODO Fazer esse exerc\'{i}cio.
\end{solution}
\question Considere a matriz $A : m \times p$ com colunas $w_i$, $i = 1, \ldots, p$ e $B : m \times (p + 1)$ com colunas $w_i$, $i = 1, \ldots, p$ iguais \`{a}s colunas de $A$ e coluna $w_{p + 1} \in \mathcal{I}(A)$.
Considere os sistemas lineares: $A x = b$ e $B y = b$, com $b \not\in \mathcal{I}(A)$.
\begin{parts}
\part Mostre que $\hat{y} = (A^\dagger b, 0)^t$ \'{e} uma solu\c{c}\~{a}o de Quadrados M\'{i}nimos para $B y = b$;
\begin{solution}
% TODO Fazer esse exerc\'{i}cio.
\end{solution}
\part $\| A^\dagger b \|_2 \geq \| B^\dagger b \|_2$? Justifique.
\begin{solution}
% TODO Fazer esse exerc\'{i}cio.
\end{solution}
\end{parts}
\question Supor que $A : m \times n$ est\'{a} particionada na forma $(A_1 , A_2)^t$ onde $A_1 : n \times n$ \'{e} n\~{a}o singular e $A_2 (m - n) \times n$. Prove que $\| A^\dagger \|_2 \leq \| A_1^{-1} \|_2$.
\begin{solution}
% TODO Fazer esse exerc\'{i}cio.
\end{solution}
\end{questions}