Content Editor: Prof. Madhu Sir

Subject Expert: Dr. BijumonContent Editor: Prof. Madhu sirPresenter: Prof. NandakumarAlgebra 24Characteristic Space of a MatrixObjectivesFrom this unit a learner is expected to achieve the following1. Recall the definitions of characteristic roots and characteristic vectors.2. Study some results associated with characteristic roots and characteristic vectors.3. Familiarize the method of finding eigen vectors of a matrix.4. Learn the definition of algebraic and geometric multiplicity.Sections1. Characteristic Space (Eigen Space) of a Matrix 2. Method of Finding Eigen Vectors of a Matrix3. Algebraic and Geometric Multiplicity1 Charecteristic Space (Eigen Space) of a Matrix If X is a characteristic vector of a matrix A with characteristic root λ, th en any scalar multiple X is also characteristic vector of A with the same characteristic root, since A(X) = AX =  λX = λ(X). More generally, any non-zero linear combination of characteristic vectors t hat share the same characteristic root λ, will itself be an characteristic vect or with characteristic root λ. Together with the zero vector, the characteris tic vectors of A with the same characteristic root form a linear subspace c alled an characteristic space. In this session we discuss characteristic space of a given matrix cor responding to a characteristic root. The method of finding characteristic ve ctors associated with a characteristic root also will be discussed. The defini tion of algebraic and geometric multiplicity will be given. Theorem 1 Corresponding to a non-zero characteristic vector X of a squ are matrix A there exists one and only one characteristic root whereas corre sponding to a characteristic root there exists more than one characteristic v ector. Proof We recall that a complex number λ is a characteristic root of a matrix A if there is a non-zero column matrix X such that AX= l X .The column matrix X is said to be a characteristic vector of the matrix A associated with the characteristic root λ. We can also say that λ is a ch aracteristic root corresponding to the non-zero characteristic vector X.2 Let us assume that there exist two distinct characteristics roots l1 and l2 corresponding to a given non-zero characteristic vector X of a square ma trix A. Then we have AX= l1 X …(1) and AX= l2 X . …(2)Subtracting (2) from (1), we obtain (l1- l 2 )X = O , the zero column matrix. Since and are distinct, we have so (3) implies that l1 l2 l1- l 2 0 X= O which is a contradiction to the fact that X is a non-zero vector. Hence corr esponding to a non-zero characteristic vector X of the square matrix A there is only one characteristic root. Again if l be the characteristic root of A, and the non-zero vector X w ill be given by AX= l X Let k be any non-zero scalar, then k( AX )= k (l X ) i.e., A( kX )= l ( kX ). Thus, kX is also characteristic vector of A corresponding to the same chara cteristic root l . Thus corresponding to a characteristic root there exists mo re than one characteristic vector. Remark If λ is a characteristic root of a matrix A then any column m atrix X such that AX= l X is called a characteristic vector of the matrix A associated with the charact eristic root λ. In view of this, zero column matrix can be regarded as a cha racteristic vector of the matrix A associated with the characteristic root λ. Hence in the proof of Theorem 1, we can also take k = 0, the zero scalar , so that kX = 0, the zero column matrix is a characteristic vector associated with the characteristic root l .Theorem 2 If X1 and X 2 are two eigen vectors corresponding to the eig en value l of a matrix A then the linear combination aX1+ b X 2 (where a and b are any scalars) is also an eigen vector corresponding to the same eig en value.3 Proof. Suppose X1 and X 2 are two eigen vectors corresponding to the eigen value l of a matrix A. Then and X1构0, X 2 0 AX1= l X 1 and Then AX2= l X 2A(a X )= a A ( X ) 1 1 …..(3) =a( lX1 ) = l ( a X 1 ) and A(b X )= b A ( X ) 2 2 …..(4) =b( lX2 ) = l ( b X 2 )Adding (3) and (4), A(a X )+ A ( b X ) 1 2 =l( aX1 ) + l ( b X 2 ) implies A(a X1+ b X 2 ) = l ( a X 1 + b X 2 ) implies aX1+ b X 2 is also an eigen v ector corresponding to the same eigen value l . Now let l be a characteristic root of an n- square matrix A. Consider the matrix equation (A-l I ) X = O .…(5)Every non-zero solution of (5) is a characteristic vector of the matrix A corresponding to the characteristic root l.Let r be the rank of the matrix (A- l I ) . Then the equation (1) posses ses a linearly independent system of (n- r ) solutions. Every non-zero line ar combination of these solutions, also is a solution of (5), and hence is a ch aracteristic vector corresponding to l.The set of these linear combinations including the zero vector is a subspace of the vector space of n 1 column matrices, and is called the characteristic space (or eigen space) of the matrix A corresponding to the characteristic root l. The above discussion gives the following result:4 Theorem 3 The characteristic space of a matrix A corresponding to a characteristic root l is the null space of the matrix (A- l I ).Method of Finding Eigen Vectors of a MatrixIn view of the Theorem 3, to determine all the eigen vectors of a square matrix A we proceed as follows:Step 1 (Determination of eigen values) Find the eigen values of the matrix A by solving the characteristic equation A  λ I  = 0 for the value of . Step 2 (Determination of eigen vectors) If λ1 is an eigen value, then the eigen vector corresponding to λ1 is determined by solving for X the matrix equation (A  λ1 I ) X = 0 .10 3 Example 1 Find the eigen vectors of the matrix A   .  4 6SolutionStep 1 (Determination of eigen values) The characteristic equation of A is A  λI  = 0. 10   3 i.e.,  0 4 6  On simplification we get λ2  16 λ + 48 = 0, which gives the eigen values λ = 4; λ = 12. Step 2 (Determination of eigen vectors ) Step 2 (i) (Determination of eigen vector corresponding to the eigen value λ = 4 ) The eigen vector X corresponding to λ = 4 is obtained by solving [A – 4 I ] X = 0 5 i.e., by solving 10  4 3   x1  0       .  4 6  4 x2  0 i.e., by solving 6 3 x1  0       . …(6) 4 2 x2  0Applying row transformations, the coefficient matrix in (6) successively gives1 1 轾6 3 轾12 轾 1 2 犏 ~犏 ~ 犏臌4 2 臌4 2 臌 0 0 and hence the corresponding matrix equation is 1 轾1 2 轾x1 轾0 犏犏 = 犏 . 臌0 0 臌x2 臌0 Since, obviously, the rank of the coefficient matrix is 1 which is less than 2, the number of unknowns, we can assign arbitrary values to n- r =2 - 1 = 1 unknown, say x1  k , and obtain an eigen vector corresponding to λ = 4 as  x1   k     . x2   2k Putting k  1, we obtain a particular eigen vector  x1  1     . x2   2 Step 2 (i) (Determination of eigen vector corresponding to the eigen value λ = 12 ) The eigen vector X corresponding to λ = 12 is obtained by solving [A – 12 I ] X = 0 i.e., by solving 10 12 3  x1  0       .  4 6 12 x2  0 i.e., by solving 2 3  x1  0       . 4  6 x2  06 Since the rank of the coefficient matrix is 1 which is less than 2, the number of unknowns, we can assign arbitrary values to n- r =2 - 1 = 1 unknown, say x2  k , and obtain an eige n vector corresponding to λ = 12 as 3  x1   2 k     . x2   k Putting k  2, we obtain a particular eigen vector  x1   3     . x2   2Example 2 Determine the characteristic roots and associated characteristic vectors for the matrix A given by 轾2 2 1 A = 犏13 1 . 犏 1 2 2 臌犏Solution The characteristic equation is 2 - l 2 1 13 - l 1= 0 1 2 2 - l or l3-7 l 2 + 11 l - 5 = 0 and the characteristic roots are l1=5, l 2 = 1, l 3 = 1When  1  5 , ( A- l I) X = 0 becomes轾 3- 2 - 1 轾x1 犏-1 2 - 1犏x2 = 0 …(7) 臌犏-1 - 2 3 臌犏x33 2  1  0 1 1  Since 1 2  1  is row equivalent to 1 0 1  , Eq.(7) is equivalent to the matrix 1  2 3  0 0 0  equation轾0 1- 1 轾x1 犏1 0- 1犏x2 = 0 臌犏0 0 0 臌犏x3A solution is given by x1 x 2  x 3 1; hence, the characteristic space associated with theT characteristic root   5 is the one-dimensional vector space spanned by the vector [1, 1, 1] .T Every vector [k, k , k] of this space is a characteristic vector of A associated with the characteristic root   5 .7 When l= l2 =1, (A - l I ) X = 0 becomes轾-1 - 2 - 1 轾x1 犏-1 - 2 - 1犏x2 = 0 犏-1 - 2 - 1 犏x3 臌臌 or x1+2 x 2 + x 3 = 0.Two linearly independent solutions are (2,- 1, 0) and (1, 0, - 1) . Thus the characteristic space associated with the characteristic root  1 is the two-dimensional vector space spanned by T T T X1 =[2, - 1,0] and X2 =[1,0, - 1] . Every vector hX1+ kX 2 =[2 h + k , - h , - k] is a characteristic of A associated with the characteristic root l =1 .Example 3 Prove that the characteristic equation of an orthogonal matrix P is a reciprocal equation.Solution We note that if an equation in the variable x remains unaltered when x is changed into its reciprocal, it is called a reciprocal equation. A property of reciprocal equation is that, if a is a 1 root of a reciprocal equation, then is also root. aLet P be an n- square orthogonal matrix. Then the characteristic equation associated with P is|P-l I | = 0.Let f( l )= |P - l I |.Then f( l ) =P - l PIPT , since PPT = I骣 1 = -lP琪 PT - I 桫 l1 = �lnP T I , since P being orthogonal, P = 1, lT 骣 1 = �ln 琪P I 桫 l1 = �ln P I , since for any matrix B, BT = B . l骣1 = ln f 琪 . 桫l8 1 1 Hence f( l )= |P - l I | = 0 if and only if f( )= |P - I | = 0 , showing that the l l characteristic equation of an orthogonal matrix P is a reciprocal equation.Theorem 4 The product of all the characteristic roots of a square matrix of order n is e qual to the determinant of the matrix (with change in sign allowed).Proof. Let A= [ aij ] be a given square matrix. Let l1, l 2 , , ln are the n characteristic ro ots of A. These roots need not be distinct. Then the characteristic equation is (l- l1)( l - l 2 )( l - ln ) = 0. Also, in terms of determinants, the characteristic equation is given by |A-l I | = 0. From the above two equations, we have A-l I =�( l - l1)( l - l 2 )( l ln ).On putting l = 0, the above gives |A |= 弊l1 l 2 l 3 ⋯ ln . This completes the proof. Example 4 Find the latent roots and latent vectors of the following matrix:轾a h g A= 犏0 b 0 犏臌犏0 0 c Solution Being an upper triangular matrix, the latent roots are the diagonal elements and they are l = a, b , c .Determination of latent root of the matrix corresponding to l = a :轾a- a h g 轾 x1 犏犏犏0b- a 0 犏 x2 = 0 犏犏臌0 0 c- a 臌 x3轾0h g 轾 x1 轾 0 犏犏犏 i.e., 犏0b- a 0 犏 x2 = 犏 0 犏犏犏臌0 0c- a 臌 x3 臌 0Here rank of the coefficient matrix is 2 and putting arbitrary value to n- r =3 - 2 = 1 vari able, say x1= k 1 , we obtain x1= k 1, x 2 = 0, x 3 = 0. Hence, the latent vector corresponding to the root l = a is given by 9 轾x1 轾 k 1 犏犏 X= x = 0 犏2 犏 (where k1 is arbitrary) 犏犏臌x3 臌0Determination of latent root of the matrix corresponding to l = b :轾a- b h g 轾 x1 轾0 轾a- b h g 轾 x1 轾0 犏犏犏 i.e., 犏犏犏犏0b- b 0 犏 x2 = 犏 0 犏0 0 0 犏x2 = 犏 0 犏犏犏犏犏犏臌0 0c- b 臌 x3 臌 0 臌0 0c- b 臌 x3 臌 0Here rank of the coefficient matrix is 2 and putting arbitrary value to n- r =3 - 2 = 1 vari able, say x1= k 2 , we obtain (a- b ) x= k, x = - k , x = 0 ( k is arbitrary) 1 2 2h 2 3 2 Hence, the latent vector corresponding to the root l = b is given by轾 k x 2 轾1 犏犏 (b- a ) X= x2 = 犏 k 2 ( k is arbitrary) 犏犏 h 2 臌犏x3 犏臌 0 Determination of latent root of the matrix corresponding to l = c :轾a- c h g 轾 x1 轾0 犏犏犏犏0b- c 0 犏 x2 = 犏 0 犏犏犏臌0 0c- c 臌 x3 臌 0轾a- c h g 轾 x1 轾0 犏犏犏 i.e., 犏0b- c 0 犏 x2 = 犏 0 犏犏犏臌0 0 0 臌x3 臌 0Here rank of the coefficient matrix is 2 and putting arbitrary value to n- r =3 - 2 = 1 vari able, say x1= k 3 , we obtain 轾轾x犏 k 1犏 3 X=犏 x = 0 . 犏2 犏犏臌犏x3 ( c- a ) 犏 k3 臌犏 g Example 5 Let X be a characteristic vector corresponding to the characteristic root l of a matrix A. Also, let f( x ) is any scalar polynomial. Show that X be a characteristic vec tor corresponding to the characteristic root f (l ) of the matrix f( A ) . Also, show that if g is a polynomial defined by10 f1( x ) g( x ) = where detf2 ( A ) 0 f2 ( x ) then X is a characteristic vector corresponding to the characteristic root g(l ) of the ma trix -1 g( A )= f1 ( A )[ f 2 ( A )] . Solution If AX= l X ThenA2 X= A( AX ) = A (l X )=l(AX ) = ll = l 2 XRepeating this process k times, we obtainAk X= l k X …(8)If 2 m then f( A ) is the matrix f( x )= a0 + a 1 x + a 2 x + + am x , 2 m 2 m f( A )= a0 I + a 1 A + a 2 A + + am A .Then f( A ) X= ( a0 I + a 1 A + a 2 A + + am A ) X m =a0 X + a 1l X + + am l X , using Eq.(8) m =(a0 + a 1l + + am l ) X = f ( l ) X So X is a characteristic vector of the matrix f( A ) asso ciated with the characteristic root f (l ) .Since |f2 ( A ) | 0, the matrix f2 ( A ) is non-singular and hence no characteristic root of is zero. Hence, if is a characteristic root of we have Since f2 ( A ) f2 (l ) f2 ( A ), f2 (l ) 0. |f ( A ) | 0, we have -1 exists. 2 [f2 ( A )] Now f1( A ) X= f 1 (l ) XAnd f2( A ) X= f 2 (l ) X …(9) From Eq. (9), 1 -1 X= [ f2 ( A )] X . f2 (l ) Hence-1 g( A ) X= f1 ( A )[ f 2 ( A )] -1 X= f1( A ){ [ f 2 ( A )] X }-1 = f1( A ){ [ f 2 (l )] X} -1 = [f2 (l )] f 1 ( A ) X11 -1 =[f2 (l )] f 1 ( l ) X = g ( l ) XHence X is a characteristic vector corresponding to the characteristic root g(l ) of the matrix g( A ).A Example 6 If l is a characteristic root of a non-singular matrix A, then prove that is l a characteristic root of adj A . SolutionBeing a characteristic root of a non-singular matrix, l 0. Also l is a characteristic root of A implies that there exists a non-zero vector X such that AX= l X . Pre-multiplying both sides by adj A , we obtain (adj A ) ( AX )= (adj A ) (l X ) implies [(adj A ) A ] X= l (adj A ) X implies A IX= l(adj A ) X implies A X= l(adj A ) XA implies X= (adj A ) X lA i.e., (adjA ) X= X lA Since X is non-zero vector, the above shows that is a characteristic root of the mat l rix adj A . Example 7 Show that the characteristic roots of an idempotent matrix are either zero or u nity. SolutionSince A is an idempotent matrix, we have A2 = A. Let X be a characteristic vector of the matrix A corresponding to the characteristic root l . Then AX= l XPre-multiplying by A, we obtain A( AX )= A (l X ) = l ( AX ) i.e., (AA ) X= l ( l X ) i.e., AX= l 2 X (since A2 = A ) 12 i.e., lX= l 2 X (since AX= l X ) 2 i.e., (l- l )X = O , the zero matrix. i.e., l2 - l = 0 (since X is a non-zero matrix) implies l=0, l = 1. 4. Algebraic and Geometric MultiplicityDefinition If l1 is a t-ple root of the characteristic equationA-l I = 0 then t is called the algebraic multiplicity of l1 and the dimension of the characteristic space of A is called geometric multiplicity of A. Example 8 0 is the characteristic root of algebraic multiplicity n of the n n zero matrix. Example 9 1 is the characteristic root of algebraic multiplicity n of the n n identity matrix. We conclude the session by stating a result. Theorem 5 The geometric multiplicity of a characteristic root cannot exceed algebraic mult iplicity of the same.SummaryIn this session we have discussed characteristic space of a given matrix corresponding to a characteristic root. The method of finding characteristic vectors associated with a charac teristic root have been described. The definition of algebraic and geometric multiplicity have been given.Assignments1. Find the eigen values and the corresponding eigen vectors of the following matrices:13 轾 3 1- 1 轾2 1- 1 轾 2- 1 1 a 犏 1 3 1 (b )犏 0 3- 2 c 犏-1 2 - 1 ( ) 犏犏 ( ) 犏臌犏-1 1 3 臌犏2 4- 3 臌犏 1- 1 2 2. Find the eigen values and the corresponding eigen vectors and eigen spaces (characteristic spaces) of the following matrices:2 1 1  2 2 0  1 1  1  (a). 1 2 1  (b). 2 2 0  (c). 1 1 0  0 0 1  0 0 1  10 1        3. Determine the characteristic roots and a basis of each of the associated characteristic spac es.1 2 2  3  9 12  3 2 2 4  0  13 4  23 2 1  (a). 2 1 (b). (c).   1 2 2  0 0 1  1 1 2 1     2 2 2 1 4. Prove that if X is a unit vector and if AX  X then XT AX = l. 5. Show that a characteristic root of every orthogonal matrix of odd order is either 1 or -1Quiz轾2 1 0 1. The characteristic space of the matrix A = 犏0 2 1 associated with the characteristic valu 臌犏0 0 2 e l = 2 is 禳镲轾a (a) 睚犏0 : a 铪镲臌犏0禳镲轾0 (b) 睚犏a: a 铪镲臌犏0禳镲轾0 (c) 睚犏0 : a 铪镲臌犏a (d) None of the aboveAns. (a)2. Fill in the blanks: If l is a characteristic root of a non-singular matrix A, then ______is a characteristic root of adj A .14 (a) l AA (b) l(c) l + A (d) l - AAns. (b) 3. Characteristic vectors corresponding to distinct characteristic roots of a real symmetric matrix are ______. (a) the same (b) one is a scalar multiple of the other (c) orthogonal (d) none of the above Ans. (c)FAQ 1. For a square matrix A, is it necessary for a scalar l to be a characteristic root that there exists non-zero vector X such that AX= l X ? Solution Yes. We note that l is a characteristic root if and only if there exists non-zero vector X suc h that AX= l X. This can be seen as follows: Let l be a characteristic root of the matrix A. Then, by definition, l must satisfy the cha racteristic equation of A. i.e., A-l I = 0.This implies that the matrix A- l I must be singular. Hence, if (A- l I ) is singular, then l is a characteristic root of the matrix A. In this case there exists non-zero vector X such that (A-l I ) X = 0. i.e., that AX= l X.Conversely, if there exists non-zero vector X such that 15 AX= l X , then(A-l I ) X = 0.Now the system (A-l I ) X = 0 possesses a non-zero vector (non-trivial matrix) X as a solution impliesA-l I = 0, and hence l is a characteristic root of the square matrix A. 2. Whether the following true: AX= l X has a non-trivial solution X if l is an eigen valu e of A. Answer Yes. This follows from definition or from the discussion just above. 3. Whether the following true: The scalar l is a characteristic root of the matrix A if and only if the matrix (A- l I ) singular. Answer Yes. This follows from the discussion above.GlossaryCharacteristic root and characteristic vector : Let A be a square matrix over , th e field of complex numbers. A complex number λ is called a characteristic root (or charact eristic value or eigen value or latent root) of a matrix A if there is a non-zero column matrix X such that AX= l X .The column matrix X is said to be a characteristic vector (or eigen vector or latent vector) of the matrix A associated with the characteristic root λ.Characteristic equation of a matrix: The equation A- xI = 0 is called the characteristic equation of the matrix A.16 Characteristic space: The characteristic space of a matrix A corresponding to a characteristic root l is the null space of the matrix (A- l I ).Algebraic multiplicity of a Characteristic root: If l1 is a t-ple root of the characteris tic equationA-l I = 0 then t is called the algebraic multiplicity of l1 . Geometric multiplicity of a Characteristic root: The dimension of the characteristic space of A is called geometric multiplicity of A.REFERENCESBooks1. I.N. Herstein, Topics in Algebra, Wiley Easten Ltd., New Delhi, 1975.2. K. B. Datta, Matrix and Linear Algebra, Prentice Hall of India Pvt. Ltd., New Delhi, 2000.3. P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, First Course in Linear Algebra, Wiley Eastern, New Delhi, 1983.4. P.B. Bhattacharya, S.K. Jain and S.R. Nagpaul, Basic Abstract Algebra (2nd Edition), Cambridge University Press, Indian Edition, 1997., New Delhi, 1983.5. S.K. Jain, A. Gunawardena and P.B. Bhattacharya, Basic Linear Algebra with MATLAB, Key College Publishing (Springer-Verlag), 2001.17 18

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