Stuck on finding dimension and basis of a solution set Consider the following system of equations:
$$
\begin{cases}
x-y-2z=0 \\
x+y=4w
\end{cases}
$$
The first part of the question requires me to write the solution set as the kernel of a matrix (matrix having $\Re$ Coefficients). The second part asks me to find the $\Re$-Dimension and $\Re$-Basis of the solution set above.
After isolating $x$ and $y$ and by substitution, I got the following:
$$
\begin{cases}
x = z + 2w \\
y = z - 2w
\end{cases}
$$
I can re-write all of this into an augmented matrix, but I don't know where to go from here.
 A: So assuming your solution is correct, the solution vector is
$$
\begin{bmatrix} x \\ y\\ z\\ w \end{bmatrix}
 = \begin{bmatrix} z + 2w \\ z - 2w\\ z\\ w \end{bmatrix}
 = \begin{bmatrix} 1 \\ 1\\ 1\\ 0 \end{bmatrix} z
 + \begin{bmatrix} 2 \\ -2\\ 0\\ 1 \end{bmatrix} w
 = \begin{bmatrix} 1 & 2 \\ 1 & -2 \\ 1 & 0\\ 0 & 1 \end{bmatrix}
   \begin{bmatrix} z \\ w \end{bmatrix}
$$
The solution is then the column space of the matrix above. Now can you find the dimension and basis of the solution set?
To have the solution set as a kernel of a matrix, can you write down the definition of a kernel and see how you can convert this definition into one?
A: Well what do you think of the coefficient matrix? $$A=\begin{bmatrix}1& -1& -2& 0\\ 1& 1& 0& -4\end{bmatrix}$$
And the associated system of equations to get its kernel space?
$$\begin{bmatrix}1& -1& -2& 0\\ 1& 1& 0& -4\end{bmatrix} \cdot \begin{bmatrix}x\\ y \\z \\w\end{bmatrix}=\mathbf{0}$$
Do you think $A$ represents a matrix such that the solution space defines its kernel? Can we calculate the rank and nullity of the matrix $A$? How do we get the kernel space basis of a matrix?
I'll leave you with this so that I do not spoil the solution for you but I do presume this answers the question.
