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LinAlgChapter6

# LinAlgChapter6 - Chapter 6 More about Eigenval— ues...

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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 non-negative 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 non-zero. 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 viva-2] 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 (ﬁn-AN: 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 = UD-lva. 126 ...
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LinAlgChapter6 - Chapter 6 More about Eigenval— ues...

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