Question

Let M= \(\begin{pmatrix} 1 & -1 & 0\\ -a & 2 & -1 \\0&-1&1\end{pmatrix}\)If a non-zero vector 𝑋=(𝑥, 𝑦, 𝑧)^T∈ℝ³ satisfies 𝑀⁶𝑋=𝑋, then a subspace of ℝ3 that contains the vector 𝑋 is

• A. {(𝑥, 𝑦, 𝑧)^𝑇∈ℝ³ ∶ 𝑥=0, 𝑦+𝑧=0}
• B. {(𝑥, 𝑦, 𝑧) ^𝑇∈ ℝ³ ∶𝑦 = 0, 𝑥+𝑧=0}
• C. {(𝑥, 𝑦, 𝑧) ^𝑇 ∈ ℝ³ ∶ 𝑧=0, 𝑥+𝑦=0}
• D. {(𝑥, 𝑦, 𝑧) ^𝑇 ∈ ℝ³ ∶𝑥=0, 𝑦−𝑧=0}

Answer 1

The Correct Option is B

We are given a matrix \( M \) and need to find a subspace containing a vector \( X \), such that \( M^6 X = X \). Let's analyze the given matrix and properties of matrix multiplication to solve this problem. 

The matrix \( M \) is given by:

\[M = \begin{pmatrix} 1 & -1 & 0 \\ -a & 2 & -1 \\ 0 & -1 & 1 \end{pmatrix}\]

This problem is asking for the subspace which consists of all vectors \( X = (x, y, z)^T \in \mathbb{R}^3 \) such that \( M^6 X = X \). This implies \( X \) must be an eigenvector of \( M^6 \) with eigenvalue 1. To solve, we need to determine eigenvectors and eigenvalues of the matrix \( M \).

Since finding \( M^6 \) analytically is complex, we aim to focus on eigenvectors and eigenvalues and the properties they imply. One property is:

If \( \lambda \) is an eigenvalue of \( M \), then \(\lambda^6\) is an eigenvalue of \( M^6 \). As \( M^6 X = X \), the eigenvalue \(\lambda\) satisfies \(\lambda^6 = 1\).

This consideration implies possible eigenvalues of \( M \) that are roots of unity, such as \( \lambda = 1, -1, \text{or complex roots}. \) The corresponding eigenvector must satisfy either \( X \in \text{Null}(M - I)\) or some symmetrized conditions due to the repeated application of \( M \).

For this problem, look into observing eigenvectors corresponding to the eigenvalue 1. As computations simplify, confirming that \( M X = X \) simplifies our eigenvector to \((y = 0, x + z = 0)\).

Thus, the subspace corresponding to these conditions is:

{(𝑥, 𝑦, 𝑧) ^𝑇∈ ℝ³ ∶𝑦 = 0, 𝑥+𝑧=0}

This option matches the description of the eigenvectors for the eigenvalue satisfying the necessary conditions for repeated applications of the matrix M.