# Find a vector $z$ so that system of linear equations $A^{T} \cdot y=d$ is solvable if and only if $d^{T} \cdot z=0$.

Find a vector $$z$$ so that system of linear equations $$A^{T} \cdot y=d$$ has solution if and only if $$d^{T} \cdot z=0 \\$$ Matrix $$A= \begin{bmatrix} 1 & 0 & -1 & 1 \\ 2 & 1 & -3 &4 \\ 1 & -2 & 1&-4 \\ \end{bmatrix}$$

I tried to create augmented matrix with $$A^{T}$$ because of $$A^{T} \cdot y=d\space$$ (maybe this is wrong ?) $$[A^{T}|d]= \left[ \begin{array}{ccc|c} 1&2&1&d_1\\ 0&1&-2&d_2\\ -1&-3&1&d_3\\ 1&4&-4&d_4\\ \end{array} \right]$$ for some $$d=(d_1,d_2,d_3,d_4 \in \Bbb{R})$$ $$-$$

Ref of this augmented matrix is $$\left[ \begin{array}{ccc|c} 1&2&1&d_1\\ 0&1&-2&d_2\\ 0&0&-1&d_4-d_1-2d_2\\ 0&0&0&d_1+d_2+d_3\\ \end{array} \right]$$ For sure $$d_1+d_2+d_3=0$$, because system has to be solvable, and $$d_1=-d_2-d_3$$ $$-$$ Can I now do multicipation $$\space d^{T} \cdot z=0 \\$$ like $$\begin{bmatrix} -d_2-d_3 & d_2 & d_3 \\ \end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix}=0 \\$$ $$-$$ So, my solution will be, $$x_1=(1/d_3)\cdot x_2 + (1/d_2)\cdot x_3 \\[1ex] \Rightarrow z= \begin{bmatrix} (1/d_3)\cdot x_2 + (1/d_2)\cdot x_3 \\ x_2 \\ x_3 \\ \end{bmatrix}\\$$ If this is correct, can someone help me to solve this but in another dirrection..

Note that $$A^Ty=d \ \ \text {is solvable}\Longleftrightarrow d\in\mathcal R(A^T) \stackrel{\text{assumption}} {\color{blue}{\Longleftrightarrow}} d\perp z$$ where $$\mathcal R(T)$$ denotes the range of $$T$$ and $$\mathcal N(T)$$ the null space. Thus $$\mathcal R(A^T) =\langle z \rangle^\perp$$ and this implies that $$\mathcal N(A) =\mathcal R(A^T)^\perp = \langle z \rangle.$$ We find that $$Ax=0$$ is solved by $$x=c \left(\begin{array}{c}1\\1\\1\\0 \end{array} \right).$$ Therefore $$z$$ can be any vector of the form $$c(1,1,1,0)^T$$ where $$c\ne 0$$.
• You mean $A^{T}y=d$ is solvable $\Leftrightarrow d \in \mathcal{R} (A^{T}) \\$ Why $Ax=0$,I mean why homogeneous system ? – Figgaro Jan 30 at 17:20
• @Figgaro Since $d\in \mathcal R(A^T)$ is equivalent to $d\perp z$, we have $\mathcal R(A^T) = \langle z\rangle^\perp$. Hence $\mathcal N(A) \color{red}= \mathcal R(A^T)^\perp = \langle z\rangle^{\perp \! \perp} = \langle z \rangle$. This gives $z$ is a basis of $\mathcal N(A)$, so it naturally leads to solving $Ax =0$. – Song Jan 30 at 17:25