Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 20E from Chapter 4.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 20E
Chapter:
Problem:
(a) find k such that Nul A is a subspace of ℝk, and (b) find k such that Col A is a subspace of ℝk. A = [1 -3 9 0 -5]
Step-by-Step Solution
Given Information
We are given with following Matrix: \[ A = \left[ \begin{array} { l l l l l } { 1 } & { - 3 } & { 9 } & { 0 } & { - 5 } \end{array} \right] \] We have to find $k$ such that Nul A is a subspace of $R^k$ and Col A is a subspace of $R^k$
Step-1: (a)
The matrix A is of order $1 \times 5$. Hence, the Null space of the matrix is a subspace of $R^5$
The k is equal to 5
Step-2: (b)
The matrix A is of order $1 \times 5$. Hence, the Column space of the matrix is a subspace of $R^1$
The k is equal to 1
We are given with following Matrix: \[ A = \left[ \begin{array} { l l l l l } { 1 } & { - 3 } & { 9 } & { 0 } & { - 5 } \end{array} \right] \] We have to find $k$ such that Nul A is a subspace of $R^k$ and Col A is a subspace of $R^k$
Step-1: (a)
The matrix A is of order $1 \times 5$. Hence, the Null space of the matrix is a subspace of $R^5$
The k is equal to 5
Step-2: (b)
The matrix A is of order $1 \times 5$. Hence, the Column space of the matrix is a subspace of $R^1$
The k is equal to 1