Science:Math Exam Resources/Courses/MATH307/April 2006/Question 01 (a)/Solution 1

This matrix A in this question is given in the form of an LU decomposition with partial pivoting where P is just the matrix which swaps the rows of A. The matrix P also would not affect R(A), N(A), R(A^T) where we need to look at the matrix U (echelon form) to find the bases.

To find a basis for the R(A), we must identify the pivot columns in matrix U, which turns out to be the first and second column.

The pivots are highlighted in red.

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

The basis for R(A) is then the corresponding columns in A to the pivot columns in U.

The pivot columns of A are highlighted in red.

$A=\left[\begin{array}{cccc}1& 0& 2& 3\\ 2& 1& 8& 8\\ 0& 1& 4& 2\end{array}\right]$

$R(A)=\text{span}(\left[\begin{array}{c}1\\ 2\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 1\end{array}\right])$

To find N(A), we can set $Ux=0$ and solve for x. This would give us 2 equations and 2 free variables.

$\begin{array}{rl}{x}_1& =-2x_3-3x_4\\ x_2& =-4x_3-2x_4\\ x_3& =x_3\\ x_4& =x_4\end{array}$

$x=x_3\left[\begin{array}{c}-2\\ -4\\ 1\\ 0\end{array}\right]+x_4\left[\begin{array}{c}-3\\ -2\\ 0\\ 1\end{array}\right]$

And therefore

$N(A)=\text{span}(\left[\begin{array}{c}-2\\ -4\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}-3\\ -2\\ 0\\ 1\end{array}\right])$

The basis of R(A^T) can be taken from the rows of U that contain pivots.

$R(A^T)=\text{span}(\left[\begin{array}{c}1\\ 0\\ 2\\ 3\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 4\\ 2\end{array}\right])$

To find the basis for N(A^T), we must row reduce A^T and solve for x in $A^{T}x=0$

From row reducing A^T, you should acquire the matrix

$\left[\begin{array}{ccc}1& 0& -2\\ 0& 1& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$

Solving for x, you end up with 2 equations and 1 free variable.

$\begin{array}{rl}{x}_1& =2x_3\\ x_2& =-x_3\\ x_3& =x_3\end{array}$

$x=x_3\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]$

And hence

$N(A^T)=\text{span}\left(\left[\begin{array}{c}2\\ -1\\ 1\end{array}\right]\right)$