0
$\begingroup$

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..

$\endgroup$
0
$\begingroup$

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 $c\ne 0$.

$\endgroup$
  • $\begingroup$ You mean $A^{T}y=d$ is solvable $\Leftrightarrow d \in \mathcal{R} (A^{T}) \\ $ Why $Ax=0$,I mean why homogeneous system ? $\endgroup$ – Figgaro Jan 30 at 17:20
  • $\begingroup$ @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 $Ax =0$. $\endgroup$ – Song Jan 30 at 17:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.