Answered: A 1 1 1 2 2

Algebra: Structure And Method, Book 1

(REV)00th Edition

ISBN:9780395977224

Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Chapter4: Polynomials

Section4.1: Exponents

Problem 34WE

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Question

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

Expert Solution

Step 1

Introduction :

A basis of a vector space is a set of vectors in that space that can be used as coordinates for it. In linear algebra, a basis for a vector space $V$ is a set of vectors in $V$ such that every vector in $V$ can be written uniquely as a finite linear combination of vectors in the basis.

Given :

$A= 1 0 0 0 0 1 1 1 0 0 1 1 1 1 2 2$

Objective :

To determine a basis for each of the subspaces $R (A^{T}), N (A), R (A)$ , and $N (A^{T})$ .

Step 2

Given matrix is $A= 1 0 0 0 0 1 1 1 0 0 1 1 1 1 2 2$ so $A^{T}=\begin{bmatrix} 1 &0&0&1 \\0 &1&0&1\\0&1&1&2\\0&1&1&2 \end{bmatrix}$

Now for the subspaces of and we will reduce matrix $A= 1 0 0 0 0 1 1 1 0 0 1 1 1 1 2 2$ to row reduced echelon form.

$A= 1 0 0 0 0 1 1 1 0 0 1 1 1 1 2 2$

Subtract row 1 from row 4 we get

$A\sim\begin{bmatrix} 1 &0&0&0 \\0 &1&1&1\\0&0&1&1\\0&1&2&2 \end{bmatrix}$

Subtract row 2 from row 4 we get

$A\sim\begin{bmatrix} 1 &0&0&0 \\0 &1&1&1\\0&0&1&1\\0&0&1&1 \end{bmatrix}$

Subtract row 3 from row 2 we get

$1 0 0 0 0 1 0 0 0 0 1 1 0 0 1 1$

Subtract row 3 from row 4 we get

$1 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0$

So we have arrived to the row reduced echelon form for the matrix A.

In the row reduced echelon form the first column, second column and third column corresponding to the original matrix A will be the basis for the i.e a basis for will be ${[1 0 0 1],[0 1 0 1], [0 11 21}$ and the fourth column corresponding to the original matrix A will be the basis for the i.e a basis for will be ${[0 1 1 2)}$ ..

Now for the subspaces of and we will reduce matrix $A^{T}=\begin{bmatrix} 1 &0&0&1 \\0 &1&0&1\\0&1&1&2\\0&1&1&2 \end{bmatrix}$ to row reduced echelon form.

$A^{T}=\begin{bmatrix} 1 &0&0&1 \\0 &1&0&1\\0&1&1&2\\0&1&1&2 \end{bmatrix}$

Subtract row 2 from row 3 and Subtract row 2 from row 4 we get

$AT 1 0 0 1 0 1 0 1 0 0 1 1 1 0 0 1 1$

Subtract row 3 from row 4 we get

$AT 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 0$

So we have arrived to the row reduced echelon form for the matrix .

In the row reduced echelon form the first column, second column and third column corresponding to the original matrix will be the basis for the i.e a basis for will be ${[1 0 0 0], [0 11 1], [0 0 1 1]}$ and the fourth column corresponding to the original matrix will be the basis for the i.e a basis for will be ${1 1 2 21}$ .

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