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Unformatted text preview: dt This is a system of two equations and two unknowns, which we can rewrite in matrix form:
� � � T3 T2 � �
�
y a1
3
2
,
= T 2
a2
0
T
2
So,
� a1
a2 � �
= T3
3
T2
2 T2
2 �−1 � T y
0 �
. Exercise 4.1 Note that for any v ∈ C m , (show this!) �v �∞ ≤ �v �2 ≤ √
m�v �∞ . (1) Therefore, for A ∈ C m×n with x ∈ Cn
�Ax�2 ≤ √
√
Ax
�∞
m�Ax�∞ → for x = 0, ��x��2 ≤ m �Ax�2 .
�
�x
2 But, from equation (1), we also know that �Ax�2 ≤
�x�2 1
�x�∞ ≥ 1
�x�2 . Thus, √
√
m�Ax�∞
m�Ax�∞ √
≤ m�A�∞ ,
≤
�x�2
�x�∞ �
Equation (2) must hold for all x = 0, therefore
3 (2) maxx=0
�
To prove the lower bound �Ax�∞ 1
√ �A�∞
n �Ax�2
�x�2 = �A�2 ≤ √
m�A�∞ . ≤ �A�2 , reconsider equation (1): �Ax�∞
�Ax�2
≤ �Ax�2 → for x �= 0,
≤
≤ �A�2 →
�x�2
�x�2 But, from equation (1) for x ∈ C n , √ n
�x�2 �Ax�∞
�x�∞
for all x = 0 including x that makes
� ≥ �x1 ∞ . So,
�
√
n�Ax�∞ √
≤
≤ n�A�2
�x�2 �Ax�∞
�x�∞ maxx=0
� √
√
n�Ax�∞
n�Ax�2 √
≤
≤ n�A�2 .
�x�2
�x�2
(3) maximum, so, �Ax�∞
�x�∞ = �A�∞ ≤ √ n�A�2 , or equivalently,
1
√ �A�∞ ≤ �A�2 .
n Exercise 4.5 Any m × n matrix A, it can be expressed as
�
�
Σ0
A=U
V �,
00
where U and V are unitary matrices. The “MoorePenrose inverse”, or pseudoinverse of A,
denoted by A+ , is then deﬁned as the n × m matrix
� −1
�
Σ
0
A+ = V
U �.
0
0
a) Now we have to show that A+ A and AA+ are symmetric, and that AA+ A = A and A+ AA+ =
A+ . Suppose that Σ is a diagonal invertible matrix with the dimension of r × r. Using the given
deﬁnitions as well a...
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This document was uploaded on 03/19/2014 for the course EECS 6.241J at MIT.
 Spring '11
 EmilioFrazzoli
 Computer Science, Electrical Engineering

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