Suppose the characteristic polynomial of linear operator T

Suppose the characteristic polynomial of linear operator T : V right arrow V is Pr(lambda) = lambda^3 (lambda - 1)^5. Show that T(T -1) = (T - 1)T. Show that T^3(T -1)^5(x) for all x element of K_0. Show that T^3(T -1)^5(x) for all x element of K_1. Show that T_3(T -1)^5 = 0. This result may be generalized: If the characteristic polynomial of a linear operator T : V right arrow V is pr(lambda) = plus minus (lambda - lambda_1)^m_1(lambda - lambda_2)^m_2...(lambda - lambda_k)^m_k then plus minus 9T - lambda_1I)^m_1(T - lambda_2I)^m_2...(T - lambda_kI)^m_k = 0

Solution

(a) T : V V is a linear operator whose characteristic equation is ³( – 1)⁵ = 0

Hence, by Cayley Hamilton Theorem, every linear operator will satisfy its own characteristic equation. Therefore, T³(T – I)⁵ = 0 where I is the identity linear operator and 0 is the zero operator.

Since, the characteristic equation of T is ³( – 1)⁵ = 0, the Eigen values of T are 0 and 1.

Let v be an Eigen vector associated with the Eigen value 0. Then, Tv = 0v = 0

So, T(T – I)v = T(Tv – v) = T(0 – v) = – Tv = 0

And, (T – I)Tv = (T – I)0 = 0

Now, let u be an Eigen vector associated with the Eigen value 1. Then, Tu = 1u = u

So, T(T – I)u = T(Tu – u) = T(u – u) = T(0) = 0

And, (T – I)Tu = (T – I)u = Tu – u = u – u = 0

Therefore, T(T – I) = (T – I)T

(d) Let A be the matrix representation of T with respect to some chosen ordered basis of a finite dimensional vector space V of dimension n. Then the characteristics equation is c₀xⁿ + c₁x^n-1 + … + c_n = 0. We need to prove that A satisfies the characteristics equation. This implies that T satisfies the characteristics equation.

Proof: det(A – xI) = c₀xⁿ + c₁x^n-1 + … + c_n

(A – xI) is a matrix polynomial in x of degree 1 and Adj(A – xI) is matrix polynomial in x of degree n – 1, since each element of Adj(A – xI) is a polynomial in x of degree n – 1 at most.

Let, Adj(A – I) = B₀x^n-1 + B₁x^n-2 + … + B_n-1 where each B_i is an n x n matrix

Now, (A – xI). Adj(A – xI) = det(A – xI).I gives

(A – xI).(B₀x^n-1 + B₁x^n-2 + … + B_n-1) = (c₀xⁿ + c₁x^n-1 + … + c_n).I

Or, A(B₀x^n-1 + B₁x^n-2 + … + B_n-1) – (B₀xⁿ + B₁x^n-1 + … + B_n-1x) = (c₀I)xⁿ + (c₁I)x^n-1 + …. +(c_nI)

Equating coefficients of like powers of x,

– B₀ = c₀I

AB₀ – B₁ = c₁I

AB₁ – B₂ = c₂I

…. … …

AB_n-2 – B_n-1 = c_n-1I

AB_n-1 = c_nI

Pre-multiplying the relations by Aⁿ, A^n-1, …, A, I respectively and adding we have

c₀Aⁿ + c₁A^n-1 + … + c_nI = 0

This implies that A satisfies the characteristics equation, i.e. T satisfies the characteristics equation.

As it was given that the characteristics equation of T is ³( – 1)⁵ = 0, T³(T – I)⁵ = 0. (proved)