Partial Differential Equations: Exercises

16 Sturm-Liouville problems, part 1

Exercise 1: Identifying self-adjoint linear operators

Recall that a linear operator \(T: \mathbb{C}^k \to \mathbb{C}^k\) is self-adjoint if

\[ \langle T(\mathbf{u}), \mathbf{v} \rangle = \langle \mathbf{u}, T(\mathbf{v}) \rangle \]

for all \(\mathbf{u}, \mathbf{v} \in \mathbb{C}^k\). If \(A\) is the matrix representation of \(T\) with respect to the standard basis, then \(T\) is self-adjoint if and only if \(A\) is equal to its conjugate transpose \(A^*\).

For the linear operators with the following matrix representations, determine whether they are self-adjoint.

1. \[\begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix}\]

2. \[\begin{bmatrix} 2 & 1 - 2i \\ 1 + 2i & -2 \end{bmatrix}\]

3. \[\begin{bmatrix} 2 & 1 & 3+2i \\ 1 & 0 & 4i \\ 3-2i & -4i & 1 \end{bmatrix}\]

4. \[\begin{bmatrix} 2i & 3 & 1 \\ 3 & 0 & 4i \\ 1 & -4i & 1 \end{bmatrix}\]

NoteSolutions.

1. Not self-adjoint.
2. Self-adjoint.
3. Self-adjoint.
4. Not self-adjoint.

Exercise 2: Finding eigenvalues and eigenvectors of linear operators

For a self-adjoint linear operator \(T: \mathbb{C}^k \to \mathbb{C}^k\), the eigenvalues are real and the eigenvectors corresponding to distinct eigenvalues are orthogonal. In contrast, for a non-self-adjoint operator, the eigenvalues can be complex and the eigenvectors may not be orthogonal.

1. For the linear operator with matrix representation \[ \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \] from the previous exercise, find the eigenvalues and corresponding eigenvectors. Compute the inner product of the eigenvectors.

2. For the linear operator with matrix representation \[ \begin{bmatrix} 2 & 1 - 2i \\ 1 + 2i & -2 \end{bmatrix} \] from the previous exercise, find the eigenvalues and corresponding eigenvectors. Compute the inner product of the eigenvectors. (This operator is self-adjoint, so you already know that the eigenvectors will be orthogonal by the Spectral Theorem, but I want you to verify this by direct computation.)

NoteSolutions.

1. Eigenvalues are \(1\) and \(2\), with corresponding eigenvectors \[ \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \] The inner product of the eigenvectors is \(1\).

2. Eigenvalues are \(-3\) and \(3\), with corresponding eigenvectors \[ \begin{bmatrix} -1+2i \\ 5 \end{bmatrix} \quad \text{and} \quad \begin{bmatrix} 1-2i \\ 1 \end{bmatrix}. \] The inner product of the eigenvectors is \[ \left\langle \begin{bmatrix} -1+2i \\ 5 \end{bmatrix}, \begin{bmatrix} 1-2i \\ 1 \end{bmatrix} \right\rangle = (-1+2i)(1+2i) + 5(1) = 0. \]

Exercise 3: Proving eigenvalues of self-adjoint operators are real

Recall that the Spectral Theorem says that the eigenvalues of a self-adjoint linear operator are real. Prove it.

NoteSolution.

Let \(T: \mathbb{C}^k \to \mathbb{C}^k\) be a self-adjoint linear operator, and let \(\lambda \in \mathbb{C}\) be an eigenvalue with corresponding (nonzero!) eigenvector \(\mathbf{v} \in \mathbb{C}^k\). Then

\[ \langle T(\mathbf{v}), \mathbf{v} \rangle = \langle \lambda \mathbf{v}, \mathbf{v} \rangle = \lambda \langle \mathbf{v}, \mathbf{v} \rangle = \lambda \|\mathbf{v}\|^2, \]

where we used linearity of the inner product in the first argument in the second equality. On the other hand, we have

\[ \langle \mathbf{v}, T(\mathbf{v}) \rangle = \langle \mathbf{v}, \lambda \mathbf{v} \rangle = \overline{\lambda} \langle \mathbf{v}, \mathbf{v} \rangle = \overline{\lambda} \|\mathbf{v}\|^2, \]

where we used conjugate linearity of the inner product in the second argument in the second equality. But since \(T\) is self-adjoint, we have \(\langle T(\mathbf{v}), \mathbf{v} \rangle = \langle \mathbf{v}, T(\mathbf{v}) \rangle\), and thus

\[ \lambda \|\mathbf{v}\|^2 = \overline{\lambda} \|\mathbf{v}\|^2. \]

But \(\|\mathbf{v}\|\) is not \(0\), and so we must have \(\lambda = \overline{\lambda}\), which means that \(\lambda\) is real.