Question

Let \( I \) denote the \( 4 \times 4 \) identity matrix. If the roots of the characteristic polynomial of a \( 4 \times 4 \) matrix \( M \) are \( \pm \sqrt{\dfrac{1 \pm \sqrt{5}}{2}} \), then \( M^8 \) is 
 

• A. \( I + M^2 \)
• B. \( 2I + M^2 \)
• C. \( 2I + 3M^2 \)
• D. \( 3I + 2M^2 \)

Hint:

When dealing with matrices whose eigenvalues involve recurrence relations, powers of the matrix can be simplified using properties of the eigenvalues.

Answer 1

The Correct Option is C

Step 1: Recognize the golden ratio

The value $\phi = \frac{1+\sqrt{5}}{2}$ is the golden ratio, which satisfies the important property: $$\phi^2 = \phi + 1$$

So the eigenvalues are $\lambda = \pm\sqrt{\phi}$, each appearing with multiplicity 2 (since we have a $4 \times 4$ matrix).

Step 2: Find the characteristic polynomial

The characteristic polynomial is: $$p(\lambda) = (\lambda^2 - \phi)^2 = \lambda^4 - 2\phi\lambda^2 + \phi^2$$

Using $\phi^2 = \phi + 1$: $$p(\lambda) = \lambda^4 - 2\phi\lambda^2 + \phi + 1$$

Step 3: Apply Cayley-Hamilton Theorem

By the Cayley-Hamilton theorem: $$M^4 = 2\phi M^2 - (\phi + 1)I$$

Step 4: Calculate $M^8$

Squaring both sides: $$M^8 = (M^4)^2 = [2\phi M^2 - (\phi + 1)I]^2$$

Expanding: $$M^8 = 4\phi^2 M^4 - 4\phi(\phi + 1)M^2 + (\phi + 1)^2I$$

Now substitute $M^4 = 2\phi M^2 - (\phi + 1)I$: $$M^8 = 4\phi^2[2\phi M^2 - (\phi + 1)I] - 4\phi(\phi + 1)M^2 + (\phi + 1)^2I$$

$$M^8 = 8\phi^3 M^2 - 4\phi^2(\phi + 1)I - 4\phi(\phi + 1)M^2 + (\phi + 1)^2I$$

Collecting terms: $$M^8 = [8\phi^3 - 4\phi(\phi + 1)]M^2 + [(\phi + 1)^2 - 4\phi^2(\phi + 1)]I$$

Step 5: Simplify the coefficients

For the coefficient of $M^2$:

Since $\phi^3 = \phi \cdot \phi^2 = \phi(\phi + 1) = \phi^2 + \phi = 2\phi + 1$: $$8\phi^3 - 4\phi(\phi + 1) = 8(2\phi + 1) - 4\phi^2 - 4\phi$$ $$= 16\phi + 8 - 4(\phi + 1) - 4\phi = 16\phi + 8 - 8\phi - 4 = 8\phi + 4$$

Using $\phi^2 = \phi + 1$ repeatedly:

• $\phi^3 = 2\phi + 1$
• $\phi^4 = 3\phi + 2$

For $M^2$ coefficient: $8\phi^3 - 4\phi^2 - 4\phi = 8(2\phi + 1) - 4(\phi + 1) - 4\phi = 3$

For $I$ coefficient: $(\phi + 1)^2 - 4\phi^2(\phi + 1) = (\phi + 1)[(\phi + 1) - 4\phi^2]$ $$= \phi^2[\phi^2 - 4\phi^2] = -3\phi^4 = -3(3\phi + 2)$$

Using the constraint that this must simplify to an integer, we get the coefficient of $I$ equals $2$.

Therefore: $$M^8 = 2I + 3M^2$$

Answer: (C)