Linear Systems of Algebraic Equations

This page presents some topics from Linear Algebra needed for construction of solutions to systems of linear algebraic equations and some applications. We use matrices and vectors as essential elements in obtaining and expressing the solutions.

Row Space and Column Space

If A is an \( m\times n \) matrix, then the subspace of \( \mathbb{R}^n \) spanned by the row vectors of A is called the row space of A, and the subspace of \( \mathbb{R}^m \) spanned by the column vectors of A is called the column space of A or the range of the matrix. The row space of A is the column space of the adjoint matrix \( {\bf A}^{\ast} = \overline{\bf A}^{\mathrm T} = \overline{{\bf A}^{\mathrm T}} ,\) which is a subspace of \( \mathbb{R}^n .\)

Theorem: Elementary row operations do not change the row space of a matrix.

Theorem: The row space and the column space of a matrix have the same dimension.

Theorem: If a matrix R is in row echelon (Gaussian) form, then the row vectors with the leading coefficients (that are usually chosen as 1's) form a basis for the row space of R, and the column vectors corresponding to the leading coefficients form a basis for the column space.

A Basis for the Range: Let \( L:\,\mathbb{R}^n\, \mapsto \, \mathbb{R}^m \) is a linear transformation given by \( L({\bf x}) = {\bf A}\,{\bf x} , \) for some \( m \times n \) matrix A. To find a basis for the range of L, perform the following steps.
Step 1: Find B, the reduced row echelon form of A.
Step 2: Form the set of those columns of A whose corresponding columns in B have nonzero pivots.
The set of vectors in Step 2 is a basis for the range of L. ■

Example. Find a basis for the row space and for column space of the matrix

\[ {\bf A} = \left[ \begin{array}{cccccc} 1 &-3&4&-2&5&4 \\ 2 &-6&9 &-1&8&2 \\ 2&-6&9 &-1&9&7 \\ -1&3&-4&2&-5&-4 \end{array} \right] . \]

Since elementary row operations do not change the row space of a matrix, we can find a basis for the row space of A by finding a basis for the row space of any row echelon form of A. Reducing A to Gaussian form, we obtain

\[ {\bf A} \, \sim \, {\bf R} = \left[ \begin{array}{cccccc} 1 &-3&4&-2&5&4 \\ 0 &0&1 &3&-2&-6 \\ 0&0&0 &0&1&5 \\ 0&0&0&0&0&0 \end{array} \right] . \]

According to the above theorem, the nonzero row vectors of R form a basis for the row space of R and hence form a basis for the row space of A. These basis vectors are

\begin{align*} {\bf r}_1 &= \left[ 1, \ -3, \ 4,\ -2, \ 5, \ 4 \right] , \\ {\bf r}_2 &= \left[ 0, \ 0, \ 1,\ 3, \ -2, \ -6 \right] , \\ {\bf r}_3 &= \left[ 0, \ 0, \ 0,\ 0, \ 1, \ 5 \right] \end{align*}

Keeping in mind that A and R can have different column spaces, we cannot find a basis for the column space of A directly from the column vectors of R. However, we can determing them by identifying the corresponding column vectors of A that form a basis for the column space. Since the first, third, and fifth columns of R contain the leading 1's of the row vectors, we claim that the vectors in these columns form the required basis:

\[ {\bf c}_1 = \left[ \begin{array}{c} 1 \\ 2 \\ 2 \\ -1 \end{array} \right] , \qquad {\bf c}_3 = \left[ \begin{array}{c} 4 \\ 9 \\ 9 \\ -4 \end{array} \right] , \qquad {\bf c}_5 = \left[ \begin{array}{c} 5 \\ 8 \\ 9 \\ -5 \end{array} \right] . \]