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Chapter 5: Q5.2-18E (page 267)

Question 18: It can be shown that the algebraic multiplicity of an eigenvalue \(\lambda \) is always greater than or equal to the dimension of the eigenspace corresponding to \(\lambda \). Find \(h\) in the matrix \(A\) below such that the eigenspace for \(\lambda = 5\) is two-dimensional:

\[A = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&h&0\\0&0&5&4\\0&0&0&1\end{array}} \right]\]

Short Answer

Expert verified

It is observed that the system requires two free variables for a two-dimensional eigenspace. This occurs only when \(h = 6\).

Step by step solution

01

Definition of the characteristic polynomial

The eigenvalue of an \(n \times n\) matrix \(A\) is a scalar \(\lambda \) such that \(\lambda \) satisfies the characteristic equation \(\det \left( {A - \lambda I} \right) = 0\).

When\(A\)is a\(n \times n\)matrix,\(\det \left( {A - \lambda I} \right)\)is thecharacteristic polynomial of\(A\)which is the polynomial of degree\(n\).

In particular, the multiplicity of an eigenvalue \(\lambda \) represents its multiplication as a root of the characteristic equation.

02

Determine h in matrix A

Write the matrix in the \[\left( {A - 5I} \right)\] form, as shown below.

\[\begin{array}{c}\left( {{\rm{A}} - 5I} \right) = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&3&{\rm{h}}&0\\0&0&5&4\\0&0&0&1\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}5&0&0&0\\0&5&0&0\\0&0&5&0\\0&0&0&5\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}5&{ - 2}&6&{ - 1}\\0&{ - 2}&{\rm{h}}&0\\0&0&0&4\\0&0&0&{ - 4}\end{array}} \right]\end{array}\]

Write the matrix as an augmented matrix \(\left( {A - 5I} \right)x = 0\), as shown below.

\( \sim \left[ {\begin{array}{*{20}{c}}0&{ - 2}&6&{ - 1}&0\\0&{ - 2}&{\rm{h}}&0&0\\0&0&0&4&0\\0&0&0&{ - 4}&0\end{array}} \right]\)

Apply row operation in the augmented matrix:

At row 2, subtract row 1 from row 2. At row 4, multiply row 4 by \( - 1\), as shown below.

\( \sim \left[ {\begin{array}{*{20}{c}}0&{ - 2}&6&{ - 1}&0\\0&0&{{\rm{h}} - 6}&1&0\\0&0&0&4&0\\0&0&0&4&0\end{array}} \right]\)

At row 4, multiply row 3 by 1 and subtract it from row 4, as shown below.

\( \sim \left[ {\begin{array}{*{20}{c}}0&{ - 2}&6&{ - 1}&0\\0&0&{{\rm{h}} - 6}&1&0\\0&0&0&4&0\\0&0&0&0&0\end{array}} \right]\)

At row 1, multiply row 1 by \( - \frac{1}{2}\) and at row 3, multiply row 3 by \(\frac{1}{4}\), as shown below.

\( \sim \left[ {\begin{array}{*{20}{c}}0&1&{ - 3}&1&0\\0&0&{{\rm{h}} - 6}&1&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\)

At row 1, subtract row 3 from row 1, and at row 2, subtract row 3 from row 2, as shown below.

\( \sim \left[ {\begin{array}{*{20}{c}}0&1&{ - 3}&0&0\\0&0&{{\rm{h}} - 6}&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\)

The above system requires two free variables for a two-dimensional eigenspace. This occurs only when \(h = 6\).

Thus, the value of \(h\) in the matrix is 6.

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Most popular questions from this chapter

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.6}&{.3}\\{.4}&{.7}\end{array}} \right)\), \({v_1} = \left( {\begin{array}{*{20}{c}}{3/7}\\{4/7}\end{array}} \right)\), \({x_0} = \left( {\begin{array}{*{20}{c}}{.5}\\{.5}\end{array}} \right)\). (Note: \(A\) is the stochastic matrix studied in Example 5 of Section 4.9.)

  1. Find a basic for \({\mathbb{R}^2}\) consisting of \({{\rm{v}}_1}\) and anther eigenvector \({{\rm{v}}_2}\) of \(A\).
  2. Verify that \({{\rm{x}}_0}\) may be written in the form \({{\rm{x}}_0} = {{\rm{v}}_1} + c{{\rm{v}}_2}\).
  3. For \(k = 1,2, \ldots \), define \({x_k} = {A^k}{x_0}\). Compute \({x_1}\) and \({x_2}\), and write a formula for \({x_k}\). Then show that \({{\bf{x}}_k} \to {{\bf{v}}_1}\) as \(k\) increases.

Let \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3}} \right\}\)be a basis for a vector space \(V\) and\(T:V \to {\mathbb{R}^2}\) be a linear transformation with the property that

\(T\left( {{x_1}{{\bf{b}}_1} + {x_2}{{\bf{b}}_2} + {x_3}{{\bf{b}}_3}} \right) = \left( {\begin{aligned}{2{x_1} - 4{x_2} + 5{x_3}}\\{ - {x_2} + 3{x_3}}\end{aligned}} \right)\)

Find the matrix for \(T\) relative to \(B\) and the standard basis for \({\mathbb{R}^2}\).

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

2. \(\left[ {\begin{array}{*{20}{c}}5&3\\3&5\end{array}} \right]\)

Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

21. Use mathematical induction to prove that for \(n \ge {\bf{2}}\),\(\begin{aligned}{c}det\left( {{C_p} - \lambda I} \right) = {\left( { - {\bf{1}}} \right)^n}\left( {{a_{\bf{0}}} + {a_{\bf{1}}}\lambda + ... + {a_{n - {\bf{1}}}}{\lambda ^{n - {\bf{1}}}} + {\lambda ^n}} \right)\\ = {\left( { - {\bf{1}}} \right)^n}p\left( \lambda \right)\end{aligned}\)

(Hint: Expanding by cofactors down the first column, show that \(det\left( {{C_p} - \lambda I} \right)\) has the form \(\left( { - \lambda B} \right) + {\left( { - {\bf{1}}} \right)^n}{a_{\bf{0}}}\) where \(B\) is a certain polynomial (by the induction assumption).)

M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set \[{{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0,0}}} \right)\] and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

19.\[A{\bf{=}}\left[{\begin{array}{*{20}{c}}{{\bf{10}}}&{\bf{7}}&{\bf{8}}&{\bf{7}}\\{\bf{7}}&{\bf{5}}&{\bf{6}}&{\bf{5}}\\{\bf{8}}&{\bf{6}}&{{\bf{10}}}&{\bf{9}}\\{\bf{7}}&{\bf{5}}&{\bf{9}}&{{\bf{10}}}\end{array}} \right]\]
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