Jordan Normal form - Speech or Presentation Example

Jordan Normal Form Problem First we need to get the eigenvalues of matrix A; in this case since matrix A is a 4*4 identity matrix whose rows have been interchanged, its eigenvalues are simply λ1=λ2=λ3=λ4=1. The dimension of the kernel of (A-4In) is 4.The kernel of a m*n matrix with coefficients in a field is the set: N (A)=Null(A)=ker (A)={x belonging to Kn: Ax=0, where 0 is a vector}The dimension of a vector space is the number of vectors of a basis of A, such that for every vector space there exists a basis and all bases of a vector has equal cardinality which is uniquely defined; A is finite dimensional since the dimension of A is finite.

Therefore the real Jordan normal matrix A is: 1 0 0 0 J= 0 1 0 0 0 0 1 4 0 0 0 1Problem 2Let A be a matrix such thatAt*A=E3=A*Ata.) A is a triangular identity matrix whose rows have been interchanged of which the determinant is got by multiplying the elements in the main diagonal; so that at any given time, its determinant is ±1i.e det A=(1*1*1)=1.b.) Eigenvalues of a 3*3 (diagonal) identity matrix is λ1=λ2=λ3=1, so that the modulus of every eigenvalue is: λ1=λ2=λ3=1 |λ1|= (12) ^1/2=1c.) The characteristic equation is: |A-λIn|=0 1 0 0 λ 0 0|A-λIn|=0= 1 0 0 - 0 λ 0 0 0 1 0 0 λ (1-λ)(1-λ)=0; λ=±1d.) Specify the real Jordan normal form of matrix A. exp (i*t) =cos (t) +i*sin (t) tThe complex numbers on the amount one are exactly the complex numbers of the above form.

Determining the real Jordan form of matrix A, so ma=(x2+1)2=(x-i)2(x+i)2.Thus p1=x=I, we have; N2,p1 and N1,p1= N(A-iI4).Thus the elementary Jordan basis for N1,p1:X11+iY11, (A-iI4)(X+iY11)=X12+iY12Resulting to AX11= -Y11+X12 and AY11=X11+Y12S…………..(1.)It’s known that; mTA,X11+iY11=(X-i)2 (A-iI4)2(X11+iY11)=0 ………………………..(2.) AX12= -Y12Thus the four real equations 1 and 2 in matrix form, with P= [X11|Y11|X12|Y12]Then P is non-singular and: 0 1 0 0P-1AP= -1 0 0 0 1 0 0 1 0 1 -1 0Problem 3a.) Nilpotent matrix is a square matrix N such that:Nk=0, for some positive integer k.

0 1For example, M= 0 1 is nilpotent since M2=0, which is to say that for any triangular matrix with 0’s along the main or leading diagonal is nilpotent.If N is nilpotent, then I+N is also nilpotent, and by extension, so should I-N, where En is a n*n identity matrix. This only applies whenever I is a n*n identity matrix, but where we have En(obtained as a result of multiplying a n*n matrix say A by its transpose (At), i.e En=A*At), the resultant matrix is an invertible matrix. Take for instance the matrix A and B in problem 4, B is the transpose of A.

Let En be 5 11En=AB= 25 11 Then En-M, is not a nilpotent matrix. Hence En-M is an invertible matrix. The inverse is given by: (En-N)-1=N+N2+N3+N4+…….=∑i=0nWhere only finitely many terms of sum is non-zero.b.) The inverse of the following matrix: 1 2 3 4 5 0 1 2 3 4 A= 0 0 1 2 3 0 0 0 1 2 0 0 0 0 1Det A=|A|= (1*1*1*1*1)=1 1 1/2 1/3 1/4 1/5 0 1 1/2 1/3 1/4Therefore A-1= 0 0 1 1/2 1/3 0 0 0 1 1/2 0 0 0 0 1Problem 4exp (A+B) ≠ exp (A)*exp (B)exp(A)=∑k=12(Ak/k!); exp(B)=∑k=12(Bk/k!)The series always converges, so the exponentials of A and B are well-defined.

If AB=BA, then: exp(A)*exp(B)=exp(A+B)For this property to hold; 1. Matrices A and B must be conformable. 2. AB=BA, otherwise the property will not hold.Let’s take for instance: 1 2 1 3A= 3 4 B= 2 4 5 11 10 14AB= 11 25 BA= 14 20Since AB≠BA, exp (A+B) ≠ exp (A)*exp (B)Work CitedBernstein, Dennis S. Matrix Mathematics: Theory, Facts, and Formulas. Princeton, N.J: Princeton University Press, 2009. Print.