Damon Vazquez

2022-09-06

Make a system with 3 equations and 3 unknowns of which the solution space V is spanned by the column vectors: $$\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right],\phantom{\rule{1em}{0ex}}\left[\begin{array}{c}1\\ 3\\ 0\end{array}\right]$$

Emilia Boyle

Beginner2022-09-07Added 10 answers

First of all, every equation system can be written as $AX=b$. The solution space is a vector space only if it is the $ker(A)$ which is a solution space of a system $AX=0$. In general the solution space is nothing but a coset of the $ker(A)$ (and the coset is not vector space unless it is equal to $ker(A)$.

So what you need now is just to find a matrix whose kernel is $V=\u27e8{v}_{1},{v}_{2}\u27e9$

We have $3$ variables, and solution space has two independent generators (in other words there are two free variables). Hence the solution space is a solution to only $one$ equation with $3$ variables. That means the other two equations should be just multiple of the first equation. Now to find the first equation we use the Gauss elimination method, but in reverse. To do this efficiently we need to bring the generating system to the form $\{(0,number,1),(1,number,0)\}$ ...etc.

To do this we can apply the elementary operations on the generators.

- Multiply one of them by a number and adding the result to another generator.

- Multiply or divide one of them by nonzero number.

For example, if

$$V=\u27e8{u}_{1}=(2,3,4),{u}_{2}=(1,2,3)\u27e9$$We multiply ${u}_{2}$ by $-2$ and add it to ${u}_{1}$, we get

$$V=\u27e8{u}_{3}=(0,-1,-2),{u}_{2}=(1,2,3)\u27e9$$

Now multiply ${u}_{3}$ with $-1$.

$$V=\u27e8{u}_{4}=(0,1,2),{u}_{2}=(1,2,3)\u27e9$$

We multiply ${u}_{4}$ by $-2$ and add it to ${u}_{2}$, we get

$$V=\u27e8{u}_{4}=(0,1,2),{u}_{5}=(1,0,-1)\u27e9$$

The system is now ready for revers Gause Elimination method. The generators in your question is much easier, and we need only to multiply by $\frac{1}{3}$ and $-1$.

The rest is easy as follows

$\begin{array}{ccc}V& =& \u27e8(1,3,0),(1,0,-1)\u27e9\\ & =& \u27e8(\frac{1}{3},1,0),(-1,0,1)\u27e9\\ & =& \{y(\frac{1}{3},1,0)+z(-1,0,1)|y,z\in \mathbb{R}\}\\ & =& \{(\frac{1}{3}y-z,y,z)|y,z\in \mathbb{R}\}\\ & =& \{(x,y,z)|x-\frac{1}{3}y+z=0,x,y,z\in \mathbb{R}\}\end{array}$

So the system desired can be for example:

$\left(\begin{array}{ccc}1& -\frac{1}{3}& 1\\ 3& -1& 3\\ 6& -2& 6\end{array}\right)$

Note that the rank of this matrix is $$1$$, therefore the dimension of the solution space is $$2$$, and ${v}_{1},{v}_{2}\in $ Solution Space, and they are linear independent, so the Solution Space is $V$.

So what you need now is just to find a matrix whose kernel is $V=\u27e8{v}_{1},{v}_{2}\u27e9$

We have $3$ variables, and solution space has two independent generators (in other words there are two free variables). Hence the solution space is a solution to only $one$ equation with $3$ variables. That means the other two equations should be just multiple of the first equation. Now to find the first equation we use the Gauss elimination method, but in reverse. To do this efficiently we need to bring the generating system to the form $\{(0,number,1),(1,number,0)\}$ ...etc.

To do this we can apply the elementary operations on the generators.

- Multiply one of them by a number and adding the result to another generator.

- Multiply or divide one of them by nonzero number.

For example, if

$$V=\u27e8{u}_{1}=(2,3,4),{u}_{2}=(1,2,3)\u27e9$$We multiply ${u}_{2}$ by $-2$ and add it to ${u}_{1}$, we get

$$V=\u27e8{u}_{3}=(0,-1,-2),{u}_{2}=(1,2,3)\u27e9$$

Now multiply ${u}_{3}$ with $-1$.

$$V=\u27e8{u}_{4}=(0,1,2),{u}_{2}=(1,2,3)\u27e9$$

We multiply ${u}_{4}$ by $-2$ and add it to ${u}_{2}$, we get

$$V=\u27e8{u}_{4}=(0,1,2),{u}_{5}=(1,0,-1)\u27e9$$

The system is now ready for revers Gause Elimination method. The generators in your question is much easier, and we need only to multiply by $\frac{1}{3}$ and $-1$.

The rest is easy as follows

$\begin{array}{ccc}V& =& \u27e8(1,3,0),(1,0,-1)\u27e9\\ & =& \u27e8(\frac{1}{3},1,0),(-1,0,1)\u27e9\\ & =& \{y(\frac{1}{3},1,0)+z(-1,0,1)|y,z\in \mathbb{R}\}\\ & =& \{(\frac{1}{3}y-z,y,z)|y,z\in \mathbb{R}\}\\ & =& \{(x,y,z)|x-\frac{1}{3}y+z=0,x,y,z\in \mathbb{R}\}\end{array}$

So the system desired can be for example:

$\left(\begin{array}{ccc}1& -\frac{1}{3}& 1\\ 3& -1& 3\\ 6& -2& 6\end{array}\right)$

Note that the rank of this matrix is $$1$$, therefore the dimension of the solution space is $$2$$, and ${v}_{1},{v}_{2}\in $ Solution Space, and they are linear independent, so the Solution Space is $V$.

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