m We can similarly replace any pair of complex conjugate eigenvalues with 2 2 real blocks. = R

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1 1 RODICA D. COSTIN Suppose that some eigenvalues are not real. Then the matrices S and are not real either, and the diagonalization of M must be done in C n. Suppose that we want to work in R n only. Recall that the nonreal eigenvalues and eigenvectors of real matrices come in pairs of complex-conjugate ones. In the complex diagonal form one can replace diagonal blocks j 0 0 j by a matrix which is not diagonal, but has real entries. To see how this is done, suppose 1 C \ R and = 1, v = v 1. Splitting into real and imaginary parts, write 1 = 1 +i 1 and v 1 = x 1 +iy 1. Then from M(x 1 + iy 1 )=( 1 + i 1 )(x 1 + iy 1 ) identifying the real and imaginary parts, we obtain Mx 1 + imy 1 =( 1 x 1y)+i( 1 x + 1 y) In the matrix S =[v 1, v,...,v n ] composed of independent eigenvectors of M, replace the first two columns v 1, v = v 1 by x 1, y 1 (which are vectors in R n ): using the matrix S =[x 1, y 1, v,...,v n ] instead of S we have M S = S where = m We can similarly replace any pair of complex conjugate eigenvalues with real blocks. Exercise. Show that each real block obtained through decomplexification has the form = R for a suitable >0 and R rotation matrix (0)..14. Jordan normal form. We noted in.1 that a matrix is similar to a diagonal matrix if and only if the dimension of each eigenspace V j equals the order of multiplicity of the eigenvalue j. Otherwise, there are fewer than n independent eigenvectors; such a matrix is called defective Jordan blocks. Defective matrices can not be diagonalized, but we will see that they are similar to block diagonal matrices, called Jordan normal forms; these are upper triangular, have the eigenvalues on the diagonal, 1 in selected placed above the diagonal, and zero in the rest. After that, in section.14. it is shown how to construct the transition matrix S, which conjugates a defective matrix to its Jordan form; its columns are made of generalized eigenvectors.

2 EIGENVALUES AND EIGENVECTORS 1 The Jordan blocks which appear on the diagonal of a Jordan normal form are as follows. 1 1 Jordan blocks are just [ ]. Jordan blocks have the form 1 (1) J ( )= 0 For example, the matrix (18) is a Jordan block J (0). Jordan blocks have the form 1 0 () J ( )= In general, a k k Jordan block, J k ( ), is a matrix having the same number,, on the diagonal, 1 above the diagonal and 0 everywhere else. Note that Jordan blocks J k ( ) have the eigenvalue with multiplicity k, and the dimension of the eigenspace is one. Example of a matrix in Jordan normal form: which is block-diagonal, having two 1 1 Jordan blocks and one Jordan block along its diagonal. The eigenvalue is simple, while has multiplicity four. The eigenspace corresponding to is two-dimensional (e and e are eigenvectors). Note how Jordan blocks act on the vectors of the basis. For (1): J ( )e 1 = e 1,soe 1 is an eigenvector. Also () J ( )e = e 1 + e which implies that (J ( ) I) e =(J ( ) I)e 1 = 0. Similarly, for (9): J ( )e 1 = e 1 so e 1 is an eigenvector. Then (4) J ( )e = e 1 + e implying that (J ( ) I) e =(J ( ) I)e 1 = 0. Finally, () J ( )e = e + e

3 18 RODICA D. COSTIN implying that (J ( ) I) e =(J ( ) I) e = 0. This illuminates the idea behind the notion of generalized eigenvectors defined in the next section The generalized eigenspace. Defective matrices are similar to a matrix which is block-diagonal, having Jordan blocks on its diagonal. An appropriate basis is formed using generalized eigenvectors: Definition 1. A generalized eigenvector of M corresponding to the eigenvalue is a vector x = 0 so that () (M I) k x = 0 for some positive integer k. Examples. 1) Eigenvectors are generalized eigenvectors (take k = 1 in ()). ) Vectors in the standard basis are generalized eigenvectors for Jordan blocks. Definition 14. The generalized eigenspace of M corresponding to the eigenvalue is the subspace E = {x (M I) k x = 0 for some k Z + } Sometimes we want to refer to only at the distinct eigenvalues of a matrix, this set is called the spectrum : (M) of a matrix M is the set of its eigen- Definition 1. The spectrum values. Theorem 1. For any n n matrix M the following hold: (i) V E ; (ii) E is a subspace; (iii) E is an invariant subspace under M; (iv) E 1 \ E =0for 1 =. (v) dim E =the multiplicity of. (vi)the set of eigenvectors and generalized eigenvectors of M span the whole space C n : (M) E = C n The proofs of (i)-(iv) are simple exercises. The proofs of (v), (vi) are not included here.

4 EIGENVALUES AND EIGENVECTORS How to find a basis for each E that can be used to conjugate a matrix to a Jordan normal form. Example 1. The matrix " # 1+a 1 () M = 1 a 1 is defective: it has eigenvalues a, a and only one independent eigenvector, (1, 1) T. It is therefore similar to J (a). To find a basis x 1, x in which the matrix takes this form, let x 1 =(1, 1) T (the eigenvector); to find x we solve (M ai)x = x 1 (as seen in () and in (4)). The solutions are x (1, 0) T + N (M ai), and any vector in this space works, for example x =(1, 0) T. For " # 1 1 (8) S =[x 1, x ]= 1 0 we have S 1 MS = J (a). Example. The matrix M = has eigenvalues,, and only one independent eigenvector v 1 =(1, 1, 1) T. Let x 1 = v 1 =(1, 1, 1) T. Solving (M I)x = x 1 we obtain x = (1, 1, 0) T (plus any vector in N (M I) =V 1 ). Next solve (M I)x = x which gives x =(0, 1, 1) T (plus any vector in the null space of M I). For S =[x 1, x, x ]wehave (9) S 1 MS = In general, if is an eigenvalue of M for which dimv is less than the multiplicity of, we do the following. Choose a basis for V. For each eigenvector v in this basis set x 1 = v and solve recursively (0) (M I)x k+1 = x k, k =1,,... Note that each x 1 satisfies () for k = 1, x satisfies () for k =, etc. At some step k 1 the system (M I)x k1 +1 = x k1 will have no solution; we found the generalized eigenvectors x 1,...,x k1 (which will give a k 1 k 1 Jordan block). We then repeat the procedure for a di erent eigenvector in the chosen basis for V, and obtain a new set of generalized eigenvectors, corresponding to a new Jordan block.

5 0 RODICA D. COSTIN Real Jordan normal form. If a real matrix has multiple complex eigenvalues and is defective, then its Jordan form can be replaced with an upper block diagonal matrix in a way similar to the diagonal case illustrated in.1., by replacing the generalized eigenvectors with their real and imaginary parts. For example, a real matrix which can be brought to the complex Jordan normal form 4 + i i i i can be conjugated (by a real matrix) to the real matrix Block matrices Multiplication of block matrices. It is sometimes convenient to work with matrices split in blocks. We have already used this when we wrote M[v 1,...,v n ]=[Mv 1,...,Mv n ] More generally, if we have two matrices M, P with dimensions that allow for multiplication (i.e. the number of columns of M equals the number of rows of P ) and they are split into blocks: M = 4 M 11 M 1, P = 4 P 11 P 1 M 1 M P 1 P then MP = 4 M 11P 11 + M 1 P 1 M 11 P 1 + M 1 P M 1 P 11 + M P 1 M 1 P 1 + M P if the number of columns of M 11 equals the number of rows of P 11. Exercise. Prove that the block multiplication formula is correct. More generally, one may split the matrices M and P into many blocks, so that the number of block-columns of M equal the number of block-rows of P and so that all products M jk P kl make sense. Then MP can be calculated using blocks by a formula similar to that using matrix elements.

6 .1.. Determinant of block matrices. EIGENVALUES AND EIGENVECTORS 1 Proposition 1. Let A, D, C, D be k k matrices. For triangular block matrices: A B A 0 M = or M = 0 D C D we have det M =deta det D. Also, 1,..., k are the eigenvalues of A, and 1,..., k are the eigenvalues of D, then the eigenvalues of M are 1,..., k, 1,..., k. The proof is left to the reader as an exercise. For a more general block matrix, with D invertible A B M = C D the identity A C B D I 0 D 1 C I = A BD 1 C B 0 D together with Proposition 1 implies that A B det C D =det(a BD 1 C)detD which, of course, equals det(ad BD 1 CD). For larger number of blocks, there are more complicated formulas.. Solutions of linear differential equations with constant coefficients In 1. we saw an example which motivated the notions of eigenvalues and eigenvectors. General linear first order systems of di erential equations with constant coe cients can be solved in a quite similar way. Consider du (1) dt = Mu where M is an m m constant matrix and u in an m-dimensional vector..1. The case when M is diagonalizable. Assume that M has m independent eigenvectors v 1,...,v m, corresponding to the eigenvalues 1,..., m. As in 1. there are m purely exponential solutions u j (t) = e jt v j for j =1,...,m. Hint: bring A, D to Jordan normal form, then M to an upper triangular form. References: J.R. Silvester, Determinants of block matrices, Math. Gaz., 84(01) (000), pp. 40-4, and P.D. Powell, Calculating Determinants of Block Matrices,