Advanced Linear Algebra: Are you ready?

Solution

• \begin{equation*} \begin{array}{r r r r} 2 \chi_0 + \amp \chi_1 + \amp \chi_2 = \amp 1 \\ -2 \chi_0 - \amp 2 \chi_1 + \amp \chi_2 = \amp -4 \\ 4 \chi_0 + \amp 4 \chi_1 + \amp \chi_2 = \amp 2 \\ \end{array} \end{equation*}

  can be represented by the appended system

  \begin{equation*} \left( \begin{array}{r r r | r} 2 \amp 1 \amp \phantom{-}1 \amp 1 \\ -2 \amp -2 \amp 1 \amp -4 \\ 4 \amp 4 \amp 1 \amp 2 \\ \end{array} \right). \end{equation*}

  Subtracting \(-1 \) times the first row from the second row and subtracting \(2 \) times the first row from the third row yields

  \begin{equation*} \left( \begin{array}{r r r | r} 2 \amp 1 \amp 1 \amp 1 \\ 0 \amp -1 \amp 2 \amp -3 \\ 0 \amp 2 \amp -1 \amp 0 \\ \end{array} \right). \end{equation*}

  Subtracting \(-2 \) times the second row from the third row yields

  \begin{equation*} \left( \begin{array}{r r r | r} 2 \amp 1 \amp 1 \amp 1 \\ 0 \amp -1 \amp 2 \amp -3 \\ 0 \amp 0 \amp 3 \amp -6 \\ \end{array} \right). \end{equation*}

  This translates to

  \begin{equation*} \begin{array}{r r r r} 2 \chi_0 + \amp \chi_1 + \amp \chi_2 = \amp 1 \\ \amp - \chi_1 + \amp 2 \chi_2 = \amp -3 \\ \amp \amp 3 \chi_2 = \amp -6 \end{array} \end{equation*}

  Solving this upper triangular system yields \(\chi_2 = -2 \text{,}\) \(\chi_1 = -1 \text{,}\) and \(\chi_0 = 2 \text{.}\)

• The LU factorization can be read off from the above "Gaussian elimination" process for reduction the system of linear equations to an upper triangular system of linear equations:

  \begin{equation*} \begin{array}[t]{c} \underbrace{ \left( \begin{array}{r r} 2 \amp 1 \amp 1 \\ -2 \amp -2 \amp 1 \\ 4 \amp 4 \amp 1 \end{array} \right) } \\ A \end{array} = \begin{array}[t]{c} \underbrace{ \left( \begin{array}{r r} 1 \amp 0 \amp 0 \\ -1 \amp 1 \amp 0 \\ 2 \amp -2 \amp 1 \end{array} \right) } \\ L \end{array} \begin{array}[t]{c} \underbrace{ \left( \begin{array}{r r} 2 \amp 1 \amp 1 \\ 0 \amp -1 \amp 2 \\ 0 \amp 0 \amp 3 \end{array} \right) } \\ U \end{array} \end{equation*}

• If \(A = L U \) then

  \begin{equation*} A x = b \end{equation*}

  can be rewritten as

  \begin{equation*} ( L U ) x = b . \end{equation*}

  Associativity of multiplication yields

  \begin{equation*} L ( U x )x = b . \end{equation*}

  If we introduce the variable \(z = U x \) we get that

  \begin{equation*} L z = b. \end{equation*}

  Thus, we can first solve the lower triangular system \(L z = b \) and then \(U x = z \text{,}\) which yields the desired solution \(x \text{.}\)

• TRUE

  We saw in the first part of this question that

  \begin{equation*} x = \left( \begin{array}{r} 2 \\ -1 \\ -2 \end{array} \right) \end{equation*}

  solves \(A x = b \text{.}\) Hence \(b \) can be written as a linear combination of the columns of \(A \text{:}\)

  \begin{equation*} \left( \begin{array}{r} 1 \\ -4 \\ 2 \end{array} \right) = (2) \left( \begin{array}{r} 2 \\ -2 \\ 4 \end{array} \right) + (-1) \left( \begin{array}{r} 1 \\ -2 \\ 4 \end{array} \right) + (-2) \left( \begin{array}{r} 1 \\ 1 \\ 1 \end{array} \right) \end{equation*}

  which means that \(b \) is in the column space of \(A \text{.}\)

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