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Discovery guide 40.2 Discovery guide

Recall.

  1. A matrix is self-adjoint if
    \begin{equation*} \inprod{\uvec{u}}{A \uvec{v}} = \inprod{A \uvec{u}}{\uvec{v}} \end{equation*}
    always, so that \(\adjoint{A} = A\text{.}\)
  2. A matrix is product-preserving (or orthogonal in the real case, unitary in the complex case) if
    \begin{equation*} \inprod{A \uvec{u}}{A \uvec{v}} = \inprod{\uvec{u}}{\uvec{v}} \end{equation*}
    always, so that \(\adjoint{A}A = I\text{.}\)

Notation.

Where necessary, in this discovery guide we will write
\begin{align*} \amp {\inprod{\blank}{\blank}}_{\R} \text{,} \amp \amp {\inprod{\blank}{\blank}}_{\C} \end{align*}
to distinguish between the real and complex dot products, respectively.

Discovery 40.2.1. Hermitian eigenvalues.

Suppose \(H\) is a Hermitian (that is, complex self-adjoint) matrix and \(\lambda\) and \(\uvec{x}\) are an eigenvalue-eigenvector pair for \(H\text{,}\) so that
\begin{equation*} H\uvec{x} = \lambda \uvec{x} \text{.} \end{equation*}
Use
\begin{equation*} {\inprod{\uvec{x}}{H \uvec{x}}}_{\C} = {\inprod{H \uvec{x}}{\uvec{x}}}_{\C} \end{equation*}
to discover something about \(\lambda\text{.}\)
Hint.
Refer to the Algebra rules of complex inner products (Proposition 36.6.2). “Simplify” each side separately, and then compare the new versions of the two sides again.

Discovery 40.2.2. Symmetric eigenvalues.

Convince yourself that a real self-adjoint matrix is also self-adjoint when considered as a complex matrix.
Based on Discovery 40.2.1, what does this mean about the eigenvalues of a symmetric matrix?

Discovery 40.2.3.

The Hermitian matrix
\begin{equation*} H = \begin{abmatrix}{rcc} 0 \amp \ci \amp 0 \\ -\ci \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \end{abmatrix} \end{equation*}
has eigenvalues \(\lambda = \pm 1\) with
\begin{align*} E_{-1}(H) \amp = \Span \left\{ \begin{abmatrix}{r} -\ci \\ 1 \\ 0 \end{abmatrix} \right\} \text{,} \amp E_1(H) \amp = \Span \left\{ \begin{bmatrix} \ci \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\} \text{.} \end{align*}

(a)

Write down a transition matrix \(P\) so that \(\inv{P} H P\) is diagonal.

(c)

What relationship between the provided eigenvectors of \(H\) was crucial to allowing Task (b) to work?
What if the two provided eigenvectors for \(\lambda = 1\) had not initially had that relationship with each other — would you have been able to “fix” it? How?

Discovery 40.2.4.

The complex matrix
\begin{equation*} A = \begin{abmatrix}{rrc} 1 \amp 0 \amp 0 \\ -2 \ci \amp -1 \amp 0 \\ -2 \amp 2 \ci \amp 1 \end{abmatrix} \end{equation*}
has eigenvalues \(\lambda = \pm 1\) with
\begin{align*} E_{-1}(A) \amp = \Span \left\{ \begin{bmatrix} 0 \\ \ci \\ 1 \end{bmatrix} \right\} \text{,} \amp E_1(A) \amp = \Span \left\{ \begin{bmatrix} \ci \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\} \text{.} \end{align*}

(a)

Write down a transition matrix \(P\) so that \(\inv{P} A P\) is diagonal.

(b)

Will you be able to write down a unitary transition matrix \(U\) so that \(\adjoint{U} A U\) is diagonal? Will trying to “fix” things as in Task (c) of Discovery 40.2.3 work?
Hint.
The columns of \(U\) must be eigenvectors to diagonalize \(A\text{!}\)
Let’s focus on what seemed to work, and why, in Discovery 40.2.3.

Discovery 40.2.5.

Suppose \(H\) is a Hermitian matrix, \(\lambda_1,\lambda_2\) are two different eigenvalues for \(H\text{,}\) and \(\uvec{x}_1,\uvec{x}_2\) are corresponding eigenvectors for those two eigenvalues, respectively.

(a)

Obtain an expression for \(\inprod{H \uvec{x}_1}{\uvec{x}_2}\) in terms of \(\lambda_1\) and \(\inprod{\uvec{x}_1}{\uvec{x}_2}\text{.}\)

(b)

Use the fact that \(H\) is self-adjoint to obtain an expression for \(\inprod{H \uvec{x}_1}{\uvec{x}_2}\) in terms of \(\lambda_2\) and \(\inprod{\uvec{x}_1}{\uvec{x}_2}\text{.}\)
Hint.
You may wish to make use of the property of eigenvalues of Hermitian matrices found in Discovery 40.2.1.

(c)

By comparing your two expressions for \(\inprod{H \uvec{x}_1}{\uvec{x}_2}\text{,}\) use the assumption that \(\lambda_1 \neq \lambda_2\) to learn something about the eigenvectors \(\uvec{x}_1,\uvec{x}_2\text{.}\)

Discovery 40.2.6.

Determine a unitary matrix \(U\) so that \(\adjoint{U} A U\) is diagonal, for matrix
\begin{equation*} A = \begin{bmatrix} 2 \amp 0 \\ 0 \amp \ci \end{bmatrix} \text{.} \end{equation*}
What is the point of this discovery activity?

Discovery 40.2.7.

(a)

Convince yourself that a diagonal complex matrix \(D\) commutes with its adjoint: \(\adjoint{D} D = D \adjoint{D}\text{.}\)

(b)

Suppose \(A\) is a unitarily diagonalizable complex matrix, so that \(\adjoint{U} A U\) is diagonal for some unitary matrix \(U\text{.}\)
Use the diagonal case from Task (a) to help verify that \(A\) commutes with its adjoint: \(\adjoint{A} A = A \adjoint{A}\text{.}\)

(c)

What do you think is the point of this discovery activity?

Discovery 40.2.8. Unitary eigenvalues.

Suppose \(A\) is a unitary complex matrix and \(\lambda\) and \(\uvec{x}\) are an eigenvalue-eigenvector pair for \(A\text{,}\) so that
\begin{equation*} A\uvec{x} = \lambda \uvec{x} \text{.} \end{equation*}
Use
\begin{equation*} {\inprod{A \uvec{x}}{A \uvec{x}}}_{\C} = {\inprod{\uvec{x}}{\uvec{x}}}_{\C} \end{equation*}
to discover something about \(\lambda\text{.}\)
Hint.
Refer to the Algebra rules of complex inner products (Proposition 36.6.2). And remember that a value can only be an eigenvalue if there exist nontrivial corresponding eigenvectors.

Discovery 40.2.9. Orthogonal eigenvalues.

Convince yourself that a real orthogonal matrix is also unitary when considered as a complex matrix.
Based on Discovery 40.2.8, what does this mean about the eigenvalues of an orthogonal matrix?
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