Definition 5.1.1.
Let \(A\) be an \(m \times n\) matrix. The rank of \(A\) is the number of leading ones in its reduced row-echelon form.
Section 5.1 Dimensions of Spans
Subsection 5.1.1 Rank
Let me remind you of a definition from Section 4.2 since it will be useful in this section.
Subsection 5.1.2 Dimensions of Spans
I can now present techniques to solve the dimension problems presented in Section 3.1 and Section 3.2. For loci and spans, I didn’t have a way of determining what information was redundant, hence what the dimensions of the objects should be. Row-reduction of matrices solves this problem.
In this section, I’ll present the solution for spans; details for loci follow in Section 5.3. Given some vectors \(\{v_1, v_2, \ldots, v_k\}\text{,}\) what is the dimension of the span? By definition, it is the number of linearly independent vectors in the set. Now I can test for linearly independent vectors.
Proposition 5.1.2.
Let \(\{v_1, v_2, \ldots, v_k\}\) be a set of vectors in \(\RR^n\text{.}\) If I make these vectors the rows of a matrix \(A\text{,}\) then the rank of the matrix \(A\) is the number of linearly independent vectors in the set. Moreover, if I do the row reduction without exchanging rows (which is always possible), then the vectors which correspond to rows with leading ones in the reduced row-echelon will form a maximal linearly independent set, i.e., a basis. Any vector corresponding to a row without a leading one is a redundant vector in the span. The set is linearly independent if and only if the rank of \(A\) is \(k\text{.}\)
Subsection 5.1.3
Throughout this section, all row reductions were calculated by Wolfram Alpha.
Example 5.1.3.
Find the dimension of this span.
\begin{equation*}
\Span \left\{
\begin{pmatrix}
5 \\ -1 \\ 0
\end{pmatrix},
\begin{pmatrix}
-2 \\ 6 \\ 1
\end{pmatrix},
\begin{pmatrix}
3 \\ 4 \\ 1
\end{pmatrix}
\right\}
\end{equation*}
To find the dimensions of the span, I put the vectors of the span into a matrix (as rows, not columns).
\begin{equation*}
\begin{pmatrix}
5 \amp -1 \amp 0 \\
-2 \amp 6 \amp 1 \\
3 \amp 4 \amp 1
\end{pmatrix}
\end{equation*}
This row reduces to the following matrix.
\begin{equation*}
\begin{pmatrix}
1 \amp 0 \amp 0 \\
0 \amp 1 \amp 0 \\
0 \amp 0 \amp 1
\end{pmatrix}
\end{equation*}
Since there are three leading ones, all three vectors are linearly independent. The rank of the matrix is three, and the dimension of the span is three.
Example 5.1.4.
Find the dimension of this span.
\begin{equation*}
\Span \left\{
\begin{pmatrix}
7 \\ -2 \\ -3
\end{pmatrix},
\begin{pmatrix}
2 \\ -2 \\ 5
\end{pmatrix},
\begin{pmatrix}
1 \\ 4 \\ -18
\end{pmatrix}
\right\}
\end{equation*}
To find the dimension of the span, I put the vectors of the span into a matrix (as rows, not columns).
\begin{equation*}
\begin{pmatrix}
7 \amp -2 \amp 3 \\
2 \amp -2 \amp 5 \\
1 \amp 4 \amp -12
\end{pmatrix}
\end{equation*}
This row reduces to the following matrix.
\begin{equation*}
\begin{pmatrix}
1 \amp 0 \amp \frac{-2}{5} \\
0 \amp 1 \amp \frac{-29}{10} \\
0 \amp 0 \amp 0
\end{pmatrix}
\end{equation*}
There are only two leading ones, so only two of the original vectors were linearly independent. The rank is two and the dimension of the span is two.
Example 5.1.5.
Find the dimension of this span.
\begin{equation*}
\Span \left\{
\begin{pmatrix}
4 \\ -1 \\ 0 \\ 2 \\ -8 \\ 3
\end{pmatrix},
\begin{pmatrix}
12 \\ 4 \\ -3 \\ 7 \\ 15 \\ 1
\end{pmatrix},
\begin{pmatrix}
0 \\ 5 \\ -2 \\ -2 \\ -14 \\ 3
\end{pmatrix},
\begin{pmatrix}
16 \\ 8 \\ -5 \\ 7 \\ -9 \\ 7
\end{pmatrix},
\begin{pmatrix}
12 \\ -6 \\ 1 \\ 11 \\ 43 \\ -5
\end{pmatrix},
\begin{pmatrix}
8 \\ 8 \\ -4 \\ 0 \\ -44 \\ 12
\end{pmatrix}
\right\}
\end{equation*}
To find the dimension of the span, I put the vectors of the span into a matrix (as rows, not columns).
\begin{equation*}
\begin{pmatrix}
4 \amp -1 \amp 0 \amp 2 \amp -8 \amp 3 \\
12 \amp 4 \amp -3 \amp 7 \amp 15 \amp 1 \\
0 \amp 5 \amp -2 \amp -2 \amp -14 \amp 3 \\
16 \amp 8 \amp -5 \amp 7 \amp -9 \amp 7 \\
12 \amp -6 \amp 1 \amp 11 \amp 43 \amp -5 \\
8 \amp 8 \amp -4 \amp 0 \amp -44 \amp 12
\end{pmatrix}
\end{equation*}
This row reduces to the following matrix.
\begin{equation*}
\begin{pmatrix}
1 \amp 0 \amp 0 \amp \frac{-3}{2} \amp 0 \amp 7 \\
0 \amp 1 \amp 0 \amp -8 \amp 0 \amp 25 \\
0 \amp 0 \amp 1 \amp -19 \amp 0 \amp 61 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0 \\
0 \amp 0 \amp 0 \amp 0 \amp 0 \amp 0
\end{pmatrix}
\end{equation*}
There are three leading ones, so only three of the original six vectors are linearly independent. The rank of the matrix is three and the dimension of the span is three.
