t Let A be a 12x9 matrix whose transpose A has bullity 4. What is the maximum number of linearly independent vectors in the column space of A?
t Let A be a 12x9 matrix whose transpose A has bullity 4. What is the maximum number of linearly independent vectors in the column space of A?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Let \( A \) be a \( 12 \times 9 \) matrix whose transpose \( A^T \) has nullity 4. What is the maximum number of linearly independent vectors in the column space of \( A \)?
**Explanation:**
To solve this problem, we need to use the fundamental theorem of linear algebra, which relates the rank and nullity of a matrix.
1. We are given that \( \text{nullity}(A^T) = 4 \).
2. The nullity of a matrix \( A \) is the dimension of the null space of \( A \), which is the number of linearly independent solutions to the equation \( A\mathbf{x} = 0 \).
For the transpose of the matrix:
\[ \text{nullity}(A^T) + \text{rank}(A^T) = \text{number of columns in } A^T \]
Since \( A \) is a \( 12 \times 9 \) matrix:
- \( A^T \) is a \( 9 \times 12 \) matrix.
- Therefore, the number of columns in \( A^T \) is 12.
So,
\[ 4 + \text{rank}(A^T) = 12 \]
Thus,
\[ \text{rank}(A^T) = 8 \]
The rank of \( A^T \) is the same as the rank of \( A \), which means:
\[ \text{rank}(A) = 8 \]
The rank of a matrix is the dimension of the column space of the matrix, which is the maximum number of linearly independent vectors in the column space.
**Answer:**
The maximum number of linearly independent vectors in the column space of \( A \) is 8.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdc37c896-f9bf-421c-983e-bcaad18451c5%2Fa134dde1-9702-41ea-ab39-08a5eeae6337%2Fu1tjlj_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Let \( A \) be a \( 12 \times 9 \) matrix whose transpose \( A^T \) has nullity 4. What is the maximum number of linearly independent vectors in the column space of \( A \)?
**Explanation:**
To solve this problem, we need to use the fundamental theorem of linear algebra, which relates the rank and nullity of a matrix.
1. We are given that \( \text{nullity}(A^T) = 4 \).
2. The nullity of a matrix \( A \) is the dimension of the null space of \( A \), which is the number of linearly independent solutions to the equation \( A\mathbf{x} = 0 \).
For the transpose of the matrix:
\[ \text{nullity}(A^T) + \text{rank}(A^T) = \text{number of columns in } A^T \]
Since \( A \) is a \( 12 \times 9 \) matrix:
- \( A^T \) is a \( 9 \times 12 \) matrix.
- Therefore, the number of columns in \( A^T \) is 12.
So,
\[ 4 + \text{rank}(A^T) = 12 \]
Thus,
\[ \text{rank}(A^T) = 8 \]
The rank of \( A^T \) is the same as the rank of \( A \), which means:
\[ \text{rank}(A) = 8 \]
The rank of a matrix is the dimension of the column space of the matrix, which is the maximum number of linearly independent vectors in the column space.
**Answer:**
The maximum number of linearly independent vectors in the column space of \( A \) is 8.
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