test tomorrow need the instructions to how to do this problem.
(1 1)
Question: A=( 0 2) find a matrix B such that C(A)=N(B)
(1 0)
any kind of help please.
Ok so the column space of $\displaystyle A, C(A)$ is the set of all linear combinations of the columns of $\displaystyle A$.
That is all vectors of the following form:
$\displaystyle c_1 \left( \begin{array}{c}
1 \\
0 \\
1 \end{array} \right) + c_2 \left( \begin{array}{c}
1 \\
2 \\
0 \end{array} \right) = \left( \begin{array}{c}
c_1+c_2 \\
2c_2 \\
c_1 \end{array} \right)$ where $\displaystyle c_1, c_2$ are scalars.
We could think about these vectors as points in 3-space with certain restrictions: $\displaystyle x = c_1+c_2, y=2c_2, z = c_1$ which leads to:
$\displaystyle z = x-\frac{y}{2}$
We want $\displaystyle B = \left( \begin{array}{ccc}
a & b & c \\
d & e & f \end{array} \right)$
such that
$\displaystyle B \left( \begin{array}{cc}
c_1+c_2 \\
2c_2 \\
c_1 \end{array} \right) = \left( \begin{array}{c}
0 \\
0 \\
\end{array} \right)$
$\displaystyle B = \left( \begin{array}{ccc}
2 & -1 & -2 \\
2 & -1 & -2 \end{array} \right)$ works I think.