Logo

AskSia

Plus

2. ( 8pts)8 \mathrm{pts}) Consider the following matrix: \[ A=\left(\begin{array}...
Apr 21, 2024
2. ( 8pts)8 \mathrm{pts}) Consider the following matrix: A=(2amp;−4amp;5amp;8amp;51amp;−2amp;2amp;3amp;14amp;−8amp;3amp;2amp;1). A=\left(\begin{array}{ccccc} 2 & -4 & 5 & 8 & 5 \\ 1 & -2 & 2 & 3 & 1 \\ 4 & -8 & 3 & 2 & 1 \end{array}\right) . Find bases for a row space, a column space, and a null space of AA, respectively.
Solution by Steps
step 1
To find a basis for the row space of matrix AA, we perform row reduction to echelon form
step 2
The row-reduced echelon form of AA is (1amp;−2amp;0amp;−1amp;00amp;0amp;1amp;2amp;00amp;0amp;0amp;0amp;1)\begin{pmatrix} 1 & -2 & 0 & -1 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}. The non-zero rows form a basis for the row space
step 3
Thus, the basis for the row space is {(1−20−10),(00120),(00001)}\left\{ \begin{pmatrix} 1 \\ -2 \\ 0 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 2 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}
step 4
To find a basis for the column space of AA, we look at the pivot columns in the original matrix corresponding to the pivot rows in the row-reduced echelon form
step 5
The pivot columns in the original matrix AA are the 1st, 3rd, and 5th columns
step 6
Thus, the basis for the column space is {(214),(523),(511)}\left\{ \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix}, \begin{pmatrix} 5 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} 5 \\ 1 \\ 1 \end{pmatrix} \right\}
step 7
To find a basis for the null space of AA, we solve the homogeneous system Ax=0Ax = 0
step 8
The null space vectors are given by the general solution (2x+y,x,−2y,y,0)(2x + y, x, -2y, y, 0), where xx and yy are free variables
step 9
Two linearly independent vectors that span the null space are (2,1,0,0,0)(2, 1, 0, 0, 0) and (1,0,−2,1,0)(1, 0, -2, 1, 0)
step 10
Thus, the basis for the null space is {(21000),(10−210)}\left\{ \begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ -2 \\ 1 \\ 0 \end{pmatrix} \right\}
Answer
Row space basis: {(1−20−10),(00120),(00001)}\left\{ \begin{pmatrix} 1 \\ -2 \\ 0 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1 \\ 2 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \right\}, Column space basis: {(214),(523),(511)}\left\{ \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix}, \begin{pmatrix} 5 \\ 2 \\ 3 \end{pmatrix}, \begin{pmatrix} 5 \\ 1 \\ 1 \end{pmatrix} \right\}, Null space basis: {(21000),(10−210)}\left\{ \begin{pmatrix} 2 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ -2 \\ 1 \\ 0 \end{pmatrix} \right\}
Key Concept
Basis for Row Space, Column Space, and Null Space
Explanation
The basis for the row space is found from the non-zero rows of the row-reduced echelon form of AA. The basis for the column space is the set of original matrix columns corresponding to the pivot positions in the row-reduced form. The basis for the null space is found by solving the homogeneous equation Ax=0Ax = 0 and extracting linearly independent vectors that span the solution space.
© 2023 AskSia.AI all rights reserved