Question

Let A be an $n\times n$ singular matrix. Describe how to construct an $n\times n$ nonzero matrix B such that $AB=0$.

Short answer

It is shown that $AB=0$.

Step-by-step solution

Explain the construction of a $n\times n$ non-zero matrix

There is a non-zero vector v in ${\mathbb{R}}^n$ such that $A\mathrm{v}=0$ because A is not invertible. Put $n$ copies of vector $\mathrm{v}$ in a $n\times n$ matrix B. It gives

$\begin{array}{c}AB=A\left(\begin{array}{ccc}\mathrm{v}& \dots & \mathrm{v}\end{array}\right)\\ =\left(\begin{array}{ccc}A\mathrm{v}& \dots & A\mathrm{v}\end{array}\right)\\ =0.\end{array}$

Thus, it is shown that $AB=0$.