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Question #41131

Difficulty: 8/10 | 🟢 Free

The Question

Let $M_2(\mathbb{R})$ be the vector space of all $2 \times 2$ real matrices and $S = \{X \in M_2(\mathbb{R}) \mid X = X^T\}$ be the subspace of symmetric matrices. For a fixed matrix $A \in M_2(\mathbb{R})$ , define linear transformations $T_1, T_2 : M_2(\mathbb{R}) \to M_2(\mathbb{R})$ by $T_1(X) = AX - XA$ $T_2(X) = AX - XA^T$ Let $k_1 = \min_{A \in M_2(\mathbb{R})} (\dim(\ker T_1))$ and $k_2 = \min_{A \in M_2(\mathbb{R})} (\dim(\ker (T_2|_S)))$ , where $T_2|_S$ is the restriction of $T_2$ to $S$ . What is the value of $k_1^2 + k_2^2$ ?

Options

A.
4
B.
6
C.
8
D.
10

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Bijectivity and Inverse|Properties of Transpose Determinant|Matrix Multiplication Non-Commutativity

Why This Question Is Tricky

Common Mistakes

This question tests the properties of linear transformations on matrix spaces, specifically focusing on bijectivity, kernel dimensions, and transpose properties.

The core trap is assuming that matrix multiplication commutativity or scalar-like properties apply, which might lead students to incorrectly conclude that $AX - XA = 0 \implies X = 0$ (yielding a kernel dimension of 0).

Furthermore, evaluating $T_2(X)$ requires recognizing that the image of a symmetric matrix under this specific transformation is always skew-symmetric.

By applying the Rank-Nullity theorem between these specific subspaces, the non-triviality of the kernel is rigorously proven, demonstrating that neither transformation can ever be injective.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Bijectivity and Inverse
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free