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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

Expert verified

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

Step by step solution

01

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

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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 \({P^{ - 1}}\left( {PA} \right)\).)

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