Consider an n×n matrix A and a non-zero n×1 vector p. Their product Ap = α² p, where {α ∈ R ∉and α -1,0,1} . Based on the given information, the Eigen value of A² is :

The correct option is α⁴ (which corresponds to Option 4, represented as "α4").

Here is the step-by-step educational explanation of why this option is correct:

1. Understanding the Definition of Eigenvalues and Eigenvectors:
Recall that for an $n\times n$ matrix $A$ and a non-zero $n\times 1$ vector $p$, if there exists a scalar $\lambda$ such that:
$Ap=\lambda p$
then $\lambda$ is an eigenvalue of the matrix $A$, and $p$ is the corresponding eigenvector.

2. Identifying the Eigenvalue of matrix A:
We are given the relation:
$Ap={\alpha }^{2}p$
Comparing this with the definition $Ap=\lambda p$, we can directly see that:
$\lambda ={\alpha }^2$
is the eigenvalue of matrix $A$ associated with the eigenvector $p$.

3. Determining the Eigenvalues of a Matrix Power:
A fundamental theorem in linear algebra states that if $\lambda$ is an eigenvalue of a matrix $A$ associated with an eigenvector $p$, then for any positive integer $k$, the matrix $A^k$ has the eigenvalue ${\lambda }^k$ associated with the same eigenvector $p$.

Let us derive this for the case of $A^2$ (where $k=2$):
We start by multiplying both sides of our initial equation by $A$ from the left:
$A\left(Ap\right)=A\left({\alpha }^{2}p\right)$

Using the associativity of matrix multiplication and the linearity property (since ${\alpha }^2$ is a scalar):
$A^{2}p={\alpha }^2\left(Ap\right)$

Now, substitute the original expression $Ap={\alpha }^{2}p$ back into the right side of the equation:
$A^{2}p={\alpha }^2\left({\alpha }^{2}p\right)$
Simplifying the scalar multiplication:
$A^{2}p={\alpha }^{4}p$

4. Conclusion:
The resulting relation $A^{2}p={\alpha }^{4}p$ fits the definition of an eigenvalue equation for the matrix $A^2$. Thus, the eigenvalue of $A^2$ is ${\alpha }^4$.