Let T be a linear operator on a finite dimensional vector space V. If m(λ) = λr + ar-1λr-1 + ... + a1λ + a0 (a0 ≠ 0) be a minimal polynomial of T thena)T is invertibleb)0 is an eigenvalue of Tc)T is singulard)0 is root of m(λ)Correct answer is option 'A'. Can you explain this answer?

Question

Let T be a linear operator on a finite dimensional vector space V. If m(&lambda;) = &lambda;r + ar-1&lambda;r-1 + ... + a1&lambda; + a0 (a0 &ne; 0) be a minimal polynomial of T thena)T is invertibleb)0 is an eigenvalue of Tc)T is singulard)0 is root of m(&lambda;)Correct answer is option 'A'. Can you explain this answer?

Answer

To prove that T is diagonalizable, we need to show that its minimal polynomial has distinct linear factors.

First, let's define the minimal polynomial of T as the monic polynomial of least degree that annihilates T. This means that the minimal polynomial p(x) satisfies p(T) = 0, where p(x) = (x - λ₁)^(m₁)(x - λ₂)^(m₂)...(x - λₖ)^(mₖ) for distinct eigenvalues λ₁, λ₂, ..., λₖ and positive integers m₁, m₂, ..., mₖ.

Now, let's assume that the minimal polynomial p(x) has repeated linear factors. Without loss of generality, assume that (x - λ)^(m) is a repeated linear factor, where λ is an eigenvalue of T and m > 1.

Since p(T) = 0, we have (T - λI)^(m) = 0, where I is the identity operator on V. This means that for any vector v in V, we have (T - λI)^(m)(v) = 0.

Now, let's consider the subspace W = Ker((T - λI)^(m-1)). Since (T - λI)^(m-1)(v) = 0 for any v in W, we have W ⊆ Ker((T - λI)^(m-1)).

Next, let's consider the vector u = (T - λI)^(m-1)(v) for some v in V. Since (T - λI)(u) = (T - λI)((T - λI)^(m-1)(v)) = (T - λI)^(m)(v) = 0, we have u in Ker(T - λI) = E(λ), the eigenspace of T corresponding to the eigenvalue λ. Therefore, u is an eigenvector of T corresponding to the eigenvalue λ.

However, since (T - λI)^(m-1)(v) = u, this means that u is a generalized eigenvector of T corresponding to the eigenvalue λ with a generalized eigenvector degree of m-1.

This contradicts the assumption that λ is diagonalizable, because a diagonalizable operator should have each eigenvalue associated with a unique eigenvector.

Therefore, our assumption that the minimal polynomial p(x) has repeated linear factors is false. This implies that the minimal polynomial of T has distinct linear factors, and thus T is diagonalizable.

Answer

First of all minimal polynomial and characteristic polynomial of a linear operator T shares same roots.. but of different multiplicity... So basically roots of minimal polynomial are eigenvalues of T... Here 0 is not a root of minimal polynomial of T as m(0) =a0 (#0)  .. so clearly 0 is not an eigenvalue of T and det(T) is non zero.. Therefore T is non singular which means invertible

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