Find a basis for the subspace of $\mathbb{R}^4$ consisting of all vectors that satisfy $x_1 + 2x_2 - x_3 = 0$

My general idea so far is:

$x_1 = -2x_2 +x_3 + 0x_4$

$x_2 = $ free

$x_3 = $ free

$x_4 = $ free

Where should I go from here? Am I even going about it correctly?

  • $\begingroup$ You are going about it correctly, just choose any basis of $\mathbb{R}^3$ (corresponding to $x_2,x_3,x_4$) and compute the $x_1$s $\endgroup$ – Plop Apr 19 '11 at 17:53

What you have is an expression for every vector in the subspace in parametric form, with three parameters:

$$\begin{align*} x_1 &= -2r + s\\ x_2 &=r\\ x_3 &=s\\ x_4 &=t \end{align*}$$ with $r,s,t\in\mathbb{R}$, arbitrary.

To get a basis for the space, for each parameter, set that parameter equal to $1$ and the other parameters equal to $0$ to obtain a vector. Each parameter gives you a vector. So setting $r=1$ and $s=t=0$ gives you one vector; setting $s=1$ and $r=t=0$ gives you a second vector; setting $t=1$ and $r=s=0$ gives you a third.

Alternatively, you can try rewriting the parametric solution in vector form: $$\left(\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{c}-2r-s\\r\\s\\t\end{array}\right) = \left(\begin{array}{r}-2\\1\\0\\0\end{array}\right)r + \cdots$$ (I'll let you finish it up).


Hint: given that you have one restriction, you expect a three-dimensional subspace. You need to find three independent vectors in that subspace. $(0,0,0,1)$ is in the subspace, but $(0,1,0,0)$ is not. How can you modify $(0,1,0,0)$ to satisfy $x_1 + 2x_2 - x_3 = 0$?


Hint: You just need to find the subspace/set of vectors orthogonal to the vector $(1,2,-1,0)^T$.


You can do orthogonal projection of the standard basis of $\mathbb{R}^4$ to $(1,2,-1,0)^{T {\perp}}$.


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