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Subsection 9.2.4 Properties of eigenvalues and vectors

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This unit reminds us of various properties of eigenvalue and eigenvectors through a sequence of homeworks.

Homework 9.2.4.1.

Show that eigenvectors are not unique.

Solution

\(x \) is an eigenvector of \(A \) if for some scalar \(\lambda \) \(A x = \lambda x \) and \(x \neq 0 \text{.}\) For any nonzero scalar \(\alpha \) we find that

\begin{equation*} A (\alpha x ) =\alpha A x = \alpha \lambda x = \lambda ( \alpha x ). \end{equation*}

Hence any (nonzero) scalar multiple of \(x \) is also an eigenvector. This again demonstrates that we care about the direction of an eigenvector rather than its length.

Homework 9.2.4.2.

Let \(\lambda \) be an eigenvalue of \(A \) and let \({\cal E}_\lambda( A ) = \{ x \in \C^m \vert A x = \lambda x \} \text{.}\) This is the set of all eigenvectors of \(A \) associated with \(\lambda \) but also includes the zero vector. Show that \({\cal E}_\lambda( A ) \) is a subspace.

Solution

A set \({\cal S} \subset \Cm \) is a subspace if and only if for all \(\alpha \in \C \) and \(x,y \in \Cm \) two conditions hold:

  • \(x \in {\cal S} \) implies that \(\alpha x \in {\cal S} \text{.}\)

  • \(x, y \in {\cal S} \) implies that \(x + y \in {\cal S} \text{.}\)

  • \(x \in {\cal E}_{\lambda}( A )\) implies \(\alpha x \in {\cal E}_{\lambda}( A )\text{:}\)

    \(x \in {\cal E}_{\lambda}(A) \) means that \(A x = \lambda x \text{.}\) If \(\alpha \in \C \) then \(\alpha A x = \alpha \lambda x \) which, by commutivity and associativity means that \(A ( \alpha x ) = \lambda ( \alpha x ) \text{.}\) Hence \((\alpha x) \in {\cal E}_{\lambda}(A) \text{.}\)

  • \(x,y \in {\cal E}_{\lambda}( A )\) implies \(x+y \in {\cal E}_{\lambda}( A )\text{:}\)

    \begin{equation*} A( x + y ) = A x + A y = \lambda A + \lambda y = \lambda( x + y ) . \end{equation*}

While there are an infinite number of eigenvectors associated with an eigenvalue, the fact that they form a subspace (provided the zero vector is added) means that they can be described by a finite number of vectors, namely a basis for that space.

Homework 9.2.4.3.

Let \(D \in \Cmxm \) be a diagonal matrix. Give all eigenvalues of \(D \text{.}\) For each eigenvalue, give a convenient eigenvector.

Solution

Let

\begin{equation*} D = \left( \begin{array}{c c c c} \delta_0 \amp 0 \amp \cdots \amp 0 \\ 0 \amp \delta_1 \amp \cdots \amp 0 \\ \vdots \amp \vdots \amp \ddots \amp \vdots \\ 0 \amp 0 \amp \cdots \amp \delta_{m-1} \end{array} \right). \end{equation*}

Then

\begin{equation*} \lambda I - D = \left( \begin{array}{c c c c} \lambda - \delta_0 \amp 0 \amp \cdots \amp 0 \\ 0 \amp \lambda - \delta_1 \amp \cdots \amp 0 \\ \vdots \amp \vdots \amp \ddots \amp \vdots \\ 0 \amp 0 \amp \cdots \amp \lambda - \delta_{m-1} \end{array} \right) \end{equation*}

is singular if and only if \(\lambda = \delta_i \) for some \(i \in \{ 0, \ldots , m-1 \} \text{.}\) Hence \(\Lambda( D ) = \{ \delta_0, \delta_1, \ldots, \delta_{m-1} \} \text{.}\)

Now,

\begin{equation*} D e_j = \mbox{ the column of } D \mbox{ indexed with } j = \delta_j e_j \end{equation*}

and hence \(e_j \) is an eigenvector associated with \(\delta_j \text{.}\)

Homework 9.2.4.4.

Let \(U \in \Cmxm \) be an upper triangular matrix. Give all eigenvalues of \(U \text{.}\) For each eigenvalue, give a convenient eigenvector.

Solution

Let

\begin{equation*} U = \left( \begin{array}{c c c c} \upsilon_{0,0} \amp \upsilon_{0,1} \amp \cdots \amp \upsilon_{0,m-1} \\ 0 \amp \upsilon_{1,1} \amp \cdots \amp \upsilon_{1,m-1} \\ \vdots \amp \vdots \amp \ddots \amp \vdots \\ 0 \amp 0 \amp \cdots \amp \upsilon_{m-1,m-1} \end{array} \right). \end{equation*}

Then

\begin{equation*} \lambda I - U = \left( \begin{array}{c c c c} \lambda - \upsilon_{0,0} \amp \upsilon_{0,1} \amp \cdots \amp \upsilon_{0,m-1} \\ 0 \amp \lambda - \upsilon_{1,1} \amp \cdots \amp \upsilon_{1,m-1} \\ \vdots \amp \vdots \amp \ddots \amp \vdots \\ 0 \amp 0 \amp \cdots \amp \lambda - \upsilon_{m-1,m-1} \end{array} \right). \end{equation*}

is singular if and only if \(\lambda = \upsilon_{i,i} \) for some \(i \in \{ 0, \ldots , m-1 \} \text{.}\) Hence \(\Lambda( D ) = \{ \upsilon_{0,0}, \upsilon_{1,1}, \ldots, \upsilon_{m-1,m-1} \} \text{.}\)

Let \(\lambda \) be an eigenvalue of \(U \text{.}\) Things get a little tricky if \(\lambda \) has multiplicity greater than one. Partition

\begin{equation*} U = \left( \begin{array}{c c c} U_{00} \amp u_{01} \amp U_{02} \\ 0 \amp \upsilon_{11} \amp u_{12}^T \\ 0 \amp 0 \amp U_{22} \end{array} \right) \end{equation*}

where \(\upsilon_{11} = \lambda \text{.}\) We are looking for \(x \neq 0 \) such that \(( \lambda I - U ) x = 0 \) or, partitioning \(x \text{,}\)

\begin{equation*} \left( \begin{array}{c c c} \upsilon_{11} I - U_{00} \amp u_{01} \amp U_{02} \\ 0 \amp 0 \amp u_{12}^T \\ 0 \amp 0 \amp \upsilon_{11} I - U_{22} \end{array} \right) \left( \begin{array}{c} x_0 \\ \chi_1 \\ x_2 \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right). \end{equation*}

If we choose \(x_2 = 0 \) and \(\chi_1 =1 \text{,}\) then

\begin{equation*} ( \upsilon I - U_{22} ) x_0 + u_{01} = 0 \end{equation*}

and hence \(x_0 \) must satisfy

\begin{equation*} ( \upsilon I - U_{22} ) x_0 = - u_{01}. \end{equation*}

If \(\upsilon I - U_{22} \) is nonsingular, then there is a unique solution to this equation, and

\begin{equation*} \left( \begin{array}{c} - ( \upsilon I - U_{22} )^{-1} u_{01} \\ 1 \\ 0 \end{array} \right) \end{equation*}

is the desired eigenvalue. HOWEVER, this means that the partitioning

\begin{equation*} U = \left( \begin{array}{c c c} U_{00} \amp u_{01} \amp U_{02} \\ 0 \amp \upsilon_{11} \amp u_{12}^T \\ 0 \amp 0 \amp U_{22} \end{array} \right) \end{equation*}

must be such that \(\upsilon_{11} \) is the FIRST diagonal element that equals \(\lambda \text{.}\)

Homework 9.2.4.5.

Let \(\lambda, \mu \in \Lambda( A ) \text{,}\) \(A x = \lambda x \text{,}\) and \(A y = \mu y \) with \(x \neq 0\) and \(y \ne 0 \text{.}\)

This should be moved to after the discussion of the Schur decomposition>