Question

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

14. Show that if \(Q\) is an invertible, then \({\mathop{\rm rank}\nolimits} AQ = {\mathop{\rm rank}\nolimits} A\). (Hint: Use Exercise 13 to study \({\mathop{\rm rank}\nolimits} {\left( {AQ} \right)^T}\).)

Short answer

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

Step-by-step solution

State the rank theorem

The rank theoremstates that the dimensions of the column space and the row space of an \(m \times n\) matrix \(A\) are equal. This common dimension, the rank of \(A\), also equals the number of pivot positions in \(A\) and satisfies the equation \({\mathop{\rm rank}\nolimits} A + \dim {\mathop{\rm Nul}\nolimits} A = n\).

Show that if \(Q\) is invertible, then \({\mathop{\rm rank}\nolimits} AQ = {\mathop{\rm rank}\nolimits} A\)

Note that \({\left( {AQ} \right)^T} = {Q^T}{A^T}\).

Suppose \({Q^T}\) is invertible. Use Exercise 13 as shown below:

\(\begin{array}{c}{\mathop{\rm rank}\nolimits} {\left( {AQ} \right)^T} = {\mathop{\rm rank}\nolimits} {Q^T}{A^T}\\ = {\mathop{\rm rank}\nolimits} {A^T}\end{array}\)

The rank of a matrix and its transpose are equal according to the rank theorem, so \({\mathop{\rm rank}\nolimits} AQ = {\mathop{\rm rank}\nolimits} A\).

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