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(ELEC560)627535 - midsol.pdf
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ELEC560 -Linear System Theory, Fall 2009-10
Reference solution to Midterm
November 11, 2009
1
Notice both Sn and Tn are closed under addition and scalar multiplication, so they are sub-spaces of Rnn .
.X Sn, LX = A.XA . X. We have
(LX). =(A.XA . X). = A.X.A . X. = A.XA . X = LX Hence, LX Sn and Sn is an invariant subspace of L.
Similarly, .Y Tn, LY = A.YA . Y .
(LY ). =(A.YA . Y ). = A.Y .A . Y . = A.(.Y )A . (.Y )= .(A.YA . Y )= .LY Therefore, LY Tn and Tn is an invariant subspace of L.
2
Let be the eigenvalue of A which has the maximal modulus, i.e., || = (A). Let u be an eigenvector corresponding to with unit length. Then
.A.2 = sup .Ax.2 .Au.2 = .u.2 = ||.u.2 = || = (A)
.x.2=1
Up to now, we have shown .A.2 (A). In the following we will show for any .> 0, a nonsingular matrix P can be found such that .P .1AP .2 <(A)+ ..
According to Schur theorem, there exists an unitary matrix U such that U.1AU = T , where T is upper triangular with diagonal entries 1,2,3,...,n which are eigenvalues of A. Let
.1.2.3
Dz = diag(z,z,z,...,z.n), where z> 0, then
..
..

.1.2.n+1
1 zt12 zt13 zt1n .1.n+2
2 zt23 zt2n 3 z.n+3
D.1
z
TDz = t3n
.
.
..
.
.
n
..
= E + F where E = diag(1,2,3,...,n) and
..
..

.1.2.n+1
0 zt12 zt13 zt1n .1.n+2
0 zt23 zt2n
.
..
F =
.... .
.
.1
tn.1,n
0
z
..
0
We know that .D.1TDz.2 = .E + F .2 .E.2 + .F .2 = (A)+ .F .2. It can be easily seen
z
..
that for any .> 0, we can make z large enough such that .F .2 <.. It follows that .D.1TDz.2 = .D.1U.1AUDz.2 (A)+ .F .2 <(A)+ .
zz
Let P = UDz, then .P .1AP .2 <(A)+ ..
The singular values of A are easy to see. Denote . ,V
. ..
.. ..
(d1)
1 1 d 0,where (d).
U
=
=.

, .1 d< 0.
. 1
(d2n)
..
then U.AV = diag(|d1|, , |d2n|) is a singular value decomposition of A. Hence the singular value of A are given by |d1|, , |d2n|.
The eigenvalues of A can be derived in several ways.
Method I
Note that
. =
. ..
. .. . ..
.. .. ..
1
1
d1 1d1 d1
..
. .
.. .
. 2
1d1 .1d1
d2n d2n
1 1
.1
..
is a similarity transformation. When d1 ./d1, the term d2n . 2= 0, by setting = .d2n1d1 is
eliminated.
When d1 = 0 we need a di.erent similarity transformation.

. =
..
. .. . .. . ..
.. .. .. ..
01 010
d2n
..
. .
... . 0
d2n
10 10
These ideas can be combined to solve the general problem. First apply the following transfor-
mation.
.
A := P AP =
..

.
d1
..
,
.
d2n
where
..
..

1 . 1 1
...
1 . n n

P =
n.1 1 . n.1
. ..
2n 1 . 2n

..
and . 0 when di d2n+1.i
i = 2n+1.i = , for i =1, , n.
1 when di <d2n+1.i
This makes sure that A.has the same eigenvalues as A but d.i d.2n+1.i and did2n+1.i = d.id.2n+1.i when i