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Unformatted text preview: Chapter 6 More about Eigenval— ues, Eigenvectors
Section 6.3 The Singular Value Decomposition Recall that if A is real symmetric, then
A = UDUT, where D is a diagonal matrix whose entries are the eigenvalues of A, and U is an orthogonal matrix,
UTU = UUT = I. There is an analogue of this for general (not necessarily square) matrices: DeﬁnitiOn Singular Value Decomposition (page 411)
Let A be a matrix. Suppose we can ﬁnd matrices V and U with VTV = I and
UT U = I, and a diagonal matrix D with nonnegative entries, such that A = VDUT. Then we call this a singular value decomposition of A. The diagonal entries of
D are called the singular values of A. Remarks
(a) Note that the singular values of a square matrix A are not necessarily the same as the eigenvalues of A. When A is symmetric, they are the absolute
values of the eigenvalues of A.
(b) The singular value decomposition is not unique, but always exists. The theory on how to ﬁnd a singular value decomposition
Assuming the decomposition, we have ATA = (VDUT)T(VDUT)
= (UT)TDTVTVTDUT
m UD(VTV)DUT
= UDIDUT
= UD2UT. This is a just our old diagonalisation of B = ATA. Note that B is symmetric,
so this diagonalisation exists. So we proceed as follows: (I) Compute B = ATA; (II) Find the eigenvalues of ATA, and its eigenvectors. Apply Gram—Schmidt
to these eigenvectors, to obtain an orthogonal matrix U such that ATA = UD2UT. 122 The singular values of A are the diagonal entries of D, and hence the square
roots of the eigenvalues of B.
(III) Assume all diagonal entries of D are nonzero. Form V = AUD‘I.
(Proof that this works when U is square: With this choice of V, VDUT
= (AUD4)DUT
= AUIUT
= AUUT=AI=AJ (IV) The singular values of A are the square roots of the eigenvalues of B. 4 Find its singular value decomposition.
Solution
Form Example
Let l—‘CDM [ONO
I_—l B = ATA=[ = The characteristic polynomial of B = ATA is O
2 HON 2 1
0 2 H HMO
I—J pB(t) = det(B—tI) = dell"? 82]) = w—nQ—a—4
=t14a+w
= u—me—n. So the eigenvalues of B = ATA are 9 and 4. Note that they are both positive. The singular values of A will be x/g = 3 and x/il = 2. Next, we ﬁnd the
eigenvectors of B. We solve (B—pI)v=0 123 by appiying row reduction to [B#I§0]= [ 5;“ 83“ 8] Eigenvectors of B for ,u = 9
Here [B#no]= [ 2 _1 0
Step 1 Row 2 +% Row 1
—4 2 0
0 0 0
Then :31 is pivotal, 3:2 is not, so set $2 = t.
Step 2 Back substitution —43:1 +2$2=0
=> —4sc1 = —2:t:2 = —2t
=>x1 =ét.
Then %t '1
[ H i ]
Set 15:2, and Wm. W—MM:[;E Step 1 Row 2 —2 Row 1 Eigenvectors of B for ,u = 4
Here 8] l 2 0 0 0 0
Then 51:1 is pivotal, $2 is not, so set :32 = 1:.
Step 2 Back substitution $1+2$2=0 =>m1 =—2:122=2t.
—2t —2
v[ t H1] 124 Then Set t = 1, and
Summary . p, = 4 is an eigenvalue of B with eigenvector V2 = _ : . NH p. = 9 is an eigenvalue of B with eigenvector v1 = [ The singular values of A are 3 and 2 and
3 0
D — i 0 2 l ' To ﬁnd the orthogonal matrix U in
ATA = UD2UT, we apply Gram—Schmidt to the eigenvectors {V1,v2}, as we did for ordinary
eigenvalues and eigenvectors. Step 1: 111
Recall that 1
‘11 _' "—V1
V1 Here V1=Vl2+22= 5,
so u iv = i 1 ] 1 x/E 1 . J5 2 Step 2: 112 Since B is symmetric, and V2 corresponds to a different eigenvalue (namely 4)
from v1 (namely 9), we already know that V2  u1 = 0, so we don’t have to
subtract out the component. All we need to do is form 1
u2=—v2.
V2
Iiere
v2 = #22 + 12 = J5,
SO
viva2]
2 «52 «5 1 ‘
Then _ _ 1N5 —2/;/3
U ‘ [mug—[2% l/x/Sl a: ﬂ 125 So
T _ 9 0 T
AA—U[0 4107. To ﬁnd the singular value decomposition of A, we use V = AUD“1 ll
l"'"'—‘I
I—IOM
o
L—...._l
mwo A
1.4
01
I"—I
[Oi—l
pal
ix: 1 l—""——1
(ﬁnAN: How “to
I
pp. G)
§u
rhth _1_.
fix/5 —12 1
8 6 = —
10 0 3‘5 Thus the singular value decomposition of A is I—"—"'_ A=VDUT,
m" r “l
0 2
1 2
= (slééﬁl)[321<ssﬂf
Remark The singular value decomposition is useful in many contexts. For example, recall
that the least squares solution of Ax = b is solved via the normal equations
ATAX =ATb. One can shew that the least squares solution is given by x = UDlva. 126 ...
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This note was uploaded on 09/07/2010 for the course CALC 1502 taught by Professor Morley during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Morley

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