# 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 $\displaystyle A, C(A)$ 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.