Advanced Linear Algebra: Are you ready?

Solution

• TRUE: \(A \) has linearly independent columns.

  You may see this immediately, since the second and third columns are equal, or you may deduce this after doing some of the remaining parts of this question.

• Reduce the appended system \(\left( \begin{array}{ c | c} A \amp b \end{array} \right)\) to row echelon form.

  \begin{equation*} \begin{array}{rcl} \left(\begin{array}{r r r | r} -1 \amp -3 \amp -3 \amp -5 \\ -1 \amp 1 \amp 1 \amp 3\\ -2 \amp -2 \amp -2 \amp -2 \end{array}\right) \amp \longrightarrow \amp \left(\begin{array}{r r r | r} -1 \amp -3 \amp -3 \amp -5 \\ 0 \amp 4 \amp 4 \amp 8\\ 0 \amp 4 \amp 4 \amp 8 \end{array}\right) \\ \amp \longrightarrow \amp \left(\begin{array}{r r r | r} -1 \amp -3 \amp -3 \amp -5 \\ 0 \amp 4 \amp 4 \amp 8\\ 0 \amp 0 \amp 0 \amp 0 \end{array}\right) \end{array} \end{equation*}

• In the row echelon form (above): Identify the pivot(s). Identify the free variable(s). Identify the dependent variable(s).

  \begin{equation*} \left(\begin{array}{r r r | r} {\color{red} {-1}} \amp -3 \amp -3 \amp -5 \\ 0 \amp {\color{red} 4} \amp 4 \amp 8\\ 0 \amp 0 \amp 0 \amp 0 \end{array}\right) \end{equation*}

  Pivots are highlighted in red. Free variable: \(\chi_2 \text{.}\) Dependent variables: \(\chi_0 \) and \(\chi_1 \text{.}\)

• Compute a specific solution.

  \begin{equation*} \left(\begin{array}{r r r} {-1} \amp -3 \amp -3 \\ 0 \amp {4} \amp 4 \end{array}\right) \left( \begin{array}{c} ? \\ ?\\ 0 \end{array}\right) = \left(\begin{array}{r} -5 \\ 8\end{array}\right), \end{equation*}

  where \(? \) is to be determined, yields

  • \(\displaystyle \chi_1 = 2 \)

  • \(-\chi_0 - 3 (2) = -5 \) so that \(\chi_0 = -1 \text{.}\)

  Solution:\(\left( \begin{array}{r} -1 \\ 2 \\ 0 \end{array}\right)\)

• Compute a basis for the null space of \(A \text{.}\)

  \begin{equation*} \left(\begin{array}{r r r} {-1} \amp -3 \amp -3 \\ 0 \amp {4} \amp 4 \end{array}\right) \left( \begin{array}{c} ? \\ ? \\ 1 \end{array}\right) = \left(\begin{array}{r} 0 \\ 0\end{array}\right) \end{equation*}

  means

  • \(4 \chi_1 + 4 = 0 \) and hence \(\chi_1 = -1 \text{.}\)

  • \(-\chi_0 - 3 (-1) - 3 = 0 \) so that \(\chi_0 = 0 \text{.}\)

  Solution:

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

  You may have solved this by examination, which is fine.

• TRUE: This matrix singular.

  There are a number of ways of justiifying this answer with what we have answered for this question so far:

  • The number of pivots is less than the number of columns of the matrix.

  • There are free variables.

  • There is a nontrivial vector in the null space.

  • \(A x = 0 \) has a nontrivial solution.

• Give a formula for the general solution.

  \(\left( \begin{array}{c} -1 \\ 2 \\ 0 \end{array}\right) + \beta \left( \begin{array}{c} 0 \\ -1 \\ 1 \end{array}\right)\) (This is only one of many possible answers.)

• Give a basis for the column space of \(A \text{.}\)

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

• Give a basis for the row space of \(A \text{.}\)

  \begin{equation*} \left(\begin{array}{r r r | r} -1 \\ -3 \\ -3 \end{array}\right) , \left(\begin{array}{r r r | r} 0 \\ 4 \\ 4 \end{array}\right). \end{equation*}

  The first two rows of original matrix is also a correct answer in this case, since no pivoting was required while reducing the matrix to row echelon form.

• What is the rank of the matrix?

  \(2 \) (the number of linearly independent columns)..

• Identify two linearly independent vectors that are orthogonal to the null space of \(A \text{.}\)

  The same vectors as are the vectors that are a basis for the row space can be used for this.