Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
We have solutions for your book!
See our solution for Question 5E from Chapter 4.6 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 5E
Chapter:
Problem:
If a 3 X 8 matrix A has rank 3, find dim Nul A, dim Row A, and rank AT...
Step-by-Step Solution
Step 1
Given Data:
Size of Matrix: $3 \times 8$
Rank of matrix: $3$
For any $ m \times n$ matrix, the relationship between its rank and number of columns is given by: \[{\rm{Rank}}\left( {\bf{A}} \right) + {\rm{Dim}}\left( {{\rm{Null}}{\bf{A}}} \right) = {\rm{n}}\]So, \[\begin{array}{l}{\rm{Rank}}\left( {\bf{A}} \right) + {\rm{Dim}}\left( {{\rm{Null}}{\bf{A}}} \right) = {\rm{n}}\\\\{\rm{Dim}}\left( {{\rm{Null}}{\bf{A}}} \right) = n - {\rm{Rank}}\left( {\bf{A}} \right)\\\\ = 8 - 3\\\\ = 5\end{array}\]
Step 2
For any matrix, the relationship between its rank and dimensions of rows is given by: \[\begin{array}{l}{\rm{Dim}}\left( {{\rm{Row}}{\bf{A}}} \right) = {\rm{Rank}}\left( {\bf{A}} \right)\\\\{\rm{Dim}}\left( {{\rm{Row}}{\bf{A}}} \right) = 3\end{array}\]
Step 3
For any matrix, the relationship between the rank of its transpose is given by: \[\begin{array}{l}{\rm{Rank}}\left( {{{\bf{A}}^T}} \right) = {\rm{Dim}}\left( {{\rm{Row}}{\bf{A}}} \right)\\\\{\rm{Rank}}\left( {{{\bf{A}}^T}} \right) = 3\end{array}\]
Step 4: ANSWERS
\[\begin{array}{l}{\rm{Dim}}\left( {{\rm{Null}}{\bf{A}}} \right) = 5\\\\{\rm{Dim}}\left( {{\rm{Row}}{\bf{A}}} \right) = 3\\\\{\rm{Rank}}\left( {{{\bf{A}}^T}} \right) = 3\end{array}\]
Given Data:
Size of Matrix: $3 \times 8$
Rank of matrix: $3$
For any $ m \times n$ matrix, the relationship between its rank and number of columns is given by: \[{\rm{Rank}}\left( {\bf{A}} \right) + {\rm{Dim}}\left( {{\rm{Null}}{\bf{A}}} \right) = {\rm{n}}\]So, \[\begin{array}{l}{\rm{Rank}}\left( {\bf{A}} \right) + {\rm{Dim}}\left( {{\rm{Null}}{\bf{A}}} \right) = {\rm{n}}\\\\{\rm{Dim}}\left( {{\rm{Null}}{\bf{A}}} \right) = n - {\rm{Rank}}\left( {\bf{A}} \right)\\\\ = 8 - 3\\\\ = 5\end{array}\]
Step 2
For any matrix, the relationship between its rank and dimensions of rows is given by: \[\begin{array}{l}{\rm{Dim}}\left( {{\rm{Row}}{\bf{A}}} \right) = {\rm{Rank}}\left( {\bf{A}} \right)\\\\{\rm{Dim}}\left( {{\rm{Row}}{\bf{A}}} \right) = 3\end{array}\]
Step 3
For any matrix, the relationship between the rank of its transpose is given by: \[\begin{array}{l}{\rm{Rank}}\left( {{{\bf{A}}^T}} \right) = {\rm{Dim}}\left( {{\rm{Row}}{\bf{A}}} \right)\\\\{\rm{Rank}}\left( {{{\bf{A}}^T}} \right) = 3\end{array}\]
Step 4: ANSWERS
\[\begin{array}{l}{\rm{Dim}}\left( {{\rm{Null}}{\bf{A}}} \right) = 5\\\\{\rm{Dim}}\left( {{\rm{Row}}{\bf{A}}} \right) = 3\\\\{\rm{Rank}}\left( {{{\bf{A}}^T}} \right) = 3\end{array}\]