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.

Relevance

Many questions in linear algebra have an infinite number of answers. For example, the answer to "for which right-hand side vectors \(b \) does \(A x = b \) have a solution?" typically is a set of vectors that has not only an infinite number of members, but an uncountable number of members. Describing that set with only a finite number of vector (the basis for the subspace that equals the set) allows one to concisely answer various questions. Thus, the notions vector subspace, basis for the subspace, and dimension of the subspace are of central importance to linear algebra.

Another imporant concept touched upon in this question is the concept of the four fundamental spaces: the column, null, row, and left-null spaces. (The question refers to three out of four of these.) This is one of the most important concepts to master before you start an advanced course in linear algebra, since it allows one to more abstractly understand when a system of linear equations has no solution, a unique solution, or an infinite number of solutions.

The process that reduces an appended matrix to row echelon form is actually not of much practical value. The reason is that when computer arithmetic is employed, roundoff error is incurred and hence the notion of a zero, central to interpreting the row echelon form, is fuzzy. In numerical linear algebra, we will employ other techiques for identifying bases for the four fundamental subspace.

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