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Let the set of solutions to the system of linear equations

$x_1-2x_2+x_3=0, 2x_1-3x_2+x_3=0$ is a subspace of $\mathbb R^3$

Find a basis for this subspace.

How do I approach this problem?

I know that dimension of $\mathbb R^3=3$ and so the dimension of the subspace is $3$ and the basis will contain $3$ vectors.

I have no clue how to find them using the given conditions. I do know that the $2$ equations in $3$ unknowns have infinitely many solutions. How do I take it from there?

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  • $\begingroup$ You may want to start solving the system. You would then need to find one or more vectors (that are linearly independent). So that any solution can be written as a linear combination of the vectors you found. You may see how to proceed once you've solved the system. $\endgroup$ – paw88789 Sep 8 '14 at 16:46
  • $\begingroup$ I used cross-multiplication to find a general solution in the form $(k,-k,k)$ but I don't know how to find a basis from here... $\endgroup$ – Diya Sep 8 '14 at 16:55
  • $\begingroup$ Actually $(k,-k,k)$ doesn't satisfy your equations. But in principle, you could think of this set as $k(1,-1,1)$. Do you see how this shows you what the basis would be? $\endgroup$ – paw88789 Sep 8 '14 at 17:05
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There is a specific algorithm to follow if you want.

  • First you row reduce the system.

$$ \left({\begin{matrix} 1 & -2 & 1 \\ 2 & -3 & 1 \\ \end{matrix}}\right) \left({\begin{matrix} x_1 \\ x_2 \\ x_2 \\ \end{matrix}}\right) = 0 \iff \left({\begin{matrix} 1 & 0 & -1 \\ 0 & 1 & - 1 \\ \end{matrix}}\right) \left({\begin{matrix} x_1 \\ x_2 \\ x_2 \\ \end{matrix}}\right) = 0 $$

  • Then you rewrite the equations with the prominent variable on the left and all that is left on the right. Say there are $k$ rows in your system of $n$ variables. Then $k$ variables will have prominent columns while $n- k$ variables won't. So send these $n - k$ variables to other side of the equations. your set of solutions is obtained by taking any values whatsover for these $n - k$ variables. Now in our example,

$$ x_1 = x_3 $$

$$ x_2 = x_3 $$

Any values you wish to choose for $x_3$ will solve the system if you take the corresponding values for $x_1$ and $x_2$ which in our case is the same. So the set of solutions is of the form,

$$ \left({\begin{matrix} x_1 \\ x_2 \\ x_2 \\ \end{matrix}}\right) = \left({\begin{matrix} x_3 \\ x_3 \\ x_3 \\ \end{matrix}}\right) = x_3\left({\begin{matrix} 1 \\ 1 \\ 1 \\ \end{matrix}}\right)$$

where $x_3$ is any scalar in $\Bbb R$. Your basis is staring at you.

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  • $\begingroup$ So the subspace has dimension 1 in this case? $\endgroup$ – Diya Sep 8 '14 at 17:06
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    $\begingroup$ @Diya: That's right. For more info, Read Linear Algebra by Hoffman, Kunze. Method is rigorously presented there. $\endgroup$ – Ishfaaq Sep 8 '14 at 17:07
  • $\begingroup$ Thank you. What if I'm given one equation though? Say $a_1-a_3-a_4=0$? I'm finding sums like that, so... $\endgroup$ – Diya Sep 8 '14 at 17:59
  • $\begingroup$ Same story. $ a_1 = a_3 + a_4 $. So all your solutions are of the form $ \left({\begin{matrix} a_3 + a_4 \\ a_2 \\ a_3 \\ a_4 \\ \end{matrix}}\right) = a_2 \left({\begin{matrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{matrix}}\right) + a_3 \left({\begin{matrix} 1 \\ 0 \\ 1 \\ 0 \\ \end{matrix}}\right) + a_4 \left({\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \\ \end{matrix}}\right) $ where $ a_2, a_3, a_4 $ are arbitrary scalars. $\endgroup$ – Ishfaaq Sep 9 '14 at 0:53

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