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December 23rd, 2017, 11:51 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 06:59 AM.

 December 24th, 2017, 05:39 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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 06:56 AM.
 January 9th, 2018, 09:30 AM #3 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 $\displaystyle A[e_{1}u_{1} u_{2}u_{3}]= [b,0,0,0]$ and $\displaystyle b$ arbitrary $\displaystyle \neq$ 0 $\displaystyle AM=B$, and $\displaystyle M$ non-sigular $\displaystyle A = BM^{-1}$
 January 9th, 2018, 12:59 PM #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, 05:18 AM #5 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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 06:18 AM.
January 10th, 2018, 06:37 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
.

,

Last edited by zylo; January 10th, 2018 at 06:51 AM.

 January 10th, 2018, 07:44 AM #7 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 whoops Last edited by zylo; January 10th, 2018 at 07:46 AM.
 January 10th, 2018, 11:12 AM #8 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115 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 11:58 AM.

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