My Math Forum find a matrix with the given null space

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 January 1st, 2017, 08:19 PM #1 Newbie   Joined: Jan 2017 From: US Posts: 1 Thanks: 0 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?
 January 4th, 2017, 06:42 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,959 Thanks: 801 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 $R^2$ of the form (x, 0)" Then (assuming for the moment that the matrix is square) we must have $\begin{pmatrix}a=&b \\ c=&d \end{pmatrix}\begin{pmatrix}x \\ 0 \end{pmatrix}= \begin{pmatrix}ax \\ cx\end{pmatrix}= \begin{pmatrix}0 \\ 0 \end{pmatrix}$. 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.)
 January 4th, 2017, 11:30 AM #3 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,230 Thanks: 93 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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