Chapter 3.5, Problem 26PS

To determine

D so that $\left[\begin{array}{l}ab\\ cd\end{array}\right]$has rank 1. Also, find basis for the row space and nullspace and show that they are perpendicular.

$d=\frac{dc}{a}$so that $\left[\begin{array}{l}ab\\ cd\end{array}\right]$has rank 1

Row space basis $\left(a,b\right)$and nullspace basis $\left(-b,a\right)$are perpendicular.

Given:

a, b, c are given with $a\ne 0$

Calculation:

In order to have rank 1, it is important that the first row would be non-zero and the second row would be multiple of the first row. Therefore, $d=\frac{dc}{a}$for $\left[\begin{array}{l}ab\\ cd\end{array}\right]$ to have rank 1.

It can be noted that the row space and the the nullspace will have dimension as 1. Thus, the first row $\left(a,b\right)$forms the basis of the row space and the nullspace would be then generated by $\left(-b,a\right)$

Multiply the vectors $\left(a,b\right)$and $\left(-b,a\right)$

$\left(a,b\right)·\left(-b,a\right)=-ab+ba\Rightarrow 0$

Therefore, the dot product of these two vectors is zero, so the row space basis $\left(a,b\right)$and nullspace basis $\left(-b,a\right)$are perpendicular.