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Discover Linear Algebra

Jeremy Sylvestre

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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 \(\adjoint{A} = A\text{.}\)
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  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 \(\adjoint{A}A = I\text{.}\)
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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.
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Discovery 40.1. Hermitian eigenvalues.

Suppose \(H\) is a Hermitian (i.e. 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 \(\lambda\text{.}\)
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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.
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Discovery 40.2. Symmetric eigenvalues.

Convince yourself that a real self-adjoint matrix is also self-adjoint when considered as a complex matrix.
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Based on Discovery 40.1, what does this mean about the eigenvalues of a symmetric matrix?
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Discovery 40.3.

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*}
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(a)

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

What relationship between the provided eigenvectors of \(H\) was crucial to allowing Task b to work?
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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?
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Discovery 40.4.

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*}
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(a)

Write down a transition matrix \(P\) so that \(\inv{P} A P\) is diagonal.
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(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 40.3 work?
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Hint.
The columns of \(U\) must be eigenvectors to diagonalize \(A\text{!}\)
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Let’s focus on what seemed to work, and why, in Discovery 40.3.
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Discovery 40.5.

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.
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(a)

Obtain an expression for \(\inprod{H \uvec{x}_1}{\uvec{x}_2}\) in terms of \(\lambda_1\) and \(\inprod{\uvec{x}_1}{\uvec{x}_2}\text{.}\)
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(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 \(\inprod{\uvec{x}_1}{\uvec{x}_2}\text{.}\)
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Hint.
You may wish to make use of the property of eigenvalues of Hermitian matrices found in Discovery 40.1.
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(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 \(\uvec{x}_1,\uvec{x}_2\text{.}\)
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Discovery 40.6.

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*}
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What is the point of this discovery activity?
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Discovery 40.7.

(a)

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

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

What do you think is the point of this discovery activity?
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Discovery 40.8. 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 \(\lambda\text{.}\)
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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.
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Discovery 40.9. Orthogonal eigenvalues.

Convince yourself that a real orthogonal matrix is also unitary when considered as a complex matrix.
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Based on Discovery 40.8, what does this mean about the eigenvalues of an orthogonal matrix?
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