Math 220 Quiz Solution

March 6, 2008

The Question

Which sets of column indices (which are integers from $1$ to $4$) correspond to maximal linearly independent subsets of the set of columns of the matrix $$M=\left(\begin{array}{cccc}\hfill 10& \hfill 0& \hfill 1& \hfill -14\\ \hfill -5& \hfill -1& \hfill 0& \hfill 7\\ \hfill 15& \hfill 0& \hfill 0& \hfill -21\end{array}\right)$$

Systematic Response

The RREF of $M$ is found to be: $$\left(\begin{array}{cccc}\hfill 1& \hfill 0& \hfill 0& \hfill -7⁄5\\ \hfill 0& \hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1& \hfill 0\end{array}\right)$$

$Mx=0$ occurs when $$\begin{array}{cccc}& \hfill x_1-\left(7⁄5\right)x_4& \hfill =\hfill & 0\hfill \\ & \hfill x_2& \hfill =\hfill & 0\hfill \\ & \hfill x_3& \hfill =\hfill & 0\hfill \end{array}$$ or $$\left(\begin{array}{c}\hfill x_1\\ \hfill x_2\\ \hfill x_3\\ \hfill x_4\end{array}\right)=t\left(\begin{array}{c}\hfill 7⁄5\\ \hfill 0\\ \hfill 0\\ \hfill 1\end{array}\right)\text{for some}t\text{.}$$ Hence, $$\frac{7}{5}{M}_1+M_4=0$$ is, up to a scalar multiple, the only linear relation among the columns. It reflects the parallelism of column $1$ and column $4$. Thus, subsets of the set of column indices corresponding to maximal linearly independent sets of columns are $$\left\{1,2,3\right\}\text{and}\left\{2,3,4\right\}\text{.}$$

Remark. The preceding also shows that the vector $\left(7⁄5,0,0,1\right)$ is a basis of the kernel of the linear map $f_M:{\mathbf{R}}^4\to {\mathbf{R}}^3$.

An Alternative: “Winging It”

This method avoids all tedious calculation, but it's only a reasonable approach in special circumstances. Here special circumstances are (1) the appearance of vectors on coordinate axes in columns $2$ and $3$ and (2) the nearly obvious fact that columns $1$ and $4$ are parallel to each other.

Columns $2$ and $3$ span the plane in ${\mathbf{R}}^3$ containing the first two axes, and, therefore, they form a basis of that plane. Neither column $1$ nor column $4$ is in the plane determined by the first two axes. Hence, the matrix has rank $3$. Moreover, $\left\{1,2,3\right\}$ and $\left\{2,3,4\right\}$ are index sets for maximal linearly independent sets of columns. There are only two other index sets of size $3$ — the size required for a maximal linearly independent subset of the set of columns in a rank $3$ matrix —, and both of the other sets contain both of the two parallel columns.