AN UPPER ESTIMATE FOR THE LARGEST SINGULAR VALUE OF A SPECIAL MATRIX, II

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1 International Journal of Pure and Applied Mathematics Volume 5 No. 4 07, ISSN: printed version); ISSN: on-line version) url: doi: 0.73/ijpam.v5i4.0 PAijpam.eu AN UPPR STIMAT FOR TH LARGST SINGULAR VALU OF A SPCIAL MATRIX, II Svetoslav I. Nenov Department of Mathematics University of Chemical Technology and Metallurgy Sofia 756, BULGARIA Abstract: Our goal is to prove an upper bound for the largest singular value of the following matrix A = D t D ) t. Here t denotes the matrix transpose, D is a non-singular matrix, and is thin matrix. AMS Subject Classification: 5A8 Key Words: singular values, largest singular value. Introduction In many applications, we have to deal with a matrix in the form t is matrix transpose) A = D t D ) t, where:. D is a symmetric, positive definite m m)-matrix.. is a full rank m n)-matrix. 3. It follows from ), n m that det t D ) 0. Received: May 4, 07 Revised: July, 07 Published: August 9, 07 c 07 Academic Publications, Ltd. url:

2 868 S.I. Nenov Our goal is to establish an upper bound for A. In many applications D is a diagonal matrix, see for example the results in [,, 4, 5, 6, 7, 0] for simple constructions of moving least squares approximations.. Main Result Lemma.. Let < n m and rank t ) = n. Then σ max A) = A kd)k ) = σ maxd) σmax ) σ min D) σmin ), where k ) resp. σ min ), σ max )) is the condition number resp. lower, largest singular value). Proof. The matrix D is a symmetric positive definite matrix. Then there exist two matrxices D and D such that: D = ), D D ) = D. We will separate the proof in several steps.. Let D = UΣV t be the singular value decomposition of D, where: a) U and V are unitary matrices. b) If σ i are the singular values of m n)-matrix D, n m, then Σ = ) Σ = 0 σ σ σ n

3 AN UPPR STIMAT FOR TH LARGST SINGULAR We have using [3, Chapter 3, P3-SVD and P5-SVD]): ) ) σ max D σ max D σ max ) = σ max) σmin D). Here σ min D) 0, because D is non-sigular matrix. 3. We have using [3, Chapter 3, P3-SVD and P7-SVD]): ) ) σ min D σ min D σ min ) = σ min) σmax D). Here σ max D) σ min D) is thin i.e. n m) matrix and rank ) = n. Then rank ) = rank t ) = n. Indeed,let = U Σ V t bethesingularvaluedecompositionofm n)- matrix, n m. Then t = U Σ V t ) t U Σ V t ) = V Σ t Σ V. Therefore rank t ) = n, because rank Σ t Σ ) = n. Moreover using thedefinitionofsingularvalue), thematrix hasmin{m,n} = nsingular values, and all non-zero singular values of are the the square roots of the non-zero eigenvalues of both t and t. But all eigenvalues of t are strictly positive. Hence 0 < σ min ) σ max ). 5. Obviously D and D are symmetric matrices) ) t D = t D )D ) t ) = D D = UΣV t) t UΣV t ) =VΣ t U t UΣV t =V Σ 0 ) Σ 0 =VΣ V t. is the singular value decomposition of t D. ) V t

4 870 S.I. Nenov 6. Using the previous item and the unitaty property of V, we have and t D ) = V Σ V t) 7. Using the previous item, we have = V t) Σ V) =V Σ V t, A =D t D ) t =D V t Σ V t. σ max A) σ max D ) σ max )σ max σ = max) σ min D)σ min Σ ), ) Σ because all singular values of an unitary matrix are indeed, the singular value decomposition of the unitaty matrix V is V = V II) and σ min Σ ) = σ min D ) > Therefore σ σ max A) max) ) σ min D)σ min Σ σmax ) σ min ) σ min D) σmax D) = σ maxd) σmax) σ min D) σmin ). ) Lemma.. Let < n m and rank t ) = n. Then σ min A) σ max A) = A.

5 AN UPPR STIMAT FOR TH LARGST SINGULAR Proof. Obviously t A = t D t D ) t = t. It follows from [3, Chapter 3, P5-SVD a)] that σ max t ) σ min A) σ max t A ) = σ max t ). Indeed t is n m)-matrix and n m. Hence σ min A). On the other hand, using [3, Chapter 3, P6-SVD], we recceive Hence σ max A). σ min t ) = σ min t A ) σ max A)σ min t ) Combining Lemma. and Lemma., we receive the following result. Theorem.. Let < n m and rank t ) = n. Then A kd)k ). Remark.. If m = n, then det) 0 and A = I. In this case, we receive the following obvious inequality σ max D) σmax t ) σ min D) σmin t ). Remark.. If n =, then the matrix has only one singular value. Hence, in this case σ max A) = A kd). Remark.3. We cannot omit the term σmaxd) σ min D) a diagonal matrix). A simple example: Let Dx) = ) 3 0, x R 0 x +, = even supposed D to be ). 3 Then Ax) = x+7 ) x 3x 9 7 and σ max Ax)) = 0 x +8 >. x+7

6 87 S.I. Nenov But the matrix has only one singular value 0) and Hence Let us mark σ max ) σ min ) =. σ max Ax)) σ maxdx)) σ min Dx)) = max{3,x} min{3,x}. min{σ max Ax)) : x [0, )} = σ max A3)) = = σ maxd3)) σ min D3)). References [] Marc Alexa, Johannes Behr, Daniel Cohen-Or, Shachar Fleishman, David Levin, Claudio T. Silva, Point-Set Surfaces, levin/ [] G. Fasshauer, Multivariate Mesh-free Approximation, fass/603 ch7.pdf [3] Faryar Jabbari, Linear System Theory II, Chapter 3: igenvalue, singular values, pseudoinverse, The Henry Samueli School of ngineering, University of California, Irvine05), fjabbari/me70b/me70b.html. [4] D. Levin, The approximation power of moving least-squares, levin/ [5] D. Levin, Mesh-independent surface interpolation, levin/ [6] D. Levin, Stable integration rules with scattered integration points, Journal of Computational and Applied Mathematics, 999), 8-87, levin/ [7] Svetoslav Nenov, Tsvetelin Tsvetkov, Matrices associated with moving least-squares approximation and corresponding inequalities, Advances in Pure Mathematics, 5 05), , doi: 0.436/apm [8] Kaare Brandt Petersen, Michael Syskind Pedersen, The Matrix Cookbook, Version: November 5, 0, download.php/374/pdf/imm374.pdf [9] Siegfried M. Rump, Verified bounds for singular values, in particular for the spectral norm of a matrix and its inverse, BIT, 5, No. 0). [0] Dicho Stratiev, Ivaylo Marinov, Rosen Dinkov, Ivelina Shishkova, Ilian Velkov, Ilshat Sharafutdinov, Svetoslav Nenov, Tsvetelin Tsvetkov, Sotir Sotirov, Magdalena Mitkova, Nikolay Rudnev, Opportunity to Improve Diesel-Fuel Cetane-Number Prediction from asily Available Physical Properties and Application of the Least-Squares Method and Artificial Neural Networks, American Chemical Society, 9, No. 3 05), doi: 0.0/ef50638c.