Let T be a linear operator on ℝ³. Let f(X) ∈ ℝ[X] denote its characteristic polynomial. Consider the following statements.

(a). Suppose T is non-zero and 0 is an eigen value of T. If we write f(X) = X g(X) in ℝ[X], then the linear operator g(T) is zero.

(b). Suppose 0 is an eigenvalue of T with at least two linearly independent eigen vectors. If we write f(X) = X g(X) in ℝ[X], then the linear operator g(T) is zero.

Which of the following is true?

Explanation:

T be a linear operator on ℝ3. So T is a 3 × 3 matrix.

(a): T is non-zero and 0 is an eigen value of T. Let other two eigenvalues are α, β, so f(x) = x(x - α)(x - β)

As f(X) = X g(X) hence g(x) = (x - α)(x - β)

If α = 2, β = -2 then g(x) = (x - 2)(x + 2) = x² - 4

So g(T) = T² - 4I 

Now, eigenvalues of T are 0, 2, -2 so eigenvalues of g(T) are 4, 0, 0

So g(T) ≠ 0 

(a) is false.

(b): 0 is an eigenvalue of T with at least two linearly independent eigen vectors.

So GM = 2 for 0

We know that AM ≥ GM

So AM ≥ 2 ⇒ AM = 2 or AM = 3 for eigenvalue 0

For the case AM = 2

Let T = \(\begin{bmatrix}0&0&0\\0&0&0\\0&0&λ\end{bmatrix}\) λ ≠ 0

So characteristic polynomial f(x) = x²(x - λ)

Therefore g(x) = x(x - λ) = x² - λx

so g(T) = T² - λT

Now, eigenvalue of T is 0, 0, λ

eigenvalue of T2 - λT is 0 - 0λ, 0 - 0λ, λ² - λ² = 0, 0, 0

so g(T) = 0

If λ = 0 then f(x) = x³ so g(x) = x² Hence g(T) = 0

(b) is correct

Option (4) is correct