Albarellak

2020-12-01

For each of the following matrices, determine a basis for each of the subspaces R(AT ), N(A), R(A), and N(AT ):

$A=\left[\begin{array}{ccc}1& 3& 1\\ 2& 4& 0\end{array}\right]$

Laaibah Pitt

Skilled2020-12-02Added 98 answers

Step 1

Given:

$A=\left[\begin{array}{ccc}1& 3& 1\\ 2& 4& 0\end{array}\right]$

The reduced row echelon form of$A=\left[\begin{array}{ccc}1& 3& 1\\ 2& 4& 0\end{array}\right]$ is $\left[\begin{array}{ccc}1& 0& -2\\ 0& 1& 1\end{array}\right]$

Since (-1,0,-2) and (0,1,1) form a basis for the row space of matrix A , we have$\{(1,0,-2{)}^{T},(0,1,1{)}^{T}\}$ form a basis for $R({A}^{T})$

When from the reduced row echelon form of matrix A , we have,

${x}_{1}-2{x}_{3}=0$

$\Rightarrow {x}_{1}=2{x}_{3}$

${x}_{1}+{x}_{3}=0$

$\Rightarrow {x}_{1}=-{x}_{3}$

Step 2

Set${x}_{3}=\alpha $ . Then N(A) consists of all vectors of the form $\alpha (2,-1,1{)}^{T}$

Therefore,$(2,-1,1{)}^{T}$ is its basis

Now${A}^{T}=\left[\begin{array}{cc}1& 2\\ 3& 4\\ 1& 0\end{array}\right]$

The reduced row echelon form of${A}^{T}=\left[\begin{array}{cc}1& 2\\ 3& 4\\ 1& 0\end{array}\right]\text{is}\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$

Since (1,0) and (0,1) form the basis for the row space of matrix${A}^{T}$ , we have $\{(1,0{)}^{T},(0,1{)}^{T}\}$ form a basis for R(A).

When$x\in N({A}^{T})$ from the reduced row echelon form of matrix ${A}^{T}$ , we have,

${x}_{1}=0\text{and}{x}_{2}=0$

Step 3 It follows that$N({A}^{T})=0$

Therefore , there is no basis for$N({A}^{T})$

Given:

The reduced row echelon form of

Since (-1,0,-2) and (0,1,1) form a basis for the row space of matrix A , we have

When from the reduced row echelon form of matrix A , we have,

Step 2

Set

Therefore,

Now

The reduced row echelon form of

Since (1,0) and (0,1) form the basis for the row space of matrix

When

Step 3 It follows that

Therefore , there is no basis for

Jeffrey Jordon

Expert2022-01-30Added 2605 answers

Answer is given below (on video)

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