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 \(A{\mathop{\rm v}\nolimits} = 0\) because A is not invertible. Put \(n\) copies of vector \({\mathop{\rm v}\nolimits} \) in a \(n \times n\) matrix B. It gives

\(\begin{array}{c}AB = A\left( {\begin{array}{*{20}{c}}{\mathop{\rm v}\nolimits} & \ldots &{\mathop{\rm v}\nolimits} \end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{A{\mathop{\rm v}\nolimits} }& \ldots &{A{\mathop{\rm v}\nolimits} }\end{array}} \right)\\ = 0.\end{array}\)

Thus, it is shown that \(AB = 0\).