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

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MATH307 April 2006

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Question 01 (a)
Let PA = LU be ${\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}}{\begin{bmatrix}1&0&2&3\\2&1&8&8\\0&1&4&2\end{bmatrix}}={\begin{bmatrix}1&0&0\\0&1&0\\2&1&1\end{bmatrix}}{\begin{bmatrix}1&0&2&3\\0&1&4&2\\0&0&0&0\end{bmatrix}}$ and $b={\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\end{bmatrix}}$ (a) Find a basis for each of the four fundamental subspaces.

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

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Solution
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. $U={\begin{bmatrix}{\color {Red}{1}}&0&2&3\\0&{\color {Red}{1}}&4&2\\0&0&0&0\end{bmatrix}}$ 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. $A={\begin{bmatrix}{\color {Red}{1}}&{\color {Red}{0}}&2&3\\{\color {Red}{2}}&{\color {Red}{1}}&8&8\\{\color {Red}{0}}&{\color {Red}{1}}&4&2\end{bmatrix}}$ $R(A)={\text{span}}\left({\begin{bmatrix}1\\2\\0\end{bmatrix}},{\begin{bmatrix}0\\1\\1\end{bmatrix}}\right)$ To find N(A), we can set $Ux=0$ and solve for x. This would give us 2 equations and 2 free variables. ${\begin{aligned}x_{1}&=-2x_{3}-3x_{4}\\x_{2}&=-4x_{3}-2x_{4}\\x_{3}&=x_{3}\\x_{4}&=x_{4}\end{aligned}}$ $x=x_{3}{\begin{bmatrix}-2\\-4\\1\\0\end{bmatrix}}+x_{4}{\begin{bmatrix}-3\\-2\\0\\1\end{bmatrix}}$ And therefore $N(A)={\text{span}}\left({\begin{bmatrix}-2\\-4\\1\\0\end{bmatrix}},{\begin{bmatrix}-3\\-2\\0\\1\end{bmatrix}}\right)$ The basis of R(A^T) can be taken from the rows of U that contain pivots. $R(A^{T})={\text{span}}\left({\begin{bmatrix}1\\0\\2\\3\end{bmatrix}},{\begin{bmatrix}0\\1\\4\\2\end{bmatrix}}\right)$ 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 ${\begin{bmatrix}1&0&-2\\0&1&1\\0&0&0\\0&0&0\end{bmatrix}}$ Solving for x, you end up with 2 equations and 1 free variable. ${\begin{aligned}x_{1}&=2x_{3}\\x_{2}&=-x_{3}\\x_{3}&=x_{3}\end{aligned}}$ $x=x_{3}{\begin{bmatrix}2\\-1\\1\end{bmatrix}}$ And hence $N(A^{T})={\text{span}}\left({\begin{bmatrix}2\\-1\\1\end{bmatrix}}\right)$