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December 23rd, 2017, 10:51 AM   #1
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how to find matrix by given null?

given null need to find - that - Can someone help?
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Last edited by skipjack; December 24th, 2017 at 05:59 AM. December 24th, 2017, 04:39 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Your images don't show on my computer but the answer to your question "How so you determine a linear transformation knowing only its null space" is "You don't. There are, in general, an infinite number of linear transformations having a given null space". Given that the null space is spanned by $\begin{pmatrix}4 \\ 1 \\ -4 \\ 5\end{pmatrix}$, $\begin{pmatrix}-2 \\ -1 \\ 0 \\ -2\end{pmatrix}$, and $\begin{pmatrix}-3 \\ 5 \\ 4 \\ 1\end{pmatrix}$, and that A can be represented by a 4 by 4 matrix, $\begin{pmatrix}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & p & q \end{pmatrix}$, we must have $\begin{pmatrix}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & p & q \end{pmatrix}\begin{pmatrix}4 \\ 1 \\ -4 \\ 5\end{pmatrix}= \begin{pmatrix}4a+ b- 4c+ 5d \\ 4e+ f- 4g+ 5h \\ 4i+ j- 4k+ 5l \\ 4m+ n- 4p+ 5q\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$. $\begin{pmatrix}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & p & q \end{pmatrix}\begin{pmatrix}-2 \\ 1 \\ 0 \\ -2\end{pmatrix}= \begin{pmatrix}-2a+ b- 2d \\ -2e+ f- 2h \\ -2i+ j- 2l\\ -2m+ n- 2q\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$. and $\begin{pmatrix}a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & p & q \end{pmatrix}\begin{pmatrix}-3 \\ 5 \\ 4 \\ 1\end{pmatrix}= \begin{pmatrix}-3a+ 5b+ 4c+ d \\ -3e+ 5f+ 4g+ h \\ -3i+ 5j+ 4k+ 1l\\ -3m+ 5+ 4p+ q\end{pmatrix}= \begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$. 12 equations to solve for 16 unknowns. Thanks from zollen Last edited by skipjack; December 24th, 2017 at 05:56 AM. January 9th, 2018, 08:30 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 $\displaystyle A[e_{1}u_{1} u_{2}u_{3}]= [b,0,0,0]$ and $\displaystyle b$ arbitrary $\displaystyle \neq$ 0 $\displaystyle AM=B$, and $\displaystyle M$ non-sigular $\displaystyle A = BM^{-1}$ January 9th, 2018, 11:59 AM #4 Member   Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 85 Thanks: 14 Math Focus: Algebra, Cryptography Good afternoon ! We can reduce the number of the unknowns in Country Boy's methode by remarking that $\displaystyle \text{rank}(A)=4-\dim N(A)=4-3=1.$Thus the matrix $\displaystyle A$ can be written as : $\displaystyle A=\begin{pmatrix}a&b&c&d\\a&b&c&d\\a&b&c&d\\a&b&c& d \end{pmatrix}.$ We will get 3 equations and 4 unknowns. Thanks from zylo January 10th, 2018, 04:18 AM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 What I originally put here was wrong. But I might as well note a slight variation of previous post: A=[a,b,c,d], a 4x1 matrix. Last edited by zylo; January 10th, 2018 at 05:18 AM. January 10th, 2018, 05:37 AM   #6
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Quote:
 Originally Posted by zylo What I originally put here was wrong. But I might as well note a slight variation of previous post: A=[a,b,c,d], a 4x1 matrix.
EDIT: A = [a,b,c,1], Au$\displaystyle _{1}$=0, Au$\displaystyle _{2}$=0, Au$\displaystyle _{3}$=0

EDIT: Damn. A has to be a 4x4 matrix. Ould Youbba was right, though 0's in last three rows might be simpler. Sorry
.

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Last edited by zylo; January 10th, 2018 at 05:51 AM. January 10th, 2018, 06:44 AM #7 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 whoops Last edited by zylo; January 10th, 2018 at 06:46 AM. January 10th, 2018, 10:12 AM #8 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 As an exersize, if $\displaystyle A = [a,b,c,1]$, derived from ould youbba observation, then $\displaystyle Au_{1}=0$, $\displaystyle Au_{2}=0$, and $\displaystyle Au_{3}=0$ become $\displaystyle 4a + b -4c = -5$ $\displaystyle -2a -b +0c = 2$ $\displaystyle -3a + 5b + 4c = -1$ Augmented matrix: $\displaystyle \begin{bmatrix} 4 & 1 & -4 &-5 \\ -2 & -1 & 0 &2 \\ -3 & 5 & 4 & -1 \end{bmatrix}$ Reduced row echelon form: $\displaystyle \begin{bmatrix} 1 & 0 & 0\ & \frac{1}{11} \\ 0 &1 &0 &- \frac{6}{11} \\ 0 & 0 & 1 & - \frac{5}{44} \end{bmatrix}$ From which $\displaystyle A = \begin{bmatrix} \frac{1}{11} &- \frac{6}{11} &-\frac{5}{44} &1 \end{bmatrix}$ To make A a 4x4 matrix, add three rows of 0's. EDIT: If I wanted to find A st Ax=0 would give u$\displaystyle _{1}$, u$\displaystyle _{2}$ and u$\displaystyle _{3}$, then that's country boy's solution. If I wanted to find A whose null space is U, (OP), then that's ould youbba's solution. If I wanted to solve A=[a,b,c,1] given for Ax=0, I would get ax1 + bx2 +cx3 + x4 = 0. x1, x2, x3 arbitrary and x4 =..., which is a 3d null space. Last edited by zylo; January 10th, 2018 at 10:58 AM. Tags find, matrix, null Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gabby88 Linear Algebra 2 January 4th, 2017 10:30 AM garb Math 0 July 3rd, 2015 01:09 AM imhotep Linear Algebra 4 February 15th, 2013 12:16 PM zorroz Abstract Algebra 0 June 5th, 2012 12:14 PM symmetry Advanced Statistics 0 April 7th, 2007 03:05 AM

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