The easiest normal systems to solve are systems of the form

The easiest normal systems to solve are systems of the form x\'(t) = Dx(t) where D is an n Times n diagonal matrix. Such a system actually consists of n uncoupled equations x\'_i (g) = d_ij x_i (t) I = 1...n. whose solution is x_i (t) = c_i e^where the c_i\'s are arbitrary constants. This raises the following question: When can we uncouple a normal system? To answer this question, we need the follow ing result from linear algebra. An n Times n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors P_1,...,P_n. Moreover, if P is the matrix whose columns are P_1...P_n, then P^-1 AP = D where D is the diagonal matrix whose entry d_ij is the eigenvalue associated with the vector p_i. Use the above result to show that the system x\'(t) = Ax(t), where A is an n Times n diagonalizable matrix, is equivalent to an uncoupled system y\'(t) = Dy(t), where y = P^-1 x and D = P^-1 AP.

Solution

Solution: Given that

x\'(t) = Ax(t)           (4)

where A is an nxn diagonalizable matrix.

Since A is diagonalizable matrix, so A has n linearly independent eigenvectors , say p₁, p₂,,.....,p_n.

Take a nonsingular matrix P whose columns are the n linearly independent eigenvectors , p₁, p₂,,.....,p_n.

Let y\'(t) = Dy(t)        (5)

where y = P^-1x, D = P^-1AP, D is the diagonal matrix whose principal diagonal entry is d_ii is the eigenvalue associated with the vector p_{i .}

Since y\'(t) = Dy(t) and y = P^-1x,

we have P^-1x\'(t) = DP^-1x(t)

implies that P(P^-1x\'(t)) = P(DP^-1x(t))   (Pre multiplying by P)

implies that (PP^-1)x\'(t) = P(P^-1AP)P^-1x(t)( As D = P^-1AP )

implies that (PP^-1)x\'(t) = (PP^-1)A(PP^-1)x(t)

implies that I .x\'(t) = I.A.I x(t) (As PP^-1 = I, identity matrix )

implies that x\'(t) = A x(t)

Hence, by the given condition we can say that the system x\'(t) = Ax(t) is equivalent to an uncoupled

system y\'(t) = Dy(t) , where y = P^-1x and D = P^-1AP. (proved)