Question

Let \(T: \mathbb{R}^4 \to \mathbb{R}^4\) be a linear transformation and the null space of \(T\) be the subspace of \(\mathbb{R}^4\) given by \[ \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : 4x_1 + 3x_2 + 2x_3 + x_4 = 0 \}. \] If \(\text{Rank}(T - 3I) = 3\), where \(I\) is the identity map on \(\mathbb{R}^4\), then the minimal polynomial of \(T\) is

• A. \(x(x - 3)\)
• B. \(x(x - 3)^3\)
• C. \(x^3(x - 3)\)
• D. \(x^2(x - 3)^2\)

Hint:

A common mistake is to misinterpret the dimension of a subspace given by an equation. In \(\mathbb{R}^n\), a single non-trivial linear equation defines a hyperplane of dimension \(n-1\). Correctly calculating the geometric multiplicities is the key step.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question relates properties of a linear transformation (like rank and nullity) to its eigenvalues, geometric multiplicities, and ultimately its minimal polynomial. Key concepts are the Rank-Nullity Theorem, eigenvalues, eigenspaces, and the condition for diagonalizability.
Step 2: Key Formula or Approach:
1. The null space of T is the eigenspace corresponding to the eigenvalue \(\lambda = 0\). Its dimension is the geometric multiplicity of \(\lambda=0\).
2. The null space of \(T-\lambda I\) is the eigenspace for the eigenvalue \(\lambda\).
3. Rank-Nullity Theorem: \(\text{Rank}(A) + \text{Nullity}(A) = \text{dimension of domain}\).
4. A linear transformation on an \(n\)-dimensional space is diagonalizable if and only if the sum of the geometric multiplicities of its eigenvalues is \(n\).
5. The minimal polynomial of a diagonalizable transformation has distinct roots (each eigenvalue appears with power 1).
Step 3: Detailed Calculation:
Analysis of Eigenvalue \(\lambda = 0\):
The null space of \(T\), denoted \(\ker(T)\), is the eigenspace for \(\lambda=0\).
We are given \(\ker(T) = \{ \mathbf{x} \in \mathbb{R}^4 : 4x_1 + 3x_2 + 2x_3 + x_4 = 0 \}\).
This is the equation of a hyperplane in \(\mathbb{R}^4\). The dimension of this subspace is \(4-1 = 3\).
So, \(\text{Nullity}(T) = \dim(\ker(T)) = 3\).
This means the geometric multiplicity of the eigenvalue \(\lambda=0\) is 3.
Analysis of Eigenvalue \(\lambda = 3\):
We are given \(\text{Rank}(T - 3I) = 3\).
Using the Rank-Nullity Theorem for the transformation \(T-3I\): \[ \text{Rank}(T - 3I) + \text{Nullity}(T - 3I) = \dim(\mathbb{R}^4) = 4 \] \[ 3 + \text{Nullity}(T - 3I) = 4 \] \[ \text{Nullity}(T - 3I) = 1 \] The null space of \(T-3I\) is the eigenspace for \(\lambda=3\). So, the geometric multiplicity of the eigenvalue \(\lambda=3\) is 1.
Check for Diagonalizability:
The sum of the geometric multiplicities of the eigenvalues is \(3 (\text{for } \lambda=0) + 1 (\text{for } \lambda=3) = 4\).
Since this sum equals the dimension of the vector space \(\mathbb{R}^4\), the linear transformation \(T\) is diagonalizable.
Determine the Minimal Polynomial:
For a diagonalizable matrix or transformation, the minimal polynomial is the product of distinct linear factors corresponding to its eigenvalues. The eigenvalues are 0 and 3.
Therefore, the minimal polynomial \(m(x)\) is: \[ m(x) = (x - 0)(x - 3) = x(x-3) \] Step 4: Final Answer:
The minimal polynomial of T is \(x(x-3)\).
Step 5: Why This is Correct:
By correctly identifying the dimensions of the eigenspaces for \(\lambda=0\) and \(\lambda=3\), we found the sum of their geometric multiplicities to be 4. This matches the dimension of the space, proving T is diagonalizable. The minimal polynomial for a diagonalizable operator contains each distinct eigenvalue factor exactly once.