PenTrik said:
Homework Statement
If V is the subspace spanned by (1,1,1) and (2,1,0), find a matrix A that has V as its row space. Find a matrix B that has V as its nullspace
Homework Equations
Ax = 0 for a nullspace
The Attempt at a Solution
So straight off the bat, I think I can solve the first part. Should be simple enough right? Matrix A that has V as its row space is just the matrix
[[1,1,1]
[2,1,0]]
But how do I find matrix B that has V as its nullspace?
The only thing I can possibly think of its to do what I have done before which is to do the rref
Do the matrix multiplication
\begin{bmatrix}a & b & c \\d & e & f\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 1 \\ -1 & 2\end{bmatrix}
and set it equal to
\begin{bmatrix}0 & 0 \\ 0 & 0\end{bmatrix}
That will give you four equations, a- c= 0, b+ 2c= 0, d- f= 0, e+ 2f= 0. That is four equations in six variables. You can solve for 4 of the variables in terms of the other 2. Notice that the problem says "find
a matrix". There are an infinite number of such matrices.
Which I think gives me
[[1,0,-1]
[0,1,2]]
Do I run linear combination on the row space from here?
Also, a similar question that I am having problems with is
Find a basis for the space of a 2x3 matrices whose nullspace contains (2,1,1)
I'm not even sure how to approach this question. My gut instinct tells me to run matrix multiplication on this one, such that
<br />
\begin{pmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}<br />
This is about as far as my understanding of nullspaces carries me.
Exactly what you needed to do! (Except that the matrix on the right should have only two rows.) When you do that multiplication you get
\begin{bmatrix} 2x_{11}+ x_{12}+ x_{13} \\ 2x_{21}+ x_{22}+ x_{23}\end{bmatrix}= \begin{bmatrix} 0 \\ 0\end{bmatrix}
which gives you two equation for the six variables: 2x_{11}+ x_{12}+ x_{13}= 0 and 2x_{21}+ x_{22}+ x_{23}= 0. Now you can solve for 2 of those in terms of the other 4 very easily: x_{12}= -2x_{11}- x_{13} and x_{22}= -x_{21}- x_{23}. Put those into the original matrix:
\begin{bmatrix} x_{11} & -2x_{11}- x_{13} & x_{13} \\ x_{21} & -2x_{21}- x_{23} & x_{23} \end{bmatrix}
= \begin{bmatrix} x_{11} & -2x_{11} & 0 \\ 0 & 0 & 0 \end{bmatrix}+ \begin{bmatrix}0 & -1 & 1 \\ 0 & 0 & 0\end{bmatrix}
+ \begin{bmatrix}0 & 0 & 0 \\ x_{21} & -2x_{21} & 0\end{bmatrix}+ \begin{bmatrix}0 & 0 & 0 \\ 0 & -x_{23} & x_{23}\end{bmatrix}
= x_{11}\begin{bmatrix}1 & -2 & 0 \\ 0 & 0 & 0\end{bmatrix}+ x_{13}\begin{bmatrix} 0 & 0 & 0 \\ 1 & -2 & 0\end{bmatrix}+ x_{23}\begin{bmatrix}0 & 0 & 0 \\ 0 & -1 & 1\end{bmatrix}
Do you get the point?
