# Finding basis of a subspace of $\mathbb R^3$

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?

• 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. – paw88789 Sep 8 '14 at 16:46
• 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... – Diya Sep 8 '14 at 16:55
• 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? – paw88789 Sep 8 '14 at 17:05

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.

• So the subspace has dimension 1 in this case? – Diya Sep 8 '14 at 17:06
• @Diya: That's right. For more info, Read Linear Algebra by Hoffman, Kunze. Method is rigorously presented there. – Ishfaaq Sep 8 '14 at 17:07
• 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... – Diya Sep 8 '14 at 17:59
• 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. – Ishfaaq Sep 9 '14 at 0:53