Let 𝑇 ∶ ℝ⁴ → ℝ⁴ be a linear transformation and the null space of 𝑇 be the subspace of ℝ⁴ given by

{(𝑥₁, 𝑥₂, 𝑥₃, 𝑥₄) ∈ ℝ⁴ ∶ 4𝑥₁ + 3𝑥₂ + 2𝑥₃ + 𝑥₄ = 0}.

If 𝑅𝑎𝑛𝑘(𝑇 − 3𝐼) = 3, where 𝐼 is the identity map on ℝ⁴ , then the minimal polynomial of 𝑇 is 

Given -

Let 𝑇 ∶ ℝ4 → ℝ4 be a linear transformation and the null space of 𝑇 be the subspace of ℝ4 given by

{(𝑥1, 𝑥2, 𝑥3, 𝑥4) ∈ ℝ4 ∶ 4𝑥1 + 3𝑥2 + 2𝑥3 + 𝑥4 = 0}.

If 𝑅𝑎𝑛𝑘(𝑇 − 3𝐼) = 3, where 𝐼 is the identity map on ℝ4

Concept -

(i) The dimension of subspace = dim (V) - number of restriction

(ii) Rank - Nullity theorem -

 η (T) + ρ (T) = n  where n is the dimension of the vector space or the order of the matrix.

(iii) The formula for Geometric multiplicity (GM) is = η(T - λ I)

(iv) AM ≥ GM

(v) If the rank of A is less than n this implies that |A| = 0

Explanation -

we have null space {(𝑥1, 𝑥2, 𝑥3, 𝑥4) ∈ ℝ4 ∶ 4𝑥1 + 3𝑥2 + 2𝑥3 + 𝑥4 = 0}

Now the dimension of null space = dim (V) - number of restriction = 4 - 1 = 3

Hence the nullity of T is 3 so this implies rank of T is 1.   [by rank - Nullity theorem]

i.e. \(ρ(T) =1 \ \ and \ \ η (T) =3\)

Now the formula for Geometric multiplicity (GM) is = η(T - λ I)

if we take λ = 0 then GM = 3 for λ = 0 and we know that AM ≥ GM then AM = 3, 4 only because it is not greater than the dimension of vector space.

But we have the another condition  𝑅𝑎𝑛𝑘(𝑇 − 3𝐼) = 3 < dim (ℝ4) then |𝑇 − 3𝐼| = 0

hence λ = 3 is another eigen value of the transformation. and we have 𝑅𝑎𝑛𝑘(𝑇 − 3𝐼) = 3 ⇒ η(𝑇 − 3𝐼) = 1

Hence the eigen values of T is 0,0,0 and 3.

Now the characteristic polynomial of T is 𝑥3 (𝑥 − 3)  and the minimal polynomial of T is x(x - 3).

Hence the option(1) is correct.