$x_1 + 2x_2 + x_3 − x_4 = 0$
$−x_1 + 2x_2 + x_3 + x_4 = 0$
$x_1 + x_3 = 0$
Consider the following matrix $A$ from the system of equations:
$$A = \left(\begin{array}{crc} 1 & 2 & 1 & -1\\ -1 & 2 & 1 & 1\\ 1 & 0 & 1 & 0\\ \end{array}\right)$$
1) Find a basis of the solution space $W (Ax=0).$
2) Find a Matrix $A$ such that the image of the transformation $T: \mathbb R^6 \to \mathbb R^4$ equals $W (\mathrm{im}(T)=W)$
3) Find a Matrix $A$ such that the kernel of the transformation $F: \mathbb R^4 \to \mathbb R^6$ equals $W (\mathrm{ker}(F)=W)$
My solution so far:
I found the solution space to be a line. I performed GJ elimination and got
$$\left(\begin{array}{crc} x_1 \\ x_2 \\ x_3 \\ x_4 \end{array}\right) = \left(\begin{array}{crc} 1 \\ 1/2 \\ -1 \\ 1 \end{array}\right) \cdot t$$ where $t$ is a free variable.
So a basis (one of the many) is the vector $$ u =\left(\begin{array}{crc} 1 \\ 1/2 \\ -1 \\ 1 \end{array}\right)$$
$W = span(u)$
2) We are looking for a matrix $B$ such that $B\cdot x = u$. The matrix $B$ will be a $4\times 6$ matrix.
Let the first row of the matrix be
$$B = \left(\begin{array}{crc} a_1 & b_1 & c_1 & d_1 & e_1 & f_1\\ .. & .. & .. & .. & .. & ..\\ .. & .. & .. & .. & .. & ..\\ .. & .. & .. & .. & .. & ..\\ \end{array}\right)$$
and this is valid:
$a_1x_1+b_1x_2+c_1x_3+d_1x_4+e_1x_5+f_1x_6 = 1$
We notice though that
$a_1=-a_3=a_4$
$b_1=-b_3=b_4$
.
.
.
$f_1=-f_3=f_4$
and that:
$a_2=\frac{1}{2}a_1$ and $b_2=\frac{1}{2} b_1$ and so on..
What I end up with is with a Matrix that looks like this:
$$B = \left(\begin{array}{crc} a_1 & b_1 & c_1 & d_1 & e_1 & f_1\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0\\ \end{array}\right)$$
which can't be right since what we get is a scalar product.
Please help me. I don't know what I am doing wrong!
Thanks :)
Happy New Year to you all! :)
1) There's an error:It is evident the third equation can't be verified if all coordinates are positive. Actually the third coordinate of $u$ is $-1$.
2) Remember the image of a linear map is generated by the column-vectors of its matrix in a given basis. So all column-vectors of A should be colinear with $u$.
3) You can take for $A$ a 6×4 matrix that's made up of two 3×4 blocks which have kernel equal to $W.$ The simplest is to take these blocks equal to the $A$ of question 1).
Indeed, since $W=\ker T\subset \mathbf R^4$, if we set $ F=T\oplus T\colon \mathbf R^3\oplus\mathbf R^3\simeq \mathbf R^6\rightarrow \mathbf R^4$, we have that $\ker F=\ker T\cap\ker T=W$.