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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.
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December 24th, 2017, 05:39 AM   #2
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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.
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Last edited by skipjack; December 24th, 2017 at 06:56 AM.
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January 9th, 2018, 09:30 AM   #3
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$\displaystyle A[e_{1}u_{1} u_{2}u_{3}]= [b,0,0,0]$ and $\displaystyle b$ arbitrary $\displaystyle \neq$ 0

$\displaystyle AM=B$, and $\displaystyle M$ non-sigular

$\displaystyle A = BM^{-1}$
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January 9th, 2018, 12:59 PM   #4
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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.
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January 10th, 2018, 05:18 AM   #5
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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.
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January 10th, 2018, 06:37 AM   #6
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Quote:
Originally Posted by zylo View Post
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.
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January 10th, 2018, 07:44 AM   #7
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whoops

Last edited by zylo; January 10th, 2018 at 07:46 AM.
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January 10th, 2018, 11:12 AM   #8
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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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