Question

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

13. Show that if \(P\) is an invertible \(m \times m\) matrix, then rank\(PA\)=rank\(A\).(Hint: Apply Exercise12 to \(PA\) and \({P^{ - 1}}\left( {PA} \right)\).)

Short answer

It is proved that \({\mathop{\rm rank}\nolimits} PA = {\mathop{\rm rank}\nolimits} A\).

Step-by-step solution

Show that \({\mathop{\rm rank}\nolimits} PA = {\mathop{\rm rank}\nolimits} A\)

Exercise 12 states that if \(B\) is \(n \times p\), then rank\(AB \le {\mathop{\rm rank}\nolimits} A\).

According to Exercise 12, \({\mathop{\rm rank}\nolimits} AB \le {\mathop{\rm rank}\nolimits} A\).

\(\begin{array}{c}{\mathop{\rm rank}\nolimits} A = {\mathop{\rm rank}\nolimits} \left( {{P^{ - 1}}P} \right)A\\ = {\mathop{\rm rank}\nolimits} {P^{ - 1}}\left( {PA} \right)\\ \le {\mathop{\rm rank}\nolimits} PA\\ = {\mathop{\rm rank}\nolimits} PA\end{array}\)

Thus, it is proved that \({\mathop{\rm rank}\nolimits} PA = {\mathop{\rm rank}\nolimits} A\).