Advanced Linear Algebra: Are you ready?

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*}

Relevance

This question checks whether you know how to find the eigenvalues and eigenvectors of a small matriix. It makes the important connection with diagonalization of matrices: Finding a nonsingular matrix \(X \) such that \(A = X \Lambda X^{-1} \text{,}\) where \(\Lambda\) is a diagonal matrix. It finally illustrates how this factorization of a matrix facilitates computing \(A^k \text{.}\)

These insights are important in practice: It says that if you look at a linear algebra problem in the right basis, the the matrix becomes a diagonal matrix. It is much easier to compute with a diagonal matrix and often this way of viewing the problem gives insight into the underlying mechanism being studied. It is also an important tool when developing practical algorithms for computing eigenvalues and eigenvectors, which are topics discussed in a typical numerical linear algebra course.

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