# Thread: Elementary linear algebra college work

1. ## Elementary linear algebra college work

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)

2. What are C(A) and N(B)?

3. C(A) is a matrix and N(B) is the north space of the matrix

4. C(A) may be a matrix but how does C operate on A? North space, I am not familiar with I am sorry, how is it defined?

5. Originally Posted by fatboy
C(A) is a matrix and N(B) is the north space of the matrix
I think the question was, "C(A) is a matrix with what relationship to A?"

-Dan

6. figured it out C(A) means the column space and N(A) is the nullspace hope that helps

7. Ok so the column space of $A, C(A)$ is the set of all linear combinations of the columns of $A$.

That is all vectors of the following form:

$c_1 \left( \begin{array}{c}
1 \\
0 \\
1 \end{array} \right) + c_2 \left( \begin{array}{c}
1 \\
2 \\
0 \end{array} \right) = \left( \begin{array}{c}
c_1+c_2 \\
2c_2 \\
c_1 \end{array} \right)$
where $c_1, c_2$ are scalars.

We could think about these vectors as points in 3-space with certain restrictions: $x = c_1+c_2, y=2c_2, z = c_1$ which leads to:

$z = x-\frac{y}{2}$

We want $B = \left( \begin{array}{ccc}
a & b & c \\
d & e & f \end{array} \right)$

such that

$B \left( \begin{array}{cc}
c_1+c_2 \\
2c_2 \\
c_1 \end{array} \right) = \left( \begin{array}{c}
0 \\
0 \\
\end{array} \right)$

$B = \left( \begin{array}{ccc}
2 & -1 & -2 \\
2 & -1 & -2 \end{array} \right)$
works I think.