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Chapter 5: Q19E (page 267)

Question: For \(A = \left( {\begin{array}{*{20}{c}}1&2&3\\1&2&3\\1&2&3\end{array}} \right)\), find the eigenvalues, with no calculation. Justify your answer.

Short Answer

Expert verified

Eigenvalues of the given matrix is 0, as the determinant of the matrix is 0, so it is non-invertible, which imply \(A{\bf{x}} = 0\) has a nontrivial solution.

Step by step solution

01

Definition

Eigenvalue: Let \(\lambda \) is a scaler, \(A\) is an \(n \times n\) matrix and \({\bf{x}}\) is an eigenvector corresponding to \(\lambda \), \(\lambda \) is said to an eigenvalue of the matrix \(A\) if there exists a nontrivial solution \({\bf{x}}\) of \(A{\bf{x}} = \lambda {\bf{x}}\).

02

Find Eigenvalues 

The given matrix is \(A = \left( {\begin{array}{*{20}{c}}1&2&3\\1&2&3\\1&2&3\end{array}} \right)\).

As per the definition of eigenvalues, if \(A{\bf{x}} = \lambda {\bf{x}}\) has a nontrivial solution, then \(\lambda \) is the eigenvalue, and the matrix \(A\) is non-invertible if and only if \(A{\bf{x}} = 0\) has nontrivial solution and 0 is the eigenvalue.

Here, the matrix is invertible if its determinant is not 0. So, check, the matrix is invertible or not by checking its determinant.

The determinant of the determinant is 0, as all the rows of the matrix are identical. Which imply the matrix is non-invertible. Hence, the eigenvalue will be 0.

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Most popular questions from this chapter

Apply the results of Exercise \({\bf{15}}\) to find the eigenvalues of the matrices \(\left( {\begin{aligned}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{1}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{1}}\end{aligned}} \right)\) and \(\left( {\begin{aligned}{*{20}{c}}{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}&{\bf{3}}\\{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{3}}&{\bf{7}}\end{aligned}} \right)\).

Question: In Exercises \({\bf{5}}\) and \({\bf{6}}\), the matrix \(A\) is factored in the form \(PD{P^{ - {\bf{1}}}}\). Use the Diagonalization Theorem to find the eigenvalues of \(A\) and a basis for each eigenspace.

5. \(\left( {\begin{array}{*{20}{c}}2&2&1\\1&3&1\\1&2&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&1&2\\1&0&{ - 1}\\1&{ - 1}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&0&0\\0&1&0\\0&0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{2}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{1}{2}}&{ - \frac{3}{4}}\\{\frac{1}{4}}&{ - \frac{1}{2}}&{\frac{1}{4}}\end{array}} \right)\)

Define\(T:{{\rm P}_3} \to {\mathbb{R}^4}\)by\(T\left( {\bf{p}} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 3} \right)}\\{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 1 \right)}\\{{\bf{p}}\left( 3 \right)}\end{aligned}} \right)\).

  1. Show that \(T\) is a linear transformation.
  2. Find the matrix for \(T\) relative to the basis \(\left\{ {1,t,{t^2},{t^3}} \right\}\)for \({{\rm P}_3}\)and the standard basis for \({\mathbb{R}^4}\).

A common misconception is that if \(A\) has a strictly dominant eigenvalue, then, for any sufficiently large value of \(k\), the vector \({A^k}{\bf{x}}\) is approximately equal to an eigenvector of \(A\). For the three matrices below, study what happens to \({A^k}{\bf{x}}\) when \({\bf{x = }}\left( {{\bf{.5,}}{\bf{.5}}} \right)\), and try to draw general conclusions (for a \({\bf{2 \times 2}}\) matrix).

a. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{{\bf{.8}}}&{\bf{0}}\\{\bf{0}}&{{\bf{.2}}}\end{aligned}} \right)\) b. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}&{\bf{0}}\\{\bf{0}}&{{\bf{.8}}}\end{aligned}} \right)\) c. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{8}}&{\bf{0}}\\{\bf{0}}&{\bf{2}}\end{aligned}} \right)\)

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).)
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