Given three degenerate eigenfunctions that are linearly independent although not necessarily orthogonal, find three linear combinations of them that are orthogonal to each other and are normalized. Are the three new combinations eigenfunctions? If so, are they degenerate?

Linear independence is a fundamental concept in linear algebra. It refers to a set of vectors where no vector can be represented as a linear combination of the others.
In simpler terms, imagine a set of vectors in space. If you cannot find any vector among them that you can create by scaling and adding up the others, they are linearly independent.
Why is this important? Because when vectors are linearly independent, they span a space uniquely.
Given our exercise, the eigenfunctions \( \phi_1, \phi_2, \phi_3 \) are linearly independent. That means you can't express \( \phi_1 \) using \( \phi_2 \) and \( \phi_3 \), and vice versa.
This property is crucial for the possibility of orthogonalization.

Eigenfunctions associated with the same eigenvalue are called degenerate eigenfunctions.
In other words, if multiple eigenfunctions correspond to the same eigenvalue, they are said to be degenerate.
In our exercise, \( \phi_1, \phi_2, \phi_3 \) are degenerate because they share the same eigenvalue.
This degeneracy implies certain symmetries or conserved quantities in the system being analyzed.
When you apply the Gram-Schmidt process to these functions, you obtain new functions \( u_1, u_2, u_3 \) that are orthogonal but still correspond to the same eigenvalue. Thus, they are also degenerate eigenfunctions.
Degeneracy is why even new combinations of these functions retain their original eigenvalue.

Normalization is the process of adjusting the length of a vector (or function) to 1 while keeping its direction constant.
Simply put, it’s making the vector “unit length.”
To normalize a vector \( u \), you divide it by its magnitude, calculated as \( \sqrt{(u, u)} \).
In our exercise, once the orthogonal functions \( u_1, u_2, u_3 \) are obtained, they are normalized:

• \( v_1 = \frac{u_1}{ \sqrt{(u_1, u_1)} } \)
• \( v_2 = \frac{u_2}{ \sqrt{(u_2, u_2)} } \)
• \(( v_3 = \frac{u_3}{ \sqrt{(u_3, u_3)} } \)

This ensures each function has a unit length and becomes an orthonormal set.
Orthonormal functions make calculations simpler and more reliable by maintaining orthogonality and unit length.

Eigenvalues are scalars that provide valuable information about the behavior of linear transformations.
For a given linear operator acting on a vector space, eigenvalues are associated with non-zero vectors that change only in scale when the operator is applied.
For an eigenvector \x \ with a corresponding eigenvalue \lambda \ under a linear operator \ A \ , the relationship is \( A x = \lambda x \).

• In the context of our exercise, given functions \( \phi_1, \phi_2, \phi_3 \) are each associated with the same eigenvalue, making them degenerate.
• After orthogonalization and normalization steps, the new combinations \( v_1, v_2, v_3 \) retain their eigenvalues.

This shows that linear transformations can be elegantly simplified using eigenvalues and their respective eigenfunctions.