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Show that if \(U\) is an orthogonal matrix, then any real eigenvalue of \(U\) must be \( \pm 1\).

Short Answer

Expert verified

It is proved that any real eigenvalues of \(U\) is \( \pm 1\).

Step by step solution

01

Statement in Theorem 7

Theorem 7states that consider that \(U\) as an \(m \times n\) matrix with orthonormal columns and assume that x and y are in \({\mathbb{R}^n}\). Then;

  1. \(\left\| {U{\bf{x}}} \right\| = \left\| {\bf{x}} \right\|\)
  2. \(\left( {U{\bf{x}}} \right) \cdot \left( {U{\bf{y}}} \right) = {\bf{x}} \cdot {\bf{y}}\)
  3. \(\left( {U{\bf{x}}} \right) \cdot \left( {U{\bf{y}}} \right) = 0\)such that if \({\bf{x}} \cdot {\bf{y}} = 0\).
02

Show that real eigenvalues of \(U\) must be \( \pm 1\) 

When \(U{\bf{x}} = \lambda {\bf{x}}\) for some \({\bf{x}} \ne 0\), then according to Theorem 7 and by a property of the norm:

\(\begin{array}{c}\left\| {\bf{x}} \right\| = \left\| {U{\bf{x}}} \right\|\\ = \left\| {\lambda {\bf{x}}} \right\|\\ = \left| \lambda \right|\left\| {\bf{x}} \right\|\end{array}\)

This demonstrates that \(\left| \lambda \right| = 1\), since \({\bf{x}} \ne 0\).

Thus, it is proved that any real eigenvalues of \(U\) is \( \pm 1\).

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Most popular questions from this chapter

In exercises 7-10, show that {u1, u2} or {u1,u2,u3} is an orthogonal basis for \({\mathbb{R}^2}\) or \({\mathbb{R}^3}\), respectively. Then express x as a linear combination of the u.

7. \[{u_1} = \left[ {\begin{align}2\\{ - 3}\end{align}} \right]\], \[{u_2} = \left[ {\begin{align}6\\4\end{align}} \right]\], and \[x = \left[ {\begin{align}9\\{ - 7}\end{align}} \right]\]

Find a polynomial \({p_{\bf{3}}}\) such that \(\left\{ {{p_{\bf{0}}},{p_{\bf{1}}},{p_{\bf{2}}},{p_{\bf{3}}}} \right\}\) (see Exercise 11) is an orthogonal basis for the subspace \({{\bf{P}}_{\bf{3}}}\) of \({{\bf{P}}_{\bf{4}}}\). Scale the polynomials \({p_{\bf{3}}}\) so that vector of values is \(\left( { - {\bf{1}},{\bf{2}},{\bf{0}}, - {\bf{2}},{\bf{1}}} \right)\).

In Exercises 11 and 12, find the closest point to\[{\bf{y}}\]in the subspace\[W\]spanned by\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\].

11.\[y = \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right]\],\[{{\bf{v}}_1} = \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\],\[{{\bf{v}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

According to Kepler’s first law, a comet should have an elliptic, parabolic, or hyperbolic orbit (with gravitational attractions from the planets ignored). In suitable polar coordinates, the position \(\left( {r,\vartheta } \right)\) of a comet satisfies an equation of the form

\(r = \beta + e\left( {r \cdot \cos \vartheta } \right)\)

Where \(\beta \) is a constant and \(e\) is the eccentricity of the orbit, with \(0 \le e < 1\) for an ellipse, \(e = 1\) for a parabola, and \(e > 1\) for a hyperbola. Suppose observations of a newly discovered comet provide the data below. Determine the type of orbit, and predict where the comet will be when \(\vartheta = 4.6\left( {{\rm{radians}}} \right)\).

\(\begin{array}{{}{}} \vartheta & & {.88}&{1.10}&{1.42}&{1.77}&{2.14} \\ \hline r& {3.00}&{2.30}&{1.65}&{1.25}&{1.01} \end{array}\)

In Exercises 3–6, verify that\[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\]is an orthogonal set, and then find the orthogonal projection of\[{\bf{y}}\]onto Span\[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\].

5.\[y = \left[ {\begin{aligned}{ - 1}\\2\\6\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}3\\{ - 1}\\2\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\{ - 2}\end{aligned}} \right]\]

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