After the completion of this section the student should recall

Chapter III Liear Algebra September 6, 7 6 CHAPTER III LINEAR ALGEBRA Objectives: After the completio of this sectio the studet should recall - the cocept of vector spaces - the operatios with vectors ad matrices - the methods of solutio of liear systems of algebraic equatios - the methods of solutio of the eigevalue problem Cotets:. Vector Spaces. Liear Combiatio, Liear Idepedece, Basis 3. Vectors 4. Matrices 5. Liear Trasformatios with the Help of Matrices 6. Determiats 7. Matri Iverse 8. Row Reductio 9. Liear Systems. Eigevalue Problem. Liear Algebra with Maple. Review questios ad eercises

6 Chapter III Liear Algebra September 6, 7. VECTOR SPACES A vector space V = { uvw,,, } is a set of objects called vectors, for which two operatios are defied: the operatio of additio of vectors u+ v ad the operatio of scalar multiplicatio ku by the scalars from the field of real umbers k, uder the coditio that they satisfy the special aioms. I the prited tet, vectors are deoted by bold letters uv,,... ad i hadwritig they are deoted by arrows above letters u,v,.... Scalars are deoted by lower case letters a,b,... Eamples of vector spaces iclude the Euclidia space (colum vectors of dimesio ), the space of all m matrices M m, the space of all a,b, the space of cotiuous fuctios f ( ) C[ a,b] all fuctios f ( ) C ( a,b) defied o the iterval [ ] cotiuously differetiable o the iterval ( ) a,b. The otio of vector spaces helps to obtai ad to study the geeral properties of solutios of algebraic ad differetial equatios ad systems of equatios. The formal defiitios of a vector space ad a vector subspace are the followig: Defiitio A liear vector space V over the field of real umbers is a set of elemets (called vectors) with two operatios: i) additio of vectors: if uv, V, the u+ v V closure for sum ii) multiplicatio by a scalar: if u V ad k, the ku V closure for scalig which satisfy the followig aioms: ) if uv, V, the u+ v = v+ u commutative law uvw, the ( + ) + = + ( + ) ) if,, V u v w u v w associative law 3) there eists a vector V, such that eistece of u+ = + u = u for ay u V zero vector 4) for ay u V there eists a vector u V eistece of u+ u = egative vector such that ( ) 5) if V u ad a,b, the a ( b ) = ( ab) uv ad k, the ( ) u ad a,b, the ( ) 6) if, V 7) if V 8) if u V, the u = u u u associative law k u+ v = ku+ kv distributive law a+ b u = au+ bu distributive law Defiitio Let V be a vector space, the ay subset U V which is a vector space itself with the operatios iduced by V is called to be a subspace of V

Chapter III Liear Algebra September 6, 7 63 Commets: ) Coditios i) ad ii) are the closure aioms. They mea that the sum of ay two vectors from the space is a vector which also belogs to this space, ad ay vector multiplied by ay scalar is a vector which also belogs to the space. Therefore, the defied operatios ca always be performed without goig outside of the vector space, because vector space V cotais all required vectors. ) Aioms -8) are the vector aioms; the set of objects which satisfies these aioms is called the vector set, ad its elemets are called vectors. 3) We defied a vector space over the field of real umbers, therefore it ca be referred to as a real vector space. Similarly, a comple vector space over the field of comple umbers also ca be defied. 4) The zero vector V is uique. It meas that if there eists a vector θ V with the same property u + θ= θ+ u = u the θ= 5) To show that some set V is a vector space, we eed to demostrate that all te aioms i-ii) ad -8) are satisfied. 6) To show that some subset U V of the vector space V is a subspace, we eed to check oly the closure aioms i) ad ii), because elemets of the subset of the vector space are already the vectors. To prove the opposite, that U V is ot a subspace, very ofte it is sufficiet to show that the zero vector V does ot belog to the subset U, ad, therefore, it caot be a subspace. 7) The egative elemet u V is uique. It meas that if there eists a vector V u+ w = the u = w w with the same property ( ) 8) u = for ay u V 9) ( ) u = u for ay u V ) a = for ay a ) a u = if ad oly if a = or u = ) Vector space ca be defied over a arbitrary field (scalars) for which operatio of multiplicatio of a vector by a scalar is defied: vector ku is defied for all u V ad k If is, the V is referred to as a real vector space. If is, the V is referred to as a comple vector space. space V subspace U

64 Chapter III Liear Algebra September 6, 7 EXAMPLES OF VECTOR SPACES Space of real umbers. The field of all real umbers with usual algebraic operatios of additio of real umbers ad multiplicatio by a real umber is a vector space. The field aioms (see Chapter I) ad algebraic rules provide satisfactio of the vector space aioms. The vector space is a particular case of a more geeral -dimesioal Euclidia vector space. Euclidea space. The elemets of the Euclidia vector space = (,,...,) i are -tuples which ca be treated as the coordiates of the poits i -dimesioal geometrical space. The operatios are defied i the followig way: + y =,,..., + y, y,..., y = + y, + y,..., + y k = The zero vector is =,,..., ( ) ( ) ( ) ( k,k,...,k ) ( ) ad the egative vector is = =,,..., ( ) ( ) It is easy to verify the satisfactio of the vector space aioms for. Euclidea space 3 is a particular case of : (,, ) = 3 i It is a physical 3-dimetioal Euclidia space which is used i mathematical modelig physical ad egieerig problems.

Chapter III Liear Algebra September 6, 7 65 Colum vectors The vector space is the set of all colum vectors = i for which operatios are defied i the similar way: y + y y + y + y = + =, y + y k k k = k The colum vectors are just a differet form of writig the real -tuples. I their tur, the colum vectors are a particular case of the m real matrices. Space of m matrices m 3. The set of all real matrices with m rows ad colums A ( aij ) a a a a a a m = i=,..,m = j=,..., a a a m m m aij with operatios a + b a + b a + b a + b a + b a + b A+ B = am + bm am + bm am + b ka ka ka ka ka ka k A = kam kam kam is a vector space with zero vector m, =

66 Chapter III Liear Algebra September 6, 7 Space of cotiuous fuctios C[ a,b ] 4. The set of all real-valued cotiuous fuctios { } [ ] = ( ) [ ] [ ] C a,b f : a,b f is cotiuous o a,b The operatios additio of vectors ad multiplicatio by a scalar are defied as the poit-wise operatios with fuctios f + g f ( ) + g( ) [ a,b] for f,g C[ a,b] kf kf ( ) [ a,b] for f C[ a,b], k Obviously, a sum of two cotiuous fuctios o [ a,b ] is a cotiuous fuctio o [ a,b ], ad a cotiuous fuctio multiplied by a scalar is still cotiuous. Therefore, the closure aioms i-ii) of a vector space are satisfied. C a,b is a fuctio idetically equal to The zero vector [ ] zero o the iterval [ a,b ]. Other vector spaces of cotiuous fuctios are: C( a,b ), C (, ) Space of differetiable fuctios C ( a,b ) 5. The set of all real valued fuctios cotiuously differetiable o the ope iterval ( a,b ) { } ( ) = ( ) ( ) ( ) C a,b f : a,b f is cotiuous o a,b with operatios idetical to operatios i the space C[ a,b ]. Space of polymoials P ( ) 6. The set of all polyomials with degree ot greater tha : ( ) = { + + + + } P a a a a a i is a vector space with operatios idetical to operatios with fuctios. Zero subspace 7. The simplest but importat eample of a vector subspace of ay space V over the field of real umbers is its subspace V, where vector V is a cosistig of oe elemet { } zero vector of the vector space V : V { } = It also is called the ull-subspace of V. V { }

Chapter III Liear Algebra September 6, 7 67 Solutio space 8. Cosider the st order liear ordiary differetial equatio y y = The set of all solutios of this equatio (provided that they eist) is a subset of the vector space C ( ) (The set of all solutios is ot empty because it obviously possess the y = ): trivial solutio ( ) { ( ) ( ) } C ( ) V = y C y y = We ca verify that the set V is a vector space. For this, accordig to commet (6), we have to verify oly the coditios i-ii) of the Defiitio : i) Assume that y,y V are solutios, the show that the sum of them y + y is also a solutio. Ideed, y + y y + y = y y + y y = ( ) ( ) ii) Assume that y V is a solutio, the show that the scalar multiple of it ky, k is also a solutio. Ideed, ky ky = ky ky = k y y = k = ( ) ( ) ( ) Therefore the set of all solutios V C ( ) is a vector space. We ivestigated the structure ad the properties of the set of all solutios of the differetial equatio ad showed that they costitute the vector space eve without solvig the equatio. I fact, the geeral solutio of this equatio is give by ( ) ce y =, c They are ifiitely may times differetiable fuctios, ad therefore, they belog to the space y( ) C ( ) C ( ). Itersectio of vector spaces Let Ui i= i V be the subspaces of the vector space V, the { u u i } U = V U, i=,..., is a subspace of V (Eercise, p.). V U U U 3

68 Chapter III Liear Algebra September 6, 7. LINEAR COMBINATION, LINEAR INDEPENDENCE, BASIS We eed a coveiet way of costructig ad describig vector spaces. I the,y. I plae ay poit ca be uiquely defied by a pair of its coordiates ( ) vector spaces the role of coordiates is played by the coefficiets i the liear combiatio represetatio of the vectors with the help of the basis vectors. For eample, the set of mooms {,, } is a basis of the vector space of all quadratic polyomials P ( ) a liear combiatio of basis vectors where each polyomial is represeted uiquely as a a a + + with coefficiets a,a,a. Therefore, the polyomials are defied just by these coefficiets. Let V be a vector space. Recall some defiitios ad facts: Liear Combiatio Let u, u,..., u V. The a vector cu + c u +... + cu is called a liear combiatio of vectors u, u,..., u with coefficiets c,c,...,c. It is obvious that ay liear combiatio of vectors from V also is a vector of V c u + c u +... + c u V Spa The set of all liear combiatios of vectors u, u,..., u V is called a spa of vectors u, u,..., u { u u u} { u + u + + u i } spa,,..., c c... c c I this defiitio coefficiets c i attai all real values geeratig ifiitely may 3 vectors. For eample, i the Euclidia space the spa spa c = c geerates a lie i the y-plae. Ad the spa spa, c c = + c,c geerates the y-plae. The spa{ si( λ ),cos ( λ ) } of two vectors si( λ ),cos ( λ) C ( ) geerates the geeral solutio of the liear ODE The { m m spa e,e } solutio of the liear ODE + =, λ. y λ y of two vectors m m e,e C ( ) geerates the geeral y m y =, m. The same geeral solutio also ca be geerated by the { ( ) ( )} spa sih m,cosh m. Because each of these liear combiatios is a vector of space V, the spa is a subset of vector space V { u u u } spa,,..., V We ca verify that the spa is a vector space itself, therefore spa u, u,..., u is a subspace of V. { }

Chapter III Liear Algebra September 6, 7 69 The questio arises, if it is possible to specify some set of vectors u, u,..., u V such that their spa geerates the whole vector space V? Ad if yes, the what is the miimal set of vectors with this property? The such a set ca be used as a basis for vector space V. Liear Idepedece The set of vectors u, u,..., u V is called liearly idepedet if their liear combiatio is equal to the zero vector with oly all zero coefficiets: cu + c u +... + c u = all ci = If a set of vectors u, u,..., u V is ot liearly idepedet, the it is called liearly depedet. Formally, it ca be stated as follows: The set of vectors u, u,..., u V is called liearly depedet if there eist coefficiets c,c,...,cc ot all equal to zero such that the liear combiatio with them is equal to the zero vector: c u + c u +... + c u = Cosider the followig useful facts about liear idepedece of vectors:. If i the set of vectors u, u,..., u V oe of vectors u k ca be represeted as a liear combiatio of the other vectors u = c u +... + c u + c u +... + c u k k k k+ k+ the the set u, u,..., u V is liearly depedet. Ideed, rewrite the equatio as c u +... + c u + u + c u +... + c u = ( ) k k k k+ k+ where at least oe coefficiet is o-zero.. If { u, u,..., u } the the set {,,..., } u u u is liearly depedet. Liear Idepedece i m a a a a Let a u =, a u =,, m u = am am am The the set u, u,..., u is liearly depedet if there eist coefficiets c,c,...,c ot all equal to zero such that cu + c u +... + c u = I the compoet form this vector equatio ca be writte as a a a a a a c + c +... + c = a a a m m m It meas that the homogeeous system of algebraic equatios for c,c,...,c a a a c a a a c = a a a c m m m has o-trivial solutios. or Ac = If the umber of vectors > m the the set u, u,..., uis liearly depedet. If = m the the set u, u,..., uis liearly idepedet oly if det A =.

7 Chapter III Liear Algebra September 6, 7 Liear idepedece of fuctios See the Sectio 5.3 Theory of Liear ODE i Chapter 5. Basis of vector space V Let V be a vector space. The the set of vectors u, u,..., u V is called a basis for the vector space V if i) the set u, u,..., u is liearly idepedet ii) { u u u } spa,,..., = V Ay vector v V is uiquely represeted as a liear combiatio of the basis vectors: v = c u + c u +... + c u That meas that the spa of u, u,..., ugeerates the space V. Theorem Every vector space possesses a basis. A basis for the vector space is ot uique, i geeral there ca be may differet bases, but the umber of vectors i the differet bases for the vector space is the same: if the set u, u,..., uis a basis for the vector space V, the ay other basis for V also cosists of vectors. The umber of vectors i the basis is called the dimesio of the vector space V. Not all vector spaces have a fiite dimesio. For eample, the space of cotiuous fuctios C[ a,b ] has a ifiite dimesio. Eamples:. The set of colum vectors z stadard e =, e =,, 3 basis i e = k = i = j = y is a stadard basis for. I particular, the stadard basis for i =, j =, k = 3 is deoted by. The set of mooms {,,,..., } is a stadard basis for the vector space of polyomials of order P ( ) : ( ) = { + + + + } P a a a... a, a i

Chapter III Liear Algebra September 6, 7 7 Fuctios defied o vector spaces U u v = f ( u) v V Let U ad V be two vector spaces. The the fuctio f :U V defied betwee these vector spaces is called a liear map or a homomorphism if for all uw, U ad c it satisfies the followig coditios: i) f ( u+ w) = f ( u) + f ( w ) ii) f ( cu) = cf ( u ) u u = u Au = v v v = v m m For eample, a liear map ca be defied by a matri as a liear trasformatio of Euclidia vector space: Let A m be a m real matri, the the trasformatio defied by m v = Au for all u is a liear map from map icludes particular trasformatios such as dilatio ad rotatio of the vector space. A liear map i the vector spaces of fuctios ca be defied by the familiar operatios such as defiite ad idefiite itegratio ad differetiatio. m to the vector space. This liear Let f :U V be a liear map (homomorphism). The Image (Rage) ( ) ( ) { } im f = f u u U is called the image (or rage) of f. It ca be show that im( f ) Rak rak ( f ) dim( im( f )) V is a subspace of V. = is called the rak of f. Rak is a dimesio of the image of f. Kerel ker ( f ) = { U f ( ) = } u u is called the kerel of f. Kerel is a subset of all vectors i space U which are mapped ito the zero vector of the space V. It ca be show that ker f U is a subspace of U. ( ) Theorem rak ( f ) = dimu dim ker( f ) The liear maps help to obtai ad to study the properties of solutios of algebraic ad differetial equatios.

7 Chapter III Liear Algebra September 6, 7 3. VECTORS Recall the defiitio of the colum-vectors i = i for which operatios of vector additio ad multiplicatio by a scalar are defied by the operatios with their compoets: y + y k y + y k + y = + =, k = y + y k The set of all colum-vectors is a vector space of dimesio with the stadard basis e =, e =,, e = 4. MATRICES Recall the vector space of all real matrices with m rows ad colums m A ( aij ) a a a a a a a a m m m, = i=,..,m = j=,..., a a a a m m mm The set of elemets with equal idices { a kk } is called the mai diagoal. aij If the umber of rows is equal to the umber of colums, = m, the the matri is called a square matri. A matri ca be iterpreted as a combiatio of its colum-vectors: aj aj Am = ( a a a ) with colum-vectotrs ai = j =,,..., a mj or their row vectors: A m A A = Am with row-vectors A ( a a a ) = i =,,...,m i i i i Colum-vectors ca be treated as the matrices. Equality A = B Equality of matrices: two matrices of the same size AB, m are equal if ad oly if all the correspodig matri elemets are equal: A = B if ( aij ) i,..,m = ( bij ) = i=,..,m j=,..., j=,..., mai diagoal

Chapter III Liear Algebra September 6, 7 73 Multiplicatio by a scalar ka Let A m, k, the ka ka ka ka ka ka k A = kam kam kam Additio A+ B Additio of two matrices of the same size AB, m A+ B m a + b a + b a + b a + b a + b a + b A+ B = am + bm am + bm am + b Traspose T A Traspose of a matri A T m A m Symmetric Matri A T T m a a a a a a T a a a a a am ( aij ) i,..,m ( aji ) = = = i=,..,m = j=,..., j=,..., a a a a a a m m m m This operatio iterchages the compoets aij aji. T Properties of traspose: ( ) T A = A ( ) T T T A+ B = A + B ( ) T T ka = ka ( ) = ( ) T T A A for ivertible matrices The square matri A is called symmetric if A T = A. The square matri A is called skew-symmetric if A T = A. Every square matri ca be separated ito a sum of a symmetric matri ad a skew-symmetric matri: T T A = As + Aa = ( A+ A ) + ( A A ) Matri multiplicatio AB Matri Multiplicatio: the product of two matrices AB is defied if the umber of the colums of the matri A is equal to the umber of rows of the matri B : for A m, B p AB mp Let A Am ( aij ) a a a a a a = = i=,..,m = j=,..., am am am b b b p b b b p p ( b B = B = ij ) i=,.., = j=,...,p b b b p ( ) = a a a ( p) = b b b

74 Chapter III Liear Algebra September 6, 7 The product AB yields a m p matri which ca be defied i oe of the followig ways akbk akb k akbkp k= k= k= k k k k k kp AB a b a b a b = k= k= k= amkbk amkb k amkb kp k= k= k= AB = ( Ab Ab Ab ) ( ) ij 3 = a b + a b +... + a b AB i j i j i j The elemet i the itersectio of the th i row ad the th j colum ( AB ) ij of the matri AB is obtaied by a scalar product of the i th row of the matri A with the th j colum of the matri B A j B j AB i = i ( AB) ij Properties of matri multiplicatio: Let A m, B p, B p, B p C pr, m,,p,r. cab = A( cb) = c( AB ). ( AB) C = A( BC ) associative law A B + B = AB + AB distributive law 3. ( ) 4. ( ) B + B C= BC+ BC distributive law 5. AB = A = or B = 6. AB = AB B = B 7. ( ) T T T AB = B A traspose of a product 8. A = zero matri 9. AI = A uit matri. AB BA do ot commute

Chapter III Liear Algebra September 6, 7 75 Eample: ( ) A B = AB 3 4 3 4 3 4 5 6 3 4 5 6 7 8 4 7 = 3 3 37 44 35 46 57 68 Particular case: product of a matri ad a vector Am = b m System of algebraic equatios: System of m algebraic equatios for ukows,,..., : matri form A = b m m a a a b a a a b = a a a b m m m m a + a +... + a = b a + a +... + a = b a + a +... + a = b m m m m Let y, be vectors. Two types of vector product ca be defied Ier product of colum-vectors (dot product) is defied as y T y y y= = y + y +... + y y The results is a umber, y. Ier product (dot product) [ ] T Outer-product y [ y y y ] Outer-product (dyadic product) of colum-vectors is defied as y y y y y y = = y y y The result is a matri, T y.

76 Chapter III Liear Algebra September 6, 7 5. LINEAR TRANSFORMATIONS WITH THE HELP OF MATRICES Let A ( a a a ) = where m m ai ai ai = ami m i =,,..., Matri multiplicatio of vectors geerates a liear trasformatio m : A A vector is trasformed to a vector A m liear trasformatio = A b m b b m The resultig vector ca be writte as A = m m b m A = ( a a a ) = a + a a m +...+ i Therefore, the image of a liear map is a set of all liear combiatios of the colums of the matri A which by the defiitio is a spa. Therefore, the m image is a subspace of the vector space ; let us call it the colum space of the matri A ad deote it i the followig way: colum space col = spa {,,..., } A a a a, cola m rak rak A = dim cola Recall the defiitio of the kerel of a liear map deoted by { } ker A = A = ker A Therefore, the kerel of a liear trasformatio defied with the help of the matri A is a set of all solutios of a homogeeous system of liear algebraic equatios A =. The colum space of the matri A is a set of all vectors m b which are images of some vectors, i.e. for which there eists a solutio of the system of o-homogeeous system of liear algebraic equatios A = b.

Chapter III Liear Algebra September 6, 7 77 Theorem 3 dim( ker A) + raka = A ( A ) rak = dim ker < m Cosider trasformatio m give by m A m. vector space m vector space trasformatio A A = m b A = b c ker A col A dim( ker A ) rak A m This importat theorem of liear algebra allows oe to make some coclusios about the solutios of the system of liear algebraic equatios: If b cola the there eists A b. such that = If c cola the there is o A c. such that = If the homogeeous system A = has oly the trivial solutio =, the the dim ker A = ad therefore rak A =. ker A = { } ad ( ) The for ay b cola there eists the uique If A is a square matri (m=), the it meas that m, ad therefore, for ay m b there eists A b. such that = col A = m is the whole space A b. such that = If colums of A are liearly idepedet the the dimesio of the vector space spaed by liearly idepedet vectors is, rak = dim ker A =. Therefore, there is oly the trivial solutio for A =. A ad ( )

78 Chapter III Liear Algebra September 6, 7 6. DETERMINANTS Determiats are defied oly for square matrices A ( matrices). The determiat of a matri A is a umber calculated accordig to special rules. The determiat is deoted i oe of the followig ways: det A Defiitio a a a a a a a a a a a a det det A = = a a a a a a m m m m m m Oe possible defiitio of the det A is based oly o its properties (it does ot show actually how to calculate the determiat): Determiat of the square matri det A : which satisfies the followig coditios:. the determiat is liear i every row det Ai + Bi = det Ai + det Bi det cai = c det Ai where the dots idicate uchaged rows. A is defied as a fuctio for every i =,,...,. If matri A cotais two equal rows Ai = Aj the det A =. Defiitio 3. Determiat of the uit matri det I = = The other defiitio demostrates the structure of the determiat ad provides the method of its calculatio: Determiat of the square matri A is a umber equal to the sum of all possible products of matri etries such that every term cotais eactly oe etry of each row ad each colum of the matri weighted by or i the followig way: a b A ad bc c d = A 33 a a a 3 a a a 3 a a a 3 3 33 = a a a a a a + a a a 33 3 3 3 3 a a a + a a a a a a 33 3 3 3 3 ( ) ( ) a ( a a a a ) = a a a a a a a a a a 33 3 3 33 3 3 + 3 3 3 a a a a a a = a a + a 3 3 3 a3 a33 a3 a33 a3 a3 = a A a A + a3 A3 = ac + ac + a3c3

Chapter III Liear Algebra September 6, 7 79 j A ij Where the mior A ij is defied as a matri obtaied from the matri A by deletig the i th row ad the th j colum i the cofactor C = ( ) i j Aij = ( ) i j det Aij ij + + The the defiitio of the determiat of the matri i the followig epasio over the st row form: A ca be formalized + j j j j j j= j= ( ) det A = a A = a C Theorem 4 (the cofactor epasio theorem) The determiat ca be calculated by the cofactor epasio over ay row: i + k ik ik ik ik k= k= ( ) det A = a A = a C i =,,..., or ay colum: k + j kj kj kj kj k= j= ( ) det A = a A = a C j =,,..., I practice, for calculatio of the determiats the cofactor epasio method is used oly for small matrices ( or 3 3) or matrices of special form (diagoal, triagle, sparse, etc). Theorem 5 (the determiat of the triagle matrices) The determiat of the upper-triagle, the low-triagle ad the diagoal matrices are equal to the product of the elemets o the mai diagoal: d d d d d d = = = d d d d d d Theorem 6 (properties of the determiat) Let AB, the. ( ) det c A = c det A det AB = det A det B. ( ) T 3. ( A ) det = det A 4. det A = raka < 5. det ( A ) = det A for ivertible A 6. If two rows of A are the same, the det A = 7. If A has a row(or a colum) all of zeroes, the det A =

8 Chapter III Liear Algebra September 6, 7 7. MATRIX INVERSE Defiitio Let matri A be a square matri. If there eists a matri C such that CA = I left iverse AC = I right iverse the the matri A is said to be ivertible, ad the matri C is called a iverse of A. It is deoted by A : iverse matri A A A = AA = I Theorem 7 If the iverse of A eists the it is uique Theorem 8 Let AB, be ivertible matrices, the a) ( ) A = A ( A is also ivertible) b) ( ) AB = B A T c) ( A ) = ( A ) T Theorem 9 If A is ivertible the A is row equivalet to I Theorem If A is ivertible the det A Theorem If A is ot ivertible the det A =. sigular matri Matri with zero determiat det A = is called sigular. Theorem If A is ivertible the for ay b the liear system A = b has a uique solutio = A b Proof: a) = A b is a solutio: b) Let u ( ) ( ) A A b = b Au = b be aother solutio AA b = b A ( Au) = A b Ib = b ( A A) u = b = b u = Iversio by row reductio: The algorithm for fidig A (for row reductio procedure see below): A I ) start with augmeted matri [ ] ) by Gaussia elimiatio reduce to I A Iverse of matri: Let a b A = c d, the d b d b A = = det A c a ad bc c a Cofactor formula for iverse: Defie the adjoit of matri A as the matri which etries are cofactors: C C C C C C adj A = C C C the the iverse of matri A is defied by adja A = det A, ij = det A C ( ) i j ij +

Chapter III Liear Algebra September 6, 7 8 Eample: Let A = 5 3 ) Row reductio (iversio of a uit matri): A I = 5 3 5 3 R 5 3 5 3 5 R 5R R R R R R R R 3 A = 5 ) det A = = 3 5 = 5 3 d b 3 3 A = = = det A c a 5 5 3) Cofactor formula for iverse: [ ] ( ) + A = 3 deta = 3 C = 3 = 3 [ ] ( ) + A = 5 deta = 5 C = 5 = 5 [ ] ( ) + A = deta = C = = [ ] ( ) + A = deta = C = = adja A = det A 3 5 3 = = det A 5

8 Chapter III Liear Algebra September 6, 7 8. ROW REDUCTION: The followig elemetary operatios with matri rows, their symbolic desigatio ad affiliated with them elemetary matrices are itroduced as: Iterchage R i Rj iterchage rows i ad j i E = j Scalig kr i multiply row i by k E = k i j Replacemet Ri + kr j Ri replace row i by the sum of row i ad row j multiplied by k i j det E = det E = k With the help of the elemetary matrices, the elemetary row operatios ca be performed by matri multiplicatio (with the correspodig actig o the determiat for square matrices): operatio iterchage operatio scalig operatio replacemet 3 E 3 = k i E A ( A) E A ( A) E A ( A) det E = det A det E = k det A det E = det A 3 det E3 = Two matrices B ad A are called row equivalet if oe of them is obtaied from the other by a sequece of elemetary operatios B = E k...e s k E k Row equivalet matrices have the same kerel. A Row Reduced Echelo Form: Row Reduced Echelo Form (RREF) is a matri which etries satisfy the followig coditios: pivot pivot rows pivot colums. All o-zero rows are above ay row with all zeroes.. Each leadig etry (first o-zero elemet) of a row is to the right of the leadig etry of the row above. 3. All etries i the colum below the leadig etry are zero. 4. All leadig etries are. 5. Leadig etry (pivot) is the oly ozero etry i the colum. Theorem 3 Every matri A is row equivalet to the uique RREF U

Chapter III Liear Algebra September 6, 7 83 Row Reduced Algorithm Gaussia Elimiatio Row operatios: Step 3 3 4 3 3 5 Fid the first pivot colum (leftmost o-zero colum) A Step 3 Select ay o-zero etry i the R R 3 3 4 pivot colum as a pivot. 3 If ecessary iterchage rows to move pivot ito pivot positio 5 Step 3 3 Use row replacemet operatios to R 3R R 9 create zeroes i all positios R3 + R R3 9 3 below the pivot R4 R R4 4 Step 4 3 Igore the pivot rows ad repeat R3 + R R3 9 steps -3 to the remaiig matri R4 R R4 R 3 9 R R 3 R 4 R 3 4 Step 5 Startig from the rightmost pivot R 3 R4 R3 use row replacemet operatios R + R4 R to create all zeroes above pivot R R4 R (back elimiatio) R + R3 R R 3R R We obtaied the RREF 3 R R R

84 Chapter III Liear Algebra September 6, 7 9. LINEAR SYSTEMS: We itroduced all ecessary foudatio ad tools to solve ad to determie the properties of a system of algebraic equatios. The liear system of m algebraic equatios for ukows,,..., ca be defied i the followig form: stadard form a + a +... + a = b a + a +... + a = b a + a +... + a = b m m m m where the system coefficiets fied real umbers. aij ad the system costats bj are System coefficiets ca be orgaized i the matri compoet form or i the colum-vector form: a a a a a a A = am am am, A = ( a a a ) m, ai ai ai = ami The vector of costats ad the vector of ukows ca be writte as: b b b = bm m = Augmeted matri (matri of coefficiets with additioal colum of costats): m augmeted matri A # a a a b a a a b = am am am b m A # = m+, ( a a a b ) o-homogeeous system A = b With the help of these otatios, the liear system ca be writte i the matri form: or i the vector form: a + a +... + a = b Associated with this system is the homogeeous system ( b = ): homogeeous system A = Solutio of the o-homogeeous system A = b is ay vector which satisfies this system. If the liear system has at least oe solutio, the it is called cosistet, otherwise it is icosistet. The homogeeous system A = has at least oe solutio = (trivial solutio), ad therefore, it is always cosistet. The o-zero vector satisfyig A = is called a o-trivial solutio. For the liear system we wat to get the aswer to the followig questios: Is the system cosistet? Ad if yes, the how may solutios does it have ad how are they foud? The mai method of solutio of the lier system is Gaussia Elimiatio: row reductio of the matri A for a homogeeous system, ad # row reductio of a augmeted matri A for a o-homogeeous system.

Chapter III Liear Algebra September 6, 7 85 Row reductio of A for homogeeous system A =. Solutio space of homogeeous system ker A. 5 If i the RREF of matri A there eists a o-pivot colum, the this colum correspods to a free variable. I this case, there eists a o-trivial solutio of the homogeeous system. For eample, let the RREF have the followig form: This row reduced echelo form represets the liear system A = : + + 3 = 5 + 4 = 3 5 4 + 5 = The the correspodig liear system with arbitrary values for free parameters ad 5 ca be writte as + + 3 = 5 = + 4 = 3 5 + = 4 5 5 = 5 Solve each equatio i terms of free variables = 3 5 = = 4 3 5 = 4 5 free varibles pivot rows pivot colums 5 = 5 ad rewrite the system i colum-vector form; ad the epad the right had side ito compoet form with free variables as coefficiets 35 = 3 = 4 = + 5 4 5 5 5 3 4 5 3 4 pivot rows basic variables 3 4,5 For ay values of free parameters,5, this equatio provides the solutio of the homogeeous system. Therefore, the solutio vector is a liear combiatio of two vectors or a spa of these two vectors. 3 spa 4, ker A = solutio space of A = c = ker A = the solutio space of the homogeeous system A = (also called a complimetary solutio) c = ker A = 3 4 5 3 = spa 4, = spa, { u u }

86 Chapter III Liear Algebra September 6, 7 This solutio set of the homogeeous system A = is a vector space which we called the kerel of the matri A. Here, we eplicitly costructed the vector space ker A. Because it is spaed by two liearly idepedet vectors, it has a dim ker A =. The accordig, to the Theorem, the dimesio two: ( ) dimesio of the set spaed by the colum vectors of the matri A is determied as ( A) + A = A ( A ) dim ker rak rak = dim ker = 5 = 3 This dimesio correspods to the umber of pivot colums i the RREF of A. If all colums i RREF are pivot colums the there are o free variables, ad the homogeeous system i this case has a trivial solutio =. Here, we demostrated the techique of eplicit costructio of the solutio space ker A of the homogeeous system A =. There are some facts about the cosistecy of liear systems: Theorem4 If the umber of ukows is greater tha the umber of equatios m, the homogeeous liear system has the o-trivial solutios: a a a a a a Let A = m am am am with m<, the there always eists = such that A = Theorem5 The dimesio of the solutio space ker A of the homogeeous liear system A = is equal to the umber of free variables i the row reduced echelo form of the matri A. If m<, the the dimesio of solutio space is at least m. Theorem6 The o-homogeeous system A = b is cosistet if ad oly if the RREF of the augmeted matri A # = ( a a a b ) has o pivot i the last colum (o row of the form ( ) : pivot colums o pivot i last colum * * * * * pivot rows The solutio set of the cosistet liear system A = b cosists of - a uique solutio if there are o free variables i RREF; - ifiitely may solutios whe there are free variables. I this case the dimesio of the solutio space is equal to the dimesio of the ker A (umber of free variables).

Chapter III Liear Algebra September 6, 7 87 Theorem7 (Solutio of the o-homogeeous system A = b ) Let the vector p be some particular solutio of A = b. The ay solutio of the o-homogeeous system A = b is represeted as = + c p where c = ker Ais the solutio of homogeous system A = (complimetary solutio). Proof: ) Show that = c + pis a solutio of A = ( ) A = A + = A + A = + b = b c p c p b. Ideed ) Show that for ay solutio of A = b, the vector c defied by the equatio = c + p is a solutio of the homogeeous system. Ideed = + c p c = p ( ) A = A = A A = b b = c p p vector space A m vector space geeral solutio of o-homogeeous system A = b = + c p A = b b particular solutio A = b p p A = c ker A = geeral solutio of homogeeous system A= col A Row reductio of the augmeted matri # A procedure for solutio of o-homogeeous system A = b : The solutio of o-homogeeous system A = b cosists of the followig steps: ) Write a augmeted matri A # = ( a a a b ) ) Use Gaussia elimiatio to obtai RREF determie if liear system is cosistet 3) Solve the system for basic variables 4) Rewrite solutio i a vector form usig free variables as the parameters

88 Chapter III Liear Algebra September 6, 7 Eample: Fid the solutio of the liear system + = 3 3 + 6 = 9 ) augmeted matri: # 3 A = 3 6 9 ) Gaussia elimiatio: 3 3 6 9 R 3R R 3 RREF basic variable free parameter 3) Equivalet system: + = 3 = Solve for basic variables: = + 3 = 4) Rewrite i a vector form: + 3 = From which the geeral solutio follows as: 3 = + The geeral solutio of the o-homogeeous system ca be writte i a vector-parametric form: = tv+ p where v 3 =, p = The solutio has a geometrical visualizatio i the -plae: v = 3 t, t = + 3 p = t = is a parametric equatio of the lie alog the vector v =, represetig the solutio set of the homogeeous system (kerel of the matri A ). 3 The by traslatio of this set by a costat vector p =, the geeral solutio of a o-homogeeous liear system is obtaied as the parametric equatio of the lie: 3 t, t = +

Chapter III Liear Algebra September 6, 7 89 LINEAR SYSTEMS WITH SQUARE MATRICES: Theorem8 Let A. The homogeeous liear system A = has a o-trivial solutio if ad oly if det A = Theorem9 If A is ivertible the for ay b the liear system A = b has the uique solutio = A b Theorem (Cramer s Rule) If A is ivertible the for ay b the uique solutio of the liear system A = b is the vector = the compoets of which are determied by the followig formula: k = det B k det A k =,,..., where matrices B k are obtaied from the matri A by replacig the colum-vector a k by the vector of costats b : [ ] B = a a b a a k =,,..., k k k+ Eample: Solve the liear system 3 + + 4 = 3 + = 3 + + 33 = Rewrite the system i the matri form: A = b where 3 4 A =, b =, = 3 3 Determiat of the matri of coefficiets det A = 5, therefore: ) The solutio of the liear system is give by 6 3 4 5 5 5 = A b = = = 3 4 7 5 5 5 ) The same solutio ca be obtaied with Cramer s Rule: 4 3 4 3 3 = = 3 4 5 3 = = 3 4, = = 3 4 5, 3 3 3 3

9 Chapter III Liear Algebra September 6, 7 INVERSE MATRIX THEOREM: The followig theorem combies some previous results ad reveals the coectio betwee differet aspects of liear algebra: Theorem (the iverse matri theorem) Let A be a square. matri, A = ( a a a ) The the followig statemets are equivalet:. A is ivertible. There eists A such that 3. A = has oly the trivial solutio = 4. For ay b there eists A A = AA = I such that A = 5. The set of colum-vectors {,,..., } 6. { } 7. spa a, a,..., a = col A = 8. raka = 9. A has pivots a a a is liearly idepedet. A is row equivalet to the idetity matri I. ker A = { }. ( A ) dim ker = 3. det A b

Chapter III Liear Algebra September 6, 7 9. EIGENVALUE PROBLEM: We have see that liear trasformatio of a vector space ca be defied with the help of matrices. Cosider the liear trasformatio A λ T ( ) : defied by a matri A as ( ) T = A. There is the followig questio cocerig the output of the liear trasformatio: are there some vectors which oly are scaled uder the give trasformatio but ot rotated: A = λ Such vectors if they eist are called the eigevectors ad the correspodig scalig coefficiets are called the eigevalues of A. They ca be foud by solutio of vector equatio A = λ. It appears that these eigevalues, i geeral, are the real or comple, λ or, ad the correspodig eigevectors are the vectors over or. Defiitio (eigevalues ad eigevectors) Let A be a matri. The value λ or for which equatio A = λ has a o-trivial solutio,, is called a eigevalue of the matri A. No-zero solutio vector is called a eigevector correspodig to eigevalue λ. It is obvious that if is a eigevector correspodig to the eigevalue λ, the ay scalar multiple c also is a eigevector correspodig to the eigevalue λ. Ideed: A c = λ c ca = cλ A = λ provided c. ( ) ( ) Ad if,..., k are the eigevectors correspodig to eigevalue λ, the ay their liear combiatio c +... + ck k also is a eigevector correspodig to eigevalue λ. Defiitio (eigespace) The set of all eigevectors correspodig to eigevalue λ completed with the zero vector (which is the trivial solutio of A = λ ) is called the eigespace of the matri A correspodig to λ : Eig A ( λ) = { A = λ } { } Eigevectors ad eigevalues of the matri A ca be foud by solvig the vector equatio A = λ A λ = A λi = eigevalue problem ( λ ) A I = This is the homogeeous liear system which has a o-trivial solutio oly if the matri of the system has the zero determiat ( A λi ) det =

9 Chapter III Liear Algebra September 6, 7 Characteristic equatio ( A I) Therefore, the eigevalues of the matri A have to satisfy this equatio, which after evaluatio of the determiat becomes a algebraic equatio i λ of degree : αλ + α λ +... + αλ + α= with real coefficiets αi Accordig to the Fudametal Theorem of Algebra there eist eactly roots of this equatio which ca be real distict, repeated, or comple cojugate. This equatio is called the characteristic equatio: a λ a a a a λ a det λ = = αλ + α λ +... + αλ + α= a a a λ Theorem The scalar λ is a eigevalue of the matri it satisfies the characteristic equatio det A λi = ( ) A if ad oly if Theorem3 If λ, λ,..., λ m are distict eigevalues of the matri A, the the correspodig eigevectors,,..., m are liearly idepedet. Theorem4 If λ is a root of multiplicity k of the characteristic equatio det A λi = ad therefore it is a eigevalue of the ( ) matri A, the the dimesio of the correspodig λ k eigespace is dim Eig ( ) A. dim EigA λ = It meas that oly oe liearly idepedet eigevector correspods to λ. So, if λ is a root of multiplicity, the ( ) Theorem5 If matri A is symmetric, real. A T = A, the its eigevalues are Theorem6 If matri A is ivertible the is ot its eigevalue. If is a eigevalue of A the det A =. PROCEDURE FOR SOLVING AN EIGENVALUE PROBLEM:. Costruct the characteristic equatio det A λi = ( ). Fid the roots of the characteristic equatio λ k eigevalues (ca be distict, repeated, comple cojugate) 3. For each eigevalue λ k solve the liear system ( ) A I u = λ k to fid eigevectors uk correspodig to λ k.

Chapter III Liear Algebra September 6, 7 93 EXAMPLES:. (distict real eigevalues) Solve the eigevalue problem for A =. 3 ) Characteristic equatio: det λ = λ 5λ + 6 = 3 λ ) The roots are: λ = λ = 3 3) Vector equatio for λ = : ( ) A I u = λ = Row reductio of matri of coefficiets: R R row reduced echelo form Correspodig equatio icludes oe basic variable ad oe free variable: + = Solve for the basic variable = Ay value of free parameter provides a solutio, choose =, the =. Therefore, the solutio vector (eigevector) is u = = Vector equatio for λ = 3 : ( ) A I u = λ = Row reductio of matri of coefficiets: R + R R R row reduced echelo form The correspodig equatio icludes oe basic variable ad oe free variable: + = Solve for the basic variables = Ay value of the free parameter provides a solutio, choose =, the =. Therefore, the solutio vector (eigevector) is u = =

94 Chapter III Liear Algebra September 6, 7. (real eigevalues, repeated eigevalue) Solve the eigevalue problem for ) Characteristic equatio: A = 3. 3 ) The roots are: λ = 4 λ det 3 λ = 3 λ λ = = eigevalue of multiplicity λ3 3) Vector equatio for λ = 4 : ( ) A I u = λ 4 3 4 = 3 4 3 = 3 Row reductio of matri of coefficiets: R + R R R R R R R 3 3 row reduced echelo form Correspodig system of equatio icludes two basic variables, ad oe free variable 3 : + 3 = + = 3 Solve for the basic variables = 3 = 3 3 Ay value of the free parameter 3 provides a solutio, choose 3 =, the = =. Therefore, the solutio vector (eigevector) is u = = 3

Chapter III Liear Algebra September 6, 7 95 Vector equatio for λ = : ( ) λ A I u = 3 = 3 3 = 3 Row reductio of matri of coefficiets: R R R R + R R 3 3 3. (comple eigevalues) row reduced echelo form The correspodig equatio icludes oe basic variable ad two free variables, 3 (the dimesio of kerel is ): + 3 = 3 = 3, It is possible to costruct two idepedet vectors with arbitrary ad = 3. The simplest o-zero vectors satisfyig these coditios are u = =, u 3 = = 3 3 Therefore, for the give matri, there are two liearly idepedet vectors correspodig to eigevalue of multiplicity two. If λ = a + bi is a eigevalue of the matri with real coefficiets A with the correspodig eigevector u, the the comple cojugate of it λ = a bi is also a eigevalue of A with the eigevector u. Ideed, let Au = λu Au = λu Au = λ u Eample: Solve the eigevalue problem for ) Characteristic equatio: λ 5 = + = 3 λ det λ 4λ 3 5 A =. 3

96 Chapter III Liear Algebra September 6, 7 ) The roots are: λ = + 3i λ = 3i 3) Vector equatio for λ = + 3i : ( ) A I u = λ k ( ) + 3i 5 = 3 ( + 3i) 3i 5 = 3i Row reductio of matri of coefficiets: ( ) 3i 5 -+3i R 3i 5 + 5i R 5 3i + 3i 3i R + R R + 3i row echelo form The correspodig equatio icludes oe basic variable ad oe free variable: + + 3i = Solve for the basic variable ( ) ( ) = 3i Ay value of free parameter provides a solutio, choose =, the = 3i. Therefore, the solutio vector (eigevector) is u 3i 3 i = = = The the eigevector correspodig to the secod eigevalue λ is 3 3 u = u = i = + i 4. (system of ODEs) For applicatio of the eigevalue problem for solutio of the System of Liear Ordiary Differetial equatios see Chapter V (Sectio 5.5).

Chapter III Liear Algebra September 6, 7 97. LINEAR ALGEBRA WITH MAPLE: May liear algebra operatios ca be performed oly usig lialg package which is dowloaded by the commad: > with(lialg): Eterig vectors ad matrices: by array: > u:=array([3,,-3]); u := [ 3,, -3] by commad: > A:=array([[-,,],[3,,]]); > v:=vector([,,-]); A := - 3 v := [,, -] Operatios with vectors: dot product: cross product: > B:=matri(3,,[,,3,4,5,6]); > w:=vector([,,]); > dotprod(u,v); > crossprod(u,v); B := 3 4 5 6 w := [,, ] [ 4,, 4 ] agle betwee two vectors: > beta:=evalf(agle(u,v));beta:=evalf(covert(beta,degrees)); Operatios with matrices: β :=.54859555 β := 9.49685 degrees Arithmetic operatios are performed by applyig commad evalm additio : > c:=evalm(u+v); scalar multiplicatio : > C:=evalm(*B); matri multiplicatio : > F:=evalm(A&*B); traspose : > traspose(c); c := [ 4, 4, -4] 4 C := 6 8 F := 4 6 4 8

98 Chapter III Liear Algebra September 6, 7 determiat: iverse: adjoit: > det(f); > iverse(f); > adjoit(f); row reduced echelo form (RREF): > rref(a); rak of matri kerel of matri: > rak(a); > kerel(a); colum space of a matri: > colspa(a); 5 8-7 6 3-3 8 5 6 - -4 {[,, -]} {[-, 3 ], [, -8]} Solutio of liear system of algebraic equatios > A:=matri(,,[,4,-,3]); > b:=array([[],[-]]); solutio with lisolver: > :=lisolve(a,b); A := 4-3 b := := solutio with iverse matri: > :=evalm(iverse(a)&*b); The Eigevalue Problem: eigevalues: := - -3-3 > G:=matri(3,3,[,,,6,-,,-,-,-]); > eigevalues(g); G := 6 - - - -, 3, -4

Chapter III Liear Algebra September 6, 7 99 > eigevectors(g); [,, {[, 6, -3]}], [-4,, {[-,, ] }], 3,, { } -, -3, with otatios: [eigevalue, multiplicity of eigevalue, {[correspodig eigevector]}] Differetiatio of matrices (each etry is differetiated): > A:=matri(,,[3*t,si(3*t),ep(*t),*t*ep(4*t)]); > Ap:=map(diff,A,t); 3 t si( 3 t ) A := e ( t) t e ( 4 t) 3 3 cos( 3 t ) Ap := e ( t ) e ( 4 t) + 8 t e ( 4 t) Itegratio of matrices (each etry is itegrated) > Ai:=map(it,A,t); Ai := 3 t cos( 3 t ) 3 t 4 t) 4 t) e( 8 e( t) e( Evaluatio of values of the fuctios iside of the matri A: > A:=matri([[si(),cos()],[ep(),*]]); si( ) cos( ) A := e > for from to do A[,]:=subs(=Pi/,A[,]) od; := si π A, A, := e > for from to do A[,]:=subs(=Pi/,A[,]) od; > evalm(a); > evalf(evalm(a)); > map(evalf,a); A, π := cos π A, si π π e := π cos π. -.53388-9 4.847738 3.459654. -.53388-9 4.847738 3.459654 π

Chapter III Liear Algebra September 6, 7. REVIEW QUESTIONS: ) What is a vector space? ) What operatios with vectors are defied i vector spaces? 3) What is the meaig of closure aioms for vector spaces i) ad ii)? 4) What is a vector subspace? 5) What coditios should be satisfied by the subset of a vector space to be a subspace? 6) Recall eamples of vector spaces. 7) Is the uio of vector subspaces of the vector space V a vector space? 9) Is the itersectio of vector subspaces of the vector space V a vector space? ) What is a liear combiatio of vectors? ) What is a spa of a set of vectors? ) What vectors are called liearly idepedet? 3) Is the set of vectors 4) What is the basis of a vector space?,, liearly idepedet? 5) What is the dimesio of a vector space? 6) What is a liear map betwee vector spaces? 7) What is the image of a liear map? 8) What is the kerel of a liear map? 9) What is the rak of a liear map? ) What is the relatioship betwee the rak ad the dimesio of the kerel of a liear map? ) Whe are two matrices equal? ) How is the sum of two matrices defied? 3) What is a symmetric matri? 4) For what matrices is the product of matrices defied? 5) Give a eample of two o-zero matrices such that AB =? 6) Show that if AB =, for, BA. AB, the ( ) = 7) How is the trasformatio of vector space defied with the help of matrices? 8) What is the determiat? 9) What is the matri iverse? 3) What are the elemetary row operatios? 3) What is the Row Reduced Echelo Form (RREF)? 3) What is a cosistet liear system? 33) Whe does a homogeeous system A = have a o-trivial solutio? 34) What is the method of solutio of liear systems of equatios? 35) What are eigevalues ad eigevectors?

Chapter III Liear Algebra September 6, 7 EXERCISES: ) Prove the followig statemets for the vector space V : a) The egative elemet u V is uique. b) Zero vector V is uique. c) u = for ay u V. ) Let Ui V be the subspaces of the vector space V. Show that the itersectio of subspaces i= i { u u i } U = V U, i=,..., is a subspace of V. 3) Let V be a vector space ad let u, u,..., u V. spa u, u,..., u is a subspace of V. Show that { } 4) Show that if { u, u,..., u } the the set {,,..., } depedet. u u u is liearly 5) Determie if the followig sets are liearly idepedet or depedet: a),, b), 6) What is the stadard basis for the vector space a b = a,b,c,d c d 7) Let V be a dimesioal vector space the (fill i the blak with >=<):,, a) if {,..., p} b) if {,..., p} c) if {,..., p} u u spas V, the p u u is a basis for V, the p u u is a liearly idepedet set i V, the p 8) Show that if { uv, } is a liearly idepedet subset of { u+ vu, v } is also liearly idepedet., the 9) Suppose that v, v,..., v m are liearly idepedet vectors i be ay vector that is ot i spa {,,..., } are liearly idepedet. ) Let A be a square matri, A. Show that ad let u v v v m. Show that m T As = ( A+ A ) is a symmetric matri ad T Aa = ( A A ) is a skew-symmetric matri (atisymmetric). v, v,..., v, u

Chapter III Liear Algebra September 6, 7 ) Fill i the blak with the best symbol or umber: a) If A is a ivertible matri, the rak A = b) If A is a ivertible matri ad det A = d A det = ) True False questios: a) If AB, are ivertible b) For ay m matri A m, the matri c) For ay m matri A m, the matri matrices, the ( ), the AB = A B T AA is symmetric T A A is symmetric d) If A is ivertible, the colums of A are liearly idepedet e) All eigevalues of a symmetric matri are real f) The set of all ivertible matrices is a subspace of the vector space of all matrices 3) Prove the followig statemet by mathematical iductio: Let A, A,..., Am be ivertible matrices. The AA Am is m = m m ivertible ad ( ) AA A A A A. 4) Let 3 4 A = ad 3 B = 3, the compute a) 3A b) A+ B c) A B d) AB e) T AB f) ( BA ) T 5) Let A = ad B = 3. Determie if it is possible to perform the followig operatios, ad if yes, fid the result: a) AB b) BA c) T BA d) T e) ( + ) A B B f) T T AB T AA 6) Let A =. Compute A = AA, A = AA. What will be 3 A?

Chapter III Liear Algebra September 6, 7 3 7) Fid the Row Reduced Echelo Form of the followig matrices: a) 3 4 5 3 5 b) 4 4 4 c) 4 3 5 3 8 5 7 d) 3 3 4 6 4 3 8) Solve the followig liear system of equatios: a) 4 + 3 = 3 + = 3 b) + 3 = 3 + = 3 c) + + = 3 4 + 3 = 3 4 + 3 = 3 4 d) 3 + 3 = + + = 3 3 + = 3 e) 3 5 + + 4 = 3 4 7 4 + + 3 = 5 3 4 5 + 7 4 6 = 3 3 4 f) 3 + 5 + 4 = 3 4 6 4 + 4 + 3 = 3 3 4 9 6 + 3 + = 4 3 4 g) 3 + 9 + + = 3 4 5 + 6 + + = 3 4 5 + 6 6 + 6 + 7 = 3 4 5 h) 4 + 5 + + 3 = 7 3 4 5 3 8 + + = 5 3 4 5 3 + 3 = 3 3 4 5 9) Fid the determiat of the followig matrices: a) c) e) 6 A = 5 b) A = 4 5 3 d) 3 4 A = 4 3 f) A = 5 9 8 5 A = 4 3 3 5 A = 5 3

4 Chapter III Liear Algebra September 6, 7 ) Fid the iverse of the followig matrices: a) c) 5 A = 3 b) A = 3 d) A = 3 A = ) Solve the followig liear systems of equatios usig matri iverse ad Cramer s Rule: a) = + = b) 4 + 5 = 3 5 + = c) + 4 = + 4 = 3 + = 3 d) + 5 = 3 3 + + 5 = 3 + = 3 ) Solve the eigevalue problem for the followig matrices: a) c) e) g) i) A = b) A = d) 3 A = 5 3 f) 3 A = 3 3 h) A = 3 5 k) A = 3 3 A = 3 3 A = 5 3 A = A = 3) Show that the matri (the rotatio matri) cosθ siθ A = siθ cosθ has comple eigevalues whe θ is ot a multiple of π

Chapter III Liear Algebra September 6, 7 5

6 Chapter III Liear Algebra September 6, 7