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January 1st, 2017, 07:19 PM   #1
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find a matrix with the given null space

I really do not know how to do this. I know what a null space is and how to find it given a matrix, but I am not sure exactly how to to work that backwards.

For example, N(A) = span([1])
0

How would I go about finding a matrix with this property?
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January 4th, 2017, 05:42 AM   #2
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If you really know what a "null space" is, then it should be obvious that there exist an infinite number of matrices having a given null space.

In particular, if the null space is "all vectors of the form (x, 0)" Then (assuming for the moment that the matrix is square) we must have
.

So we must have ax= 0 for all x, cx= 0 for all x. What does that tell you?

(If the matrix is not required to be square, then we can have many more rows with the first column 0.)
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January 4th, 2017, 10:30 AM   #3
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Consider a system TX=0 in rref.

$\displaystyle \begin{vmatrix}
1 & 0 & a & b & c \\ 0 & 1 & e & f & g \end{vmatrix} \begin{vmatrix} x1\\ x2\\ x3\\ x4\\ x5 \end{vmatrix} = 0$

x1 + ax3 + bx4 + cx5 = 0
x2 + dx3 + ex4 + fx5 = 0

$\displaystyle X=\begin{vmatrix}
-ax3 -bx4 -cx5\\ -dx3 -ex5 -fx6\\ x3\\ x4\\ x5
\end{vmatrix}=x3 \begin{vmatrix} -a\\ -d\\ 1\\ 0\\
0 \end{vmatrix}+ x4\begin{vmatrix} -b\\ -e\\ 0\\ 1\\
0 \end{vmatrix} +x5\begin{vmatrix} -c\\ -f\\ 0\\ 0\\
1 \end{vmatrix}
$

For Example, the 3d Null Space

$\displaystyle X=\alpha \begin{vmatrix} 2\\ 3\\ 1\\ 0\\ 0\end{vmatrix} + \beta
\begin{vmatrix} 1\\ 5\\ 0\\ 1\\
0 \end{vmatrix} + \gamma \begin{vmatrix} 4\\ 2\\ 0\\ 0\\
1 \end{vmatrix}
$

is the solution of the system

$\displaystyle \begin{vmatrix}
1 & 0 & -2 & -1 & -4\\
0 & 1 & -3 & -5 & -2 \end{vmatrix}X=0$
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