CANONICAL FORM. University of Minnesota University of Illinois. Computer Science Dept. Electrical Eng. Dept. Abstract

Transcription

1 PLAING ZEROES and the KRONEKER ANONIAL FORM D L oley P Van Dooren University of Minnesota University of Illinois omputer Science Dept Electrical Eng Dept Minneapolis, MN USA Urbana, IL 8 USA boleymailcsumnedu vdoorenuicslcsluiucedu Keyords: transmission zeroes, zero assignment, control, Kronecker form, matrix pencils Abstract Given a linear time-invariant control system, it is ell knon that the transmission zeroes are the generalized eigenvalues of a matrix pencil Adding outputs to place additional zeroes is equivalent to appending ros to this pencil to place ne generalized eigenvalues Adding inputs is likeise equivalent to appending columns Since both problems are dual to each other, e only sho in this paper ho to choose the ne ros to place the ne zeroes in any desired locations The process involves the extraction of the individual right Kronecker blocks of the pencil, accomplished entirely ith unitary transformations In particular, hen adding one ne output, ie appending a single ro, the maximum number of ne zeroes that can be placed is exactly the largest right Kronecker index Introduction The placement of transmission of zeroes via synthesis of ne outputs and/or inputs has been studied from the point of vie of system theory, and certain algorithms have been developed [8, 9,, ] As for the assignment of zeroes via feedback design, the assignment of zeroes via output synthesis can be analyzed in terms of the theory of matrix pencils, so that a complete characterization of the number of zeroes that can be placed in any given case can be obtained In [8, 9] an algorithm to synthesize outputs to assign the zeroes as proposed based on the desired form of the transfer function In [], a method as proposed to assign zeroes for a SISO system as ell as to assign zeroes to the input/output maps from individual inputs to individual outputs In this paper, e study this problem using the Kronecker theory of pencils [] Specically, e study the problem of zero placement for an arbitrary matrix pencil by the addition of ne ros or columns in terms of the structure of the Kronecker anonical Form (KF) We sho ho additional ros or columns can be appended to a pencil to place as many zeroes as possible, and discuss the limits on this placement Like [, ], our approach ends up reducing the problem to a problem of pole placement, for hich several algorithms are available (see eg [, ]) Hoever, e are { {

2 able to synthesize outputs to place the zeroes for an entire MIMO system, as ell as determining the limits on the number of ne zeroes that can be placed We treat this problem by studying the problem of assigning the generalized eigenvalues of a general matrix pencil The zero placement problem ill then be a special case The methods in this paper are all based on the transformation of the original pencil to one hich can be partitioned into the various components of the Kronecker canonical form The transformations are carried out entirely using unitary transformations, and hence enjoys some good numerical stability properties The computations are based on the so-called staircase algorithm of [], hich separates the left and right Kronecker parts and computes the values of the individual Kronecker indices We propose a ne extension to this algorithm, still based on unitary transformations, hich can actually extract the individual Kronecker blocks Once the individual Kronecker blocks have been extracted, the zeroes may be placed ithin each Kronecker block in a manner very similar in spirit to that of [] This paper is organized as follos First e describe the basic theory that relates the Kronecker theory of matrix pencils to the problem of placing zeroes or more generally placing the generalized eigenvalues for a pencil Next e describe our computational procedure for extracting the Kronecker blocks and placing the zeroes We leave in an Appendix a step by step description of the ne process used to extract the individual Kronecker blocks asic Theory onsider a linear time-invariant generalized state-space system of dimension n : hich is irreducible, ie here _Ex(t) = Ax(t) + u(t); y(t) = x(t) + Du(t); () ( A? E ) and A? E both have full rank n for all nite (this also means reachable and observable at nite points), and here E ( E ) and both have full rank n (this also means reachable and observable at innity) It is ell knon that its transmission zeroes are also the zeroes of the matrix pencil [,, ] : A? E G? F = : () D We ould like to add outputs or inputs to () to place ne zeroes in desired locations in the complex plane This corresponds to appending ros or columns, respectively, to () to place the zeroes of the embedded system We discuss the general question of ho many zeroes can be placed by appending ros and outline a procedure to compute the ros to append to place the zeroes at given locations To x ideas, e analyze ho many zeroes e can place by a judicious choice of additional ros appended to () We rite the Generalized Schur Form for () []: P (G? F )Q = G r? F r G reg? F reg G l? F l { { A ; ()

3 here P and Q are unitary (orthogonal in the real case) and G r? F r contains the right (short fat) Kronecker blocks, G l? F l contains the left (tall thin) Kronecker blocks, and G reg? F reg is the regular part The blocks are characterized by the properties that G r? F r has full ro rank and G l? F l has full column rank for all values of in the complex plane (including innity), and G reg? F reg is square and nonsingular except at a nite number of isolated values of hich are called the eigenvalues of the pencil The nite eigenvalues are the nite zeroes of the pencil The innite eigenvalues correspond to the innite zeroes of the pencil, except that each k k Jordan block I k? J at innity has only k? zeroes at innity (but k innite eigenvalues) [] Notice that this denition implies also that the total number of zeroes equals rank(f reg ) [] In the Kronecker anonical Form P; Q are nonsingular matrices, the entries are zero, and G r? F r has the block \diagonal" form G r? F r = R () A ; () R k () here each R i () is s i (s i + ), has full ro rank for all, and represents a single Kronecker block The fs i g's are the right Kronecker indices of the pencil, and e assume ithout loss of generality that these indices are in nondecreasing order In the sequel, e ill sho ho to obtain an upper triangular version of the overall form (), but ith nonzero entries above the diagonal blocks, using only unitary transformations In any case, G l? F l ill have a similar upper triangular form but ith rectangular diagonal blocks ith one more ro than column and full column rank We append some number p of ne ros to () to obtain P I G F? Q = Z G r? F r G reg? F reg G l? F l Z r Z reg Z l A : () The object is to choose these p ne ros so as to place as many zeroes as possible The rightmost block column (; ; G T l? Fl T ; ZT l )T has full column rank regardless of the choice of Z l The middle block column (; G T reg? F reg T ; ; ZT reg) T can lose rank only at values of here G reg? F reg already loses rank (ie only at existing generalized eigenvalues) and only for certain choices of Z reg The entry Z reg can sometimes be chosen so that the middle block column does not lose rank (or loses less rank than does G reg? F reg alone) at any particular existing eigenvalue In the presence of an G l? F l block, the result may be that the existing eigenvalue disappears from the augmented pencil () (or the eigenvalue remains ith a smaller multiplicity) ut in any case, neither Z reg nor Z l can be used to place any ne zeroes or to increase the multiplicity of any existing zeroes Some of these eects are explained in more detail in the next subsections Hence, only Z r can be used to place ne zeroes The choice of Z r is independent of Z reg ; Z l, so e may set the latter to zero Actually, if e don't set those to zero, there ill be coupling beteen the parts The eect of this coupling is discussed in the next subsections, but in general it ill not aect nely placed zeroes, unless they happen to coincide ith zeroes already present in G reg? F reg We need only consider the sub-pencil of () Gr? F r Z r = R () R k () Z Z k A ; () { {

4 here each R i is s i (s i + ) and represents a single Kronecker block It ill be seen that the zeroes can be placed by choosing the p ros to be appended, Z r = (Z ; ; Z k ), to have the form z T k?p+ z T k?p+ Z r = ; () z T k here each ro vector z T Ri () i is computed so that the individual (s i + ) (s i + ) pencil z T has a i subset of the desired zeroes, for i = k? p + ; ; k The rest of this section and the next section are devoted to lling in many of the theoretical and computational details behind the zero-placement algorithm, respectively A Ho Many Zeroes an e Placed? We discuss the specic question: ho many zeroes can be placed by appending one ro or multiple ros For this e rst need to recall some basic results on pencils and their polynomial null spaces def Let r be the normal rank of the m n pencil G? F, then it has n r = n? r right null vectors v i () def and m r = m? r left null vectors u j (), hich can be chosen to be polynomial ollecting these vectors in a n n r polynomial matrix V () and in a m m r polynomial matrix U() e thus have : (G? F )V () = ; U T ()(G? F ) = : (8) No the columns of V () and U() are said to be a minimal basis for the respective null spaces if their column degrees are minimal This is the case if and only if [, p8]: V (), respectively U(), has full column rank or all nite the highest column degree coecient matrix of V (), respectively U(), has full column rank One proves [] that if V (), respectively U(), is minimal then its column degrees are (up to a permutation)) equal to the right Kronecker indices fs i g, respectively left Kronecker indices ft i g, of G? F Moreover, the minimality of the bases refers to the fact that any other polynomials basis for these null spaces must have higher column degrees A consequence of all this is also that the number k of right Kronecker indices is equal to n r, and the number of left Kronecker indices is equal to m r One denes then the orders o r and o l of the right and left null spaces to be the sum of the column degrees of their minimal bases, ie o r = P n r i= s i and o l = P m r G r? F r has dimension o r (o r + n r ) G l? F l has dimension (o l + m r ) o l i= t i A simple consequence of this is (see []) : In order to use this for bounding the number of assignable zeros hen appending ros or columns e need the folloing result, proved in [] : Lemma Let G? F be a pencil ith null space orders o l and o r and number of nite and innite zeros o f and o (multiplicities counted), then rank(f ) = o l + o r + o f + o { {

5 Notice that in the above result e count zeroes at innity, not eigenvalues, to be compatible ith their system theoretic interpretation (see text above () or []) Since no appending constant ros or columns does not change rank(f ) e can only increase the number of zeroes by minimizing the null space orders We no give certain inequalities hich ill lead to the main result Theorem Let G? F be a m n pencil ith normal rank r and ith Kronecker indices and null space orders o r = P n r i= s i and o l = P m r j= t j Then appending p constant ros and denoting this pencil by G? F yields ne normal rank and null space orders r, o r and o l satisfying : r r r + p; o l o l ; ith equality only if r = r + p; ix j= s j ix j= s j for i n? r : Proof: The rst result is trivial since the normal rank of a pencil is its rank for almost any value of and appending p ros in a constant matrix then immediately gives the bounds r r r + p For the second bound e start from () and perform a generalized Schur decomposition on the subpencil consisting of the upper left part to get : ^P I Z r Z reg Z l G r? F r G reg? F reg G l? F l ^Q = I Since the subpencil ^Gl? ^F l G l? F l ^G r? ^F r ^Greg? ^F reg ^Gl? ^F l G l? F l has full column rank for all values of (including innity) its number of columns equals the ne left null space order o l and hence o l o l Moreover, equality is only met hen ^G l? ^F l is void ut then e also have that the ne dimension of the left null space equals the old one, ie m + p? r = m? r, hich yields the required result For the last inequality, let V () be a (m + p) (m + p? r ) minimal basis for the right null space of G? F Then obviously, e also have (G? F )V () = hich implies that the right null space V () of (G? F ) is a subspace of the right null space V () of (G? F ) As a subspace, it then follos from the theory of minimal bases [, x] that there exists a polynomial matrix M() such that V () = V () M(): Let V h and V h be the coecient matrices of the highest column degrees in V () and V () (these are the column coecients of s i and s j, respectively) Since V () and V () are minimal bases, e kno that V h and V h both have full column rank From this it follos that element m j;i () of the matrix M() can def not have degree larger than d j;i = s i? s j Moreover, the coecient matrix M h ith the coecient of s i?s j as (j; i)-th entry, has also full column rank, since [] : V h = V h M h : A { {

6 For every nonzero element m h j;i in M h e kno that s i = s j +d j;i s j since d j;i must be non-negative and no cancellation can occur beteen columns in this matrix product due to the linear independence of the columns in V h, V h and M h Since M h has full column rank, there must exist distinct indices j ; : : :; j n?r such that m h j i ;i = and hence n?r X i= X s n?r i Since the sequence fs j g is increasing, this implies that n?r X i= i= X s n?r i Finally, the same reasoning can be applied for the rst i columns of the matrices V h and M h, yielding the desired third bound i= s ji : This then automatically leads to the folloing main result Theorem Let G? F be a pencil ith right Kronecker indices s s nr Suppose e append p ros to obtain G F? : Z The maximum number of ne zeroes that can be placed is s i : s nr?p+ + ::: + s nr ; and a matrix Z can be found to place that many zeroes at any previously chosen locations in the complex plane This can be achieved by embedding the p largest Kronecker blocks only The other right Kronecker indices s ; ; s nr?p of the augmented pencil ill then be unchanged Proof: From the inequalities in the previous theorem it is clear that o n?r X l + o r o l + j= s j = o l + o r? n?r X j=n?r + Since rank(f ) is not aected by the embedding, it follos from Lemma that the maximum increase in number of zeroes satises o f + o? (o f + o ) n?r X j=n?r + The right hand side of this inequality is maximized by taking as many terms as possible, ie by taking r = r + p and hence : o f + o? (o f + o ) n?r X j=n?r?p+ Moreover, by using the embedding suggested in (,,) ith Z l = = Z r, this upper bound is actually Ri () met Indeed, each block P i () def = ith z T i = is regular and has s i zeroes After a permutation z T i of ros in () e have that each P i () appears on diagonal and becomes part of the ne regular part G reg? J reg of G? F The blocks for hich z T i = decouple, and the corresponding right Kronecker blocks remains intact in the augmented pencil This can be seen by noting that the right annihilating vectors corresponding to { { s j : s j : s j :

7 these blocks in the original pencil () (or its upper triangular equivalent) remain so for the augmented pencil (), ith the same degrees in Similarly, the left Kronecker blocks remain unaected for the same reason To place the zeroes for an individual Kronecker block, suppose that R() = (b; A)? (; U) (ith A; U square) is a single right Kronecker block and hence has full ro rank for all Suppose e add R() the single ro z T = (; y T ) Then observe that the nite zeroes of P () def = are exactly the eigenvalues of the pencil A + b? y T? U We can choose = and choose y T by standard pole placement techniques [, ] The vector y T alays exists and is generally unique When e choose =, P () has at least one innite zero In fact, as long as z T = the number of trailing zeros in that ro indicates the number of innite zeroes in P () In order to compute the proper ros Z r, it is necessary to extract the individual right Kronecker blocks The procedure to do this is described in detail in the next section, but a brief outline is as follos We rst apply the staircase algorithm [] to extract the right Kronecker part and compute the corresponding indices We then permute the ros and columns to extract the smallest right Kronecker block into the upper left corner and decouple this block from the rest of the pencil On the remaining collection of right Kronecker blocks e repeat this step to extract the next smallest right Kronecker block, until all the right Kronecker blocks have been extracted At each step, to decouple the upper left from the loer right, e annihilate the entries in the loer left block { in a very particular order hich ends up completely lling in the upper right block All the transformations applied are unitary transformations, and the result ill be an upper triangular version of the pencil (), here s s s l The Eect of oupling In this section, e illustrate some of the variations in the Kronecker structure that can occur hen a ro is appended The scheme suggested above implies that the matrix Z r has a decoupled form as in (), hich is not strictly required Since the method proposed belo adds one block P i () at a time to the regular part, let us analyze hat happens hen e append a single ro z T = (z T ; zt ) to a right Kronecker block R() and a regular block G reg? F reg as in : P () = R() G reg? F reg A : (9) z T z T If one of z T or zt is zero, then the to parts decouple, so let us assume that zt = and zt = Let be a zero of P () ith corresponding left eigenvector u T = (u T ; ut ; ) Then (ut ; ) must be a left R() eigenvector of corresponding to eigenvalue Once this condition is satised, one can alays z T nd a u T satisfying ut (G reg? F reg ) + z T =, to make ut the left eigenvector of P ( ), assuming is not an eigenvalue of (G reg? F reg ) For then (G reg? F reg ) has full column rank, and a solution alays exists If ; u T is an eigenvalue and left eigenvector of (G reg? F reg ), then a left eigenvector of P () R() corresponding to is (; u T ; ) regardless of hether or not the pencil also has an eigenvalue We illustrate this case ith the folloing example: P () = R() G reg? F reg z T { { A = ( ) z T z T A : z T

8 Regardless of the choice of z T ;, the entirepencil has an eigenvalue ith left eigenvector (; ; ) If R() e set z T = (; ) so that the pencil also has an eigenvalue, and e set = to couple the to parts together, the resulting pencil is z T P () = A : This is a regular pencil ith characteristic polynomial detp () =? It has a double, defective, eigenvalue at zero If the to parts are decoupled by setting =, the characteristic polynomial remains unchanged, but the double eigenvalue at zero becomes nondefective If instead e set z T = (; ), the characteristic polynomial becomes?, yielding simple eigenvalues at and, independent of the R() choice of So if both individual pencils and (G reg? F reg ) have a common eigenvalue, that z T eigenvalue may remain in the full pencil P () ith a Jordan chain combined from the Jordan chains from the individual pencils, or else the common eigenvalue may have a ne independent Jordan chain This implies that hen placing ne zeroes, it is best to avoid any existing zeroes if one ants to keep Jordan chains as short as possible We can summarize some of this discussion ith the folloing Theorem onsider the augmented pencil () As long as the nely placed zeroes do not coincide ith any zeroes already existing in G reg? F reg, the block Z r may be computed to place those zeroes independent of Z reg ; Z l If ne zeroes are placed over existing ones, their Jordan chains might or might not coalesce Ri () The same comment applies henever common zeros are chosen beteen added blocks P i () = since coupling is likely to occur via the nonzero elements above diagonal in () as ell Since this paper focuses only on the placement of zeroes and not their Jordan structure, e do not pursue this discussion here z T i omputational Procedure We sketch an algorithm to place the zeroes The algorithm consists of three stages In the rst stage, e essentially compute the generalized Schur form (), separating the left Kronecker part from the regular part and the right Kronecker part This can be carried out using the staircase algorithm [, ] and e ill not discuss this process in detail In the second stage, e extract the individual Kronecker blocks, via a ne elimination procedure proposed in this paper Finally, in the third stage e compute the ros to be appended in order to place the zeroes The entire extraction process (the rst to stages) is carried out using only unitary transformations, hence it enjoys a backard stability property The zero placing part also enjoys a backard stability property, hence the numerical stability of the hole process is favorable The broad steps of the process is as follos Algorithm Zero Placement Use the staircase algorithm [, ] to extract the left and right Kronecker part from the pencil Assume the left Kronecker part of the pencil is the m n pencil G l? F l, ith n = m + k hoose p k as the number of ros to append { 8 {

9 Extract the individual Kronecker blocks from the staircase form To the largest Kronecker blocks compute the ro(s) that must be appended in order to place the desired zeroes Append the computed ros and back-transform through all the accumulated basis transformations back to the original basis for the given pencil G? F We ll in some of the details for each step in turn Staircase Form The staircase form is a form that can be obtained from the original pencil that exposes the various Kronecker indices and orders of a pencil, as ell as exposing any regular part In fact, the presence of a regular part can be determined by the staircase algorithm, except that it may not be as numerically reliable as the approach in [] (This is still an open area of research) The algorithm used to extract the left and right Kronecker part is the variant described in [] to hich e refer the reader for all the details, particularly on ho to handle the presence of a regular part or left Kronecker blocks The result of this algorithm is of the folloing typical form : G l? F l = G ; G ; G ; G ;k G ;k+ G ; G ; G ;k G ;k+ G ; G ;k G ;k+ G k;k G k;k+? Figure Typical Staircase Form F ; F ; F ; F ;k+ F ; F ; F ;k+ F ; F ;k+ ; () F k;k+ here the matrices F i;i+ are n i n i nonsingular and the matrices G i;i are n i n i? and of full ro rank n i Therefore, the sequence fn i ; i = ; : : :; kg is nonincreasing and its dual sequence are the right Kronecker indices fs i ; i = ; : : :; n r g, ie : there are n i+? n i indices equal to i for i = ; : : :; k here e have assumed n k+ = Notice that this implies that if the smallest Kronecker index is s = q, then the rst q matrices G i;i for i = ; : : :; q are square invertible as ell Finally, e need to use a variant of the above form here the square matrices F i;i+ are uppertriangular and the rectangular matrices G i;i have leading zero columns and a trailing upper-triangular matrix This form can alays be obtained as explained in [] and ill be exploited in the subsequent steps Extract Individual Kronecker locks The next step is to extract the individual Kronecker blocks We do this by permuting toard the upper left the entries of the matrix corresponding to the Kronecker block of interest, and then annihilating the coupling elements in the appropriate o-diagonal block Due to the nature of the Kronecker structure, it is necessary to extract rst the smallest Kronecker block, then extract the next smallest from hat is left, and so on { 9 {

10 It is easier to illustrate the process and then describe it formally Let q be the number of leading diagonal G blocks that are square That is, let q be the number such that G ; ; G qq in () are square, but G q+;q+ is not Then the smallest Kronecker index is s = q In the example orked out belo e assumed there are such blocks, so the smallest Kronecker index is s = In this particular example, e must thus form a block We form this block from the, entries of those rst square G blocks together ith the, entries of the corresponding F blocks That is, e permute the ros and columns of G; F to collect together the upper left entries of all the leading square G blocks This ill form a leading submatrix Then e must decouple this leading submatrix by eliminating the coupling entries To sho ho this orks in more detail, e partition all the blocks of Figure to expose the, scalar entries, shoing the result as Figure We denote scalar entries by g; f, column vectors by g; f, ro vectors by g ; f (completely unrelated to the transpose of any corresponding column vector), and submatrices by G; F Note that potentially the ro, column and matrix blocks could be empty g g g g g g g g g G g G g G g G G g g g g g g g G g G g G G g g g g g G g G G G G G Figure - Partitioned Staircase Form f f f f f f f F f F f F F f f f f f F f F F f f f F F F We permute the leading, entries into the upper left position to obtain Figure in hich the block is exposed: g g g g g g g g g g g g g g g g g g g g g g g g G G G G G g g G G G G g G G G G G G Figure - Permuted Form f f f f f f f f f f f f f f f f f F F F F f F F F F F F We then decouple the Kronecker block from the rest by annihilating one by one the column entries in the loer left part This is done ith (almost) alternating left and right unitary transformations The entries in the loer left are annihilating in a very particular order, starting ith the \outer" diagonal strip In this example, the rst items eliminated are the entries in the outer diagonal (marked ith in Figure ), eliminated in order: g ; f ; g ; f ; f Then the next diagonal entries are eliminated in order: g ; f ; g Then nally g is eliminated The entries in G are eliminated using unitary transformations from the right, and the entries in F using transformations from the left, as illustrated in the appendix Each elimination results in a ll in the corresponding position in the upper right block This example is suciently general to sho the pattern of lls in the g; f part for the general case { {

11 The result is shon in Figure In Figure, ^ denotes entries that ere modied from Figure, ^ denotes entries that ere purposely eliminated, and ^ denotes entries that ere lled in during this process g ^g ^g ^g ^g ^g ^g g g ^g ^g ^g ^ ^g ^g ^g ^g ^g ^g ^ ^ ^g ^g ^g ^ ^ ^ ^G ^G ^G ^G ^G ^ ^ ^G ^G ^G ^G ^ ^G G G G G G ^f ^f ^f ^ ^f ^f ^f ^f ^ ^ ^f ^f ^ ^ ^ ^f f f ^f ^f ^f ^ ^ ^F ^F ^F ^F ^ ^F ^F ^F F F F Figure - Result from Extraction of Smallest Kronecker lock We summarize the process as follos Algorithm Kronecker lock Extraction Start ith an m n pencil G? F in staircase form, ith k = n? m Let q = s ( q k) be the number of leading diagonal G blocks that are square in the staircase form (Figure ) Permute ros and columns of the pencil so that the \," entries of the blocks G ij ; F ij, i = ; ; q, j = ; ; q +, are in the upper left, as in Figure Denote the partitioned m n pencil by G (;) G (;) F (;) F (;) G (;) G (;)? F (;) F (;) ; here each block is partitioned as in Figure The leading block G (;)? F (;) is q (q + ) In the partitioning of Figure for i q, G ii are k k, so that G (;) ii, G (;) ii, G (;) ii, G (;) ii are, respectively,, (k? ), (k? ), (k? ) (k? ) Eliminate the entries G (;) ij ; F (;) ij, in the permuted matrices This modies parts of all four blocks (,), (,), (,), (,) For i = q; ; ; : For j = i; ; ; : Push G (;) j;j++k?i right into G(;) j;j If j >, Push F (;) j?;j++k?i up into F (;) j+k?i;j++k?i The resulting modied \(,)" block is not further modied by this algorithm, so e denote it by G [] []? F The modied \(,)" block is still is staircase form The modied \(,)" block is full and the modied \(,)" block is all zero If k, apply this algorithm recursively to the (m? k) (m? k? ) pencil G (;)? F (;) Apply all the resulting unitary transformations from the right also to the block G (;)? F (;) In the above algorithm description, e use the short hand \push right" and \push up" to mean the folloing { {

12 Push G (;) right into G (;) means: Find a unitary transformation Q such that (G (;) (; G e(;) ), here G e(;) all, compute ( G e(;) ; G e(;) ) = (G (;) ; G (;) )Q, and replace G (;), G (;) ith G e(;) likeise ith F (;), F (;) Push F (;) up into F (;) ; G(;) )Q = is upper triangular Then apply transformation to entire pencil: ie, for, G e(;) Do means: Find a unitary transformation Q 8 such that! ef (;) = : F (;) Q 8 F (;) Then apply transformation to entire pencil: ie, for all compute!! ef (;) F (;) = ef (;) Q 8 F (;) ; and replace F (;), F (;) ith e F (;), e F (;), respectively Do likeise on the F (;), F (;) The nal form of this process is Q [] L (G[]? F [] )Q [] R = (G[]? F [] ) = G [] G [] k G [] kk A? F [] F [] k F [] kk A ; () here for each i = ; ; k, G [] ii is s i (s i +), upper trapezoidal, and F [] ii = (; U i ) ith U i s i s i, upper triangular, here s i s k+?i are the right Kronecker indices dened as in Theorem, in nondecreasing order The s i (s i + ) pencil G [] ii? F [] ii has full ro rank s i for all values Of course, this is not the Kronecker anonical Form, but it is an analogous form achievable via unitary transformations Each diagonal block is equivalent to a single Kronecker block For most any purpose for hich the Kronecker form ould be required one can make use of this form just as ell We remark that in this extraction process, all the entries that are annihilated are never lled in during subsequent steps Assume e use Givens rotations to annihilate the entries Applying each rotation costs O(n) operations, including the cost of accumulating the Givens rotations At most O(n ) such rotations are generated, since there are no more than O(n ) entries to annihilate (e can't annihilate more entries than there are in the hole matrix!) Hence the entire extraction process takes O(n ) operations Of course, a more precise analysis is possible, but it is dicult because the exact cost ill range from free to O(n ) depending on the exact distribution of the Kronecker indices Place Zeroes To the form () e can compute the p ros needed to place s def = s nr + +s nr?p+ zeroes Let ; ; s be the given set of ne zeroes to be placed Then the ros to append have the folloing form Z [] = z T n r?p+ z T n r?p+ z T n r { { A ; ()

13 here e have n? s = s + + s nr?p leading columns of zeroes, and the nonzero entries are computed as described belo For each i = n r ; n r? ; ; n r? p +, each z T i is an s i -vector chosen so that the square (s i + ) (s i + ) pencil! G [] [] n r+?i;n r +?i F z T? i n r +?i;n r +?i has zeroes f^si +j g s i Recall from () that the pencil () has the general form (b j= i; A i )? (; U i ) here U i is square, upper triangular, and nonsingular Let z T i = (; yi T ) Then observe that the zeroes of bi A i? U i are exactly the eigenvalues of the pencil A i +b i? i yi T?U i We can choose i = and i choose y T i y T i by standard pole placement techniques [, ] In this case, the ne ro is generally unique To see that the fz T i g chosen in this ay places the zeroes for the entire pencil, e can permute the ros to put the ne regular part of the pencil in the loer right and the remaining right Kronecker structure in the upper left: eg e G here eg? e F =! G [] n r?p+;n r?p+ z T n r?p+ is regular ith the desired zeroes and G? F = e G G [] n r;n r z T n r has the right Kronecker structure left over G [] G [] ;n r?p G [] n r?p;n r?p ef e F? F e ; A? A? F [] n r?p+;n r?p+ F [] n r ;n r F [] F [] ;n r?p F [] n r?p;n r?p A ; () A Returning to the original pencil G? F, our original problem as to compute a p m matrix Z to place s zeroes We collect together all the transformations to obtain the formula for Z: [] Q L Q[] L F [] so that I G Z? here H denotes the conjugate transpose of F Q [] R Q[] R = G [] Z []? Z def = Z [] (Q [] R )H (Q [] R )H ; onclusion We have examined the general problem of placing the generalized eigenvalues to an arbitrary matrix pencil by the addition of ne ros of constant coecients We found that the number of zeroes that can be placed is limited to the order of the right Kronecker part (the sum of all the right Kronecker indices) { {

14 In addition, hen p ros are added and there are multiple right Kronecker indices, the number of zeroes that can be placed is limited to the sum of the p largest right Kronecker indices When the number of ros added is just right to make the system square, then the number of zeroes that can be placed is equal to the sum of all the right Kronecker indices We have outlined a ne method based entirely on unitary transformations to compute the right Kronecker indices and to extract the individual Kronecker blocks y combining this procedure ith pole placement algorithms in the literature, e arrive at a complete method for assigning the generalized eigenvalues for a pencil, hich enjoys good numerical stability properties due to the use of unitary transformations From a control point of vie, this method places the transmission zeroes by the synthesis of ne outputs It could just as easily be used to synthesize inputs instead The ne decomposition in hich the individual Kronecker blocks are extracted represents a unitary analog to the Kronecker canonical form in much the same spirit as the Schur decomposition is a unitary analog to the Jordan canonical form Acknoledgement Part of this research as performed hile the authors ere visiting the Institute of Mathematics and Applications of the University of Minnesota, Minneapolis, during the summer quarter of the Applied Linear Algebra Year organized there We greatly appreciated the hospitality and the productive atmosphere of that institute P Van Dooren as partially supported by the National Science Foundation (Grant R 999) References [] W A erger, R J Perry, and H H Sun An algorithm for the assignment of system zeroes Automatica, ():{, 99 [] Th eelen, P Van Dooren, An improved algorithm for the computation of Kronecker's canonical form of a singular pencil Linear Algebra & Applications, :9{, 988 [] D L oley Estimating the sensitivity of the algebraic structure of matrix pencils ith simple eigenvalue estimates SIAM J Matrix Anal, :{, 99 [] A Emami-Naeini and P Van Dooren omputation of zeros of linear multivariable systems Automatica, 8:{, 98 [] F R Gantmacher Theory of Matrices, volume helsea, Ne York, 99 [] T K Kailath Linear Systems Prentice Hall, 98 [] J Kautsky, N K Nichols, and P Van Dooren Robust pole assignment in linear state feedback Int J ontrol, :9{, 98 [8] Kouvaritakis and A G J MacFarlane Geometric approach to the analysis and synthesis of system zeroes, part I Int J ontr, :9{, 9 [9] Kouvaritakis and A G J MacFarlane Geometric approach to the analysis and synthesis of system zeroes, part II: Non-square systems Int J ontr, :{8, 9 { {

15 [] G Miminis and Paige A direct method for pole assignment of time-invariant multi-input linear systems Automatica, ():{, 988 [] P Misra A computational algorithm for squaring up part I: Zero input-output matrix In Proc onf Dec ontr, pages 9{, December 99 [] P Van Dooren The computation of Kronecker's canonical form of a singular pencil Lin Alg & Appl, :{, 99 [] G Verghese, P Van Dooren, and T Kailath Properties of the system matrix of a generalized state-space system Int J ontr, :{, 99 Appendix We sho the process to decouple the Kronecker block from the rest by annihilating one by one the entries in the loer left, starting ith the state of Figure This is done ith almost alternating left and right unitary transformations The entries in the loer left are annihilating in a very particular order, starting ith the \outer" diagonal strip (marked ith an overbar \ " in Figure ) We sho the step by step annihilations for this particular example, here the ros and columns aected by the individual Givens rotation are marked ith arros, the entries used to construct the rotation are enclosed in boxes (eg e denotes the entry being annihilated), and all the other entries modied in that step are marked ith a ide tilde \e" Fills (zeroes made nonzero) are denoted by \g " hen they occur and \ ^" in subsequent steps Likeise, entries set to zero are denoted \ e " the rst time and \^" in subsequent steps The hat \^" denotes entries changed from Figure In this particular example, steps - zero out the outer diagonal strip of G; F blocks (marked \ " in Figure ), steps -8 zero out the next strip of each, and step 9 zeroes out the last, resulting in the situation of Figure Note also that the G ; F blocks and all the entries belo and to their right remain unchanged through this hole process Situation of Figure : g g g g g g g g g g g g g g g g g g g g g g g g G G G G G g g G G G G g G G G G G G f f f f f f f f f f f f f f f f f F F F F f F F F F F F { {

16 Step : Rotate from right: g g g fg g g fg g g f g g fg g f g g g f f f f ff f f g fg g f g g f f f f f f g g fg G G G g G G f f f f f g fg G G g f G G f f F F f F F e G g G G f f f F F F F G G F F G Step : Rotate from left: g g g ^g g g ^g g g g g ^g g ^g g g fg fg f f g f g f g (= g g ^g G G ^G G G fg fg g G g G g G g G (= ^ ^G G G G G G Step : Rotate from right: f f ^f f ^f f f f ^f ^f f f f f e f f f f (= f ^f F ^F F F e f F f F f F (= F F F g g fg ^g g fg ^g g g f g fg ^g g f ^g g g f f ^f ff ^f f f fg ^g e ^g ^g ^g f f ^f f ^f f f g fg ^g G G g ^G G G ^f ^ ^f ^f e ^g G g ^G ^G ^G f f ^f F f ^F F F ^ ^ ^G G G ^F ^F ^F F G G F F G Step : Rotate from left: g g ^g ^g g ^g ^g g g fg fg fg f g f g f g f g f (= ^g ^g ^ ^g ^g ^g fg fg fg G g G g G g G G g (= ^ ^g ^G ^G ^G ^G ^ ^G G G G G G f ^f ^f ^f ^f f f f f f f e f f f f f f ^f ^ ^f (= ^f e f f f F f F f F f F (= ^ ^F ^F ^F F F F { {

17 Step : Rotate from right: g fg ^g ^g f g ^g ^g g g fg ^g ^g e ^g ^g ^g ^g ^g ^g ^ ^g ^g ^g e ^g ^g g G ^G ^G G ^G ^ ^g ^G ^G ^G ^G ^ ^G G G G G G Step : Rotate from right: ff ^f ^f f ^f ^f ^f ^f ^f ^f ^ ^f ^f ^f ^f ^ ^f ^f ^ ^f ^F ^F ^F ^F ^ ^F ^F ^F F F F g ^g ^g fg ^g fg ^g g g ^f ^f f ^g ^g fg ^ g f ^g ^g ^g f ^ ff ^f ^f ^f ^g fg e ^g ^g ^g ^f f f e ^f ^f ^f ^ ^g fg ^G G g ^G G ^G f f f ^ ^f ^f ^ ^ e G g ^G ^G ^G f f F f ^F ^F ^F ^ ^ ^G G G ^F ^F ^F F G G F F G Step : Rotate from left: g ^g ^g ^g ^g ^g ^g g g ^g ^g ^g ^ ^g ^g ^g ^g fg fg f e f g f g f g (= ^ fg fg g G g G g G g G g G (= ^ ^ ^G ^G ^G ^G ^ ^G G G G G G Step 8: Rotate from right: ^f ^f ^f ^ ^f ^f ^f ^f ^ ^f ^f ^f ^f ^f f f e e f f f f (= ^ e f F f F f F f F (= ^ ^F ^F ^F F F F g ^g fg ^g g f ^g ^g g g ^f f f ^f e ^g fg ^g e ^g ^g ^g ^g fg ^g e ^ ^g ^g ^g ^f ^f ^f ^f f f ^f f ^ ^f ^f ^f ^ e ^g G g ^G ^G ^G ^G ^f ^ ^ ^f ^f ^ ^ ^G ^G ^G ^G ^ ^ ^F ^F ^F ^F ^ ^G G G ^ ^F ^F ^F G G F F F G { {

18 Step 9: Rotate from right, yielding situation of Figure : e ^e ^e fe g f ^g ^g g g ^f ^f f f e ^g ^g fg e ^g ^g ^g ^g ^g fg e ^ ^g ^g ^g ^f ^f ^f ^f ^f f f e ^ ^f ^f ^f ^ ^ e G g ^G ^G ^G ^G f f f ^ ^ ^f ^f ^ ^ ^G ^G ^G ^G ^ ^ ^F ^F ^F ^F ^ ^G G G ^ ^F ^F ^F G G F F F G Figure A - Annihilating Steps { 8 {