Consider an n×n orthogonal matrix A with real entries and each column having unit Euclidean norm. Which of the following statements is/are correct?

The correct statements are:
1. The value of the determinant of A is either +1 or −1
2. The eigenvalues of A have modulus 1
3. ∥Ax∥ = ∥x∥, for all x ∈ ℝⁿ, and (Ax)^T(Ay) = x^Ty, for all x,y ∈ ℝⁿ

Step-by-step Explanation:

By definition, an $n\times n$ matrix $A$ with real entries is orthogonal if its columns are mutually orthogonal unit vectors. This translates to the matrix equation:
$$A^{T}A=I$$
where $A^T$ is the transpose of $A$ and $I$ is the $n\times n$ identity matrix.

1. Determinant of A:
Taking the determinant on both sides of the relation $A^{T}A=I$:
$$det\left(A^{T}A\right)=det\left(I\right)$$
Using the properties of determinants, specifically $det\left(BC\right)=det\left(B\right)det\left(C\right)$ and $det\left(A^T\right)=det\left(A\right)$, we have:
$$det\left(A^T\right)·det\left(A\right)=1$$
$${[det(A\left)\right]}^2=1$$
Taking the square root of both sides gives:
$$det\left(A\right)=\pm 1$$
Thus, the value of the determinant of A is indeed either +1 or −1.

2. Preservation of Euclidean Norm and Inner Product:
For any vectors $x,y\in {ℝ}^n$, the inner product (dot product) of $Ax$ and $Ay$ is:
$${\left(Ax\right)}^T\left(Ay\right)=x^T{A}^{T}Ay$$
Since $A^{T}A=I$, we substitute this in to get:
$${\left(Ax\right)}^T\left(Ay\right)=x^{T}Iy=x^{T}y$$
By setting $y=x$, we obtain the relation for the squared Euclidean norm:
$${\parallel Ax\parallel }^2={\left(Ax\right)}^T\left(Ax\right)=x^{T}x={\parallel x\parallel }^2$$
Taking the square root yields:
$$\parallel Ax\parallel =\parallel x\parallel$$
This proves that $\parallel Ax\parallel =\parallel x\parallel$ for all $x\in {ℝ}^n$ and ${\left(Ax\right)}^T\left(Ay\right)=x^{T}y$ for all $x,y\in {ℝ}^n$.

3. Eigenvalues of A:
Let $\lambda$ be an eigenvalue of $A$ (which may be a complex number since real matrices can have complex eigenvalues), and let $v$ be its corresponding non-zero eigenvector. Thus:
$$Av=\lambda v$$
Taking the conjugate transpose (denoted by $*$) on both sides of the equation:
$${\left(Av\right)}^{*}={\left(\lambda v\right)}^{*}\Rightarrow v^{*}{A}^{*}=\overline{\lambda }{v}^{*}$$
Multiplying this conjugate transpose equation by the original eigenvalue equation:
$$\left(v^{*}{A}^{*}\right)\left(Av\right)=\left(\overline{\lambda }{v}^{*}\right)\left(\lambda v\right)$$
Since $A$ is a real matrix, $A^{*}=A^T$. Because $A$ is orthogonal, $A^{*}A=A^{T}A=I$. Therefore, the left side simplifies to:
$$v^{*}\left(A^{*}A\right)v=v^{*}Iv=v^{*}v$$
The right side simplifies to:
$$\overline{\lambda }\lambda v^{*}v={\left|\lambda \right|}^2{v}^{*}v$$
Equating both sides:
$$v^{*}v={\left|\lambda \right|}^2{v}^{*}v$$
Since $v$ is a non-zero vector, we know that $v^{*}v>0$. We can divide both sides by $v^{*}v$ to get:
$${\left|\lambda \right|}^2=1$$
Thus, taking the positive square root:
$$\left|\lambda \right|=1$$
Consequently, all eigenvalues of A have a modulus of 1.