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Subsection 2.12 Question

Question 12.

Consider \(A = \left(\begin{array}{r r r} 7 \amp -10 \\ 5 \amp -8 \end{array}\right) \text{.}\)

  • Compute the eigenvalues and corresponding eigenvalues of \(A \text{.}\) (In other words, compute eigenpairs.)

  • Give the matrix \(X \) that diagonalizes \(A \text{:}\)

    \begin{equation*} \begin{array}[t]{c} \underbrace{ \left( \begin{array}{c c c} ~~~~ \amp ~~~~\\ ~~~~\amp ~~~~ \end{array} \right) } \\ X \end{array} ^{-1} \left(\begin{array}{r r r} 7 \amp -10 \\ 5 \amp -8 \end{array}\right) \begin{array}[t]{c} \underbrace{ \left( \begin{array}{c c c} ~~~~ \amp ~~~~ \\ ~~~~ \amp ~~~~ \end{array} \right) } \\ X \end{array} =\begin{array}[t]{c} \underbrace{ \left( \begin{array}{c c c} ~~~~\amp 0 \\ 0 \amp ~~~~ \end{array} \right) } \\ D \end{array} \end{equation*}

  • Compute \(X^{-1} A^4 X \text{.}\)

Solution

Solution

  • Compute the eigenvalues and corresponding eigenvalues of \(A \text{.}\) (In other words, compute eigenpairs.)

    \begin{equation*} \begin{array}{rcl} {\rm det}(A- \lambda I) \amp=\amp {\rm det}(\left(\begin{array}{cc} 7-\lambda \amp -10 \\ 5 \amp -8-\lambda \end{array}\right)) = ( 7 - \lambda)( -8 - \lambda) - (5)(-10)\\ \amp =\amp -56 + \lambda +\lambda^2 +50 = \lambda^2 + \lambda -6 = (\lambda -2)( \lambda+3) \end{array} \end{equation*}

    So, the eigenvalues are \(2 \) and \(-3 \text{.}\)

    • \(\lambda = 2 \text{:}\)

      \begin{equation*} \left(\begin{array}{cc} 7-2 \amp -10 \\ 5 \amp -8-2 \end{array}\right) \left( \begin{array}{c} \chi_0 \\ \chi_1 \end{array}\right) = \left( \begin{array}{c} 0 \\ 0 \end{array}\right) \end{equation*}

      or, equivalently,

      \begin{equation*} \left(\begin{array}{r r r} 5 \amp -10 \\ 5 \amp -10 \end{array}\right) \left( \begin{array}{c} \chi_0 \\ \chi_1 \end{array}\right) = \left( \begin{array}{c} 0 \\ 0 \end{array}\right) \end{equation*}

      By examination, an eigenvector equals

      \begin{equation*} \left( \begin{array}{c} 2 \\ 1 \end{array}\right) \end{equation*}

    • \(\lambda = -3 \text{:}\)

      \begin{equation*} \left(\begin{array}{cc} 7-(-3) \amp -10 \\ 5 \amp -8-(-3) \end{array}\right) \left( \begin{array}{c} \chi_0 \\ \chi_1 \end{array}\right) = \left( \begin{array}{c} 0 \\ 0 \end{array}\right) \end{equation*}

      or, equivalently

      \begin{equation*} \left(\begin{array}{cc} 10 \amp -10 \\ 5 \amp -5 \end{array}\right) \left( \begin{array}{c} \chi_0 \\ \chi_1 \end{array}\right) = \left( \begin{array}{c} 0 \\ 0 \end{array}\right) \end{equation*}

      By examination, an eigenvector equals

      \begin{equation*} \left( \begin{array}{c} 1 \\ 1 \end{array}\right) \end{equation*}

  • Give the matrix \(X \) that diagonalizes \(A \text{:}\)

    \begin{equation*} \begin{array}[t]{c} \underbrace{ \left( \begin{array}{rr} 1\amp1\\ -1\amp1 \end{array} \right) } \\ X \end{array} ^{-1} \left(\begin{array}{r r } 7 \amp -10 \\ 5 \amp -8 \end{array}\right) \begin{array}[t]{c} \underbrace{ \left( \begin{array}{rr} 1\amp1\\ -1\amp1 \end{array} \right) } \\ X \end{array} = \left( \begin{array}{rr} 2 \amp 0 \\ 0 \amp -3 \end{array} \right) \end{equation*}

    Notice: answer is non-unique and is likely consistent with answer to previous part.

  • Compute \(X^{-1} A^4 X \text{.}\)

    \begin{equation*} \begin{array}{rcl} X^{-1} A^4 X \amp = \amp X^{-1} ( X D X^{-1} )^4 X \\ \amp = \amp X^{-1} ( X D X^{-1}) ( X D X^{-1}) ( X D X^{-1} )( X D X^{-1} ) X \\ \amp = \amp (X^{-1} X) D (X^{-1} X) D (X^{-1} X) D (X^{-1} X) D (X^{-1} X) \\ \amp = \amp D^4 = \left( \begin{array}{rr} 2\amp-0\\ 0\amp-3 \end{array} \right) ^4 = \left( \begin{array}{rr} 16\amp 0\\ 0\amp 81 \end{array} \right) . \end{array} \end{equation*}