Question

Let \[ M = \begin{bmatrix} 1 & 1 \\ 2 & 4 \end{bmatrix} \quad \text{and} \quad x = \begin{bmatrix} 3 \\ 4 \end{bmatrix}. \] Then \[ \lim_{n \to \infty} M^n x \]

• A. does not exist
• B. is \( \begin{bmatrix} 2 \\ 4 \end{bmatrix} \)
• C. is \( \begin{bmatrix} 1 \\ 2 \end{bmatrix} \)
• D. is \( \begin{bmatrix} 3 \\ 4 \end{bmatrix} \)

Hint:

For limits of matrix powers, find the eigenvalues and eigenvectors to determine the long-term behavior of the system.

Answer 1

The Correct Option is B

Step 1: Find the dominant eigenvalue of M.
Matrix M = [[1, 1], [2, 4]]. The characteristic polynomial is:
λ² − 5λ + 2 = 0.
The eigenvalues are:
λ₁ = (5 + √17)/2 ≈ 4.56 (dominant), λ₂ = (5 − √17)/2 ≈ 0.44.

Step 2: The limit of Mⁿx (directional behavior).
Since the dominant eigenvalue λ₁ > 1, the vector Mⁿx grows without bound in magnitude. However, its direction approaches the eigenvector associated with λ₁.

Step 3: Find the eigenvector corresponding to λ₁.
Solve (M − λ₁I)v = 0.
This yields a direction proportional to the vector [2, 4]. Any scalar multiple represents the same limiting direction.

Step 4: Interpret the meaning of the limit.
Although ‖Mⁿx‖ → ∞, the direction of Mⁿx stabilizes and approaches the eigenvector corresponding to λ₁. Thus the “limit vector” in direction (ignoring magnitude) is proportional to [2, 4].

Final Answer:
The limit is the vector pointing in the direction \[ \begin{bmatrix} 2 \\ 4 \end{bmatrix}. \]