\[ \begin{bmatrix} 2 \\ 4 \\ 0 \end{bmatrix} , \begin{bmatrix} -1\\ 6 \\ 2 \end{bmatrix} , \begin{bmatrix} 1 \\ 5 \\ -3 \end{bmatrix} , \begin{bmatrix} 7 \\ 2 \\ 3 \end{bmatrix} \]
The question aims to find the dimension of the subspace spanned by the given column vectors.
The background concepts needed for this question include the column space of the vector, the row-reduced echelon form of the matrix, and the dimension of the vector.
Expert Answer
The dimension of the subspace spanned by the column vectors can be found by making a combined matrix of all these column matrices, then finding the row-reduced echelon form to find the dimension of the subspace of these given vectors.
The combined matrix $A$ with these column vectors is given as:
\[ \begin{bmatrix} 2 & -1 & 1 & 7 \\ 4 & 6 & 5 & 2 \\ 0 & 2 & -3 & 3 \end{bmatrix} \]
The row-reduced echelon form of the matrix $A$ is given as:
\[ R_1 = \dfrac{R_2}{2} \]
\[ \begin{bmatrix} 1 & -1/2 & 1/2 & 7/2 \\ 4 & 6 & 5 & 2 \\ 0 & 2 & -3 & 3 \end{bmatrix} \]
\[ R_2 = R_2\ -\ 4R_1 \]
\[ \begin{bmatrix} 1 & -1/2 & 1/2 & 7/2 \\ 0 & 8 & 3 & -12 \\ 0 & 2 & -3 & 3 \end{bmatrix} \]
\[ R_2 = \dfrac{R_2}{8} \]
\[ \begin{bmatrix} 1 & -1/2 & 1/2 & 7/2 \\ 0 & 1 & 3/8 & -3/2 \\ 0 & 2 & -3 & 3 \end{bmatrix} \]
\[ R_1 = R_1 + \dfrac{R_2}{2} \]
\[ \begin{bmatrix} 1 & 0 & 11/16 & 11/4 \\ 0 & 1 & 3/8 & -3/2 \\ 0 & 2 & -3 & 3 \end{bmatrix} \]
\[ R_3 = R_3\ -\ 2R_2 \]
\[ \begin{bmatrix} 1 & 0 & 11/16 & 11/4 \\ 0 & 1 & 3/8 & -3/2 \\ 0 & 0 & -15/4 & 6 \end{bmatrix} \]
\[ R_3 = – \dfrac{4R_3}{15} \]
\[ \begin{bmatrix} 1 & 0 & 11/16 & 11/4 \\ 0 & 1 & 3/8 & -3/2 \\ 0 & 0 & 1 & -8/5 \end{bmatrix} \]
\[ R_1 = R_1\ -\ \dfrac{11R_3}{16} \]
\[ \begin{bmatrix} 1 & 0 & 0 & 77/20 \\ 0 & 1 & 3/8 & -3/2 \\ 0 & 0 & 1 & -8/5 \end{bmatrix} \]
\[ R_2 =R_2\ -\ \dfrac{3R_3}{8} \]
\[ \begin{bmatrix} 1 & 0 & 0 & 77/20 \\ 0 & 1 & 0 & -9/10 \\ 0 & 0 & 1 & -8/5 \end{bmatrix} \]
Numerical Result:
The pivot columns of the row-reduced echelon form of matrix $A$ is the dimension of the subspace spanned by these vectors, which is $3$.
Example
Find the dimension of the subspace spanned by the given matrix which consists of $3$ vectors expressed as columns of the vector. The matrix is given as:
\[ \begin{bmatrix} 1 & -1 & 1 \\ 2 & 3 & 5 \end{bmatrix} \]
The row-reduced echelon form of the matrix $A$ is given as:
\[ R_2 = R_2\ -\ 2R_1 \longrightarrow R_2 = \dfrac{R_2}{5} \longrightarrow R_1 = R_1 + R_2 \]
\[ \begin{bmatrix} 1 & 0 & 8/5 \\ 0 & 1 & 3/5 \end{bmatrix} \]
There are only $2$ pivot columns in the row-reduced echelon form of the matrix $A$. Therefore, the dimension of the subspace spanned by these vectors is $2$.