Let the eigenvalues of a 2 × 2 matrix A be 1, -2 with eigenvectors x₁ and x₂ respectively. Then the eigenvalues and eigenvectors of the matrix A³ would, respectively, be

Understanding Eigenvalues and Eigenvectors of Matrix Powers

This question tests our understanding of a fundamental property relating the eigenvalues and eigenvectors of a matrix $A$ to those of its powers, such as $A^3$.

Key Property of Eigenvalues and Eigenvectors

If a matrix $A$ has an eigenvalue $\lambda$ with a corresponding eigenvector $x$, it means that when $A$ acts on $x$, the result is simply $\lambda$ times $x$. Mathematically, this is expressed as:

$$Ax=\lambda x$$

Now, let's consider the matrix $A^2$. If we apply $A$ twice to the eigenvector $x$:

$$A^{2}x=A(Ax)=A(\lambda x)$$

Since $\lambda$ is a scalar, we can pull it out:

$$A^{2}x=\lambda (Ax)$$

We know that $Ax=\lambda x$, so substituting this back:

$$A^{2}x=\lambda (\lambda x)={\lambda }^{2}x$$

This shows that ${\lambda }^2$ is an eigenvalue of $A^2$ with the same eigenvector $x$.

Extending this to $A^3$:

$$A^{3}x=A(A^{2}x)=A({\lambda }^{2}x)$$

Again, pulling out the scalar ${\lambda }^2$:

$$A^{3}x={\lambda }^2(Ax)$$

Substituting $Ax=\lambda x$:

$$A^{3}x={\lambda }^2(\lambda x)={\lambda }^{3}x$$

So, if $\lambda$ is an eigenvalue of $A$ with eigenvector $x$, then ${\lambda }^k$ is an eigenvalue of $A^k$ with the same eigenvector $x$. This property holds true for any positive integer $k$.

Applying the Property to the Given Problem

We are given that the eigenvalues of the 2 × 2 matrix $A$ are ${\lambda }_1=1$ and ${\lambda }_2=-2$. The corresponding eigenvectors are $x_1$ and $x_2$, respectively.

We need to find the eigenvalues and eigenvectors of $A^3$. Using the property derived above:

**Calculating Eigenvalues for $A^3$ **

- For the first eigenvalue ${\lambda }_1=1$: - The corresponding eigenvalue for $A^3$ will be ${\lambda }_1^3=1^3=1$. - The eigenvector remains the same: $x_1$.
- For the second eigenvalue ${\lambda }_2=-2$: - The corresponding eigenvalue for $A^3$ will be ${\lambda }_2^3=(-2{)}^3=-8$. - The eigenvector remains the same: $x_2$.

Conclusion

Therefore, the eigenvalues of the matrix $A^3$ are 1 and -8, and their corresponding eigenvectors are $x_1$ and $x_2$, respectively. This matches the first option provided.