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Just stuck on a small problem:

The set

$V = {{(x_1, x_2, x_3, x_4) \in R^4 \ | \ x_1 -2x_2 + 3x_3 +x_4=0}}$

is a subspace of $R^4$.

Find a basis ${(v_1, v_2, v_3, v_4)}$ for $R^4$ such that ${(v_1, v_2, v_3)}$ is a basis for V and $v_4$ is a vector orthogonal to V (i.e $v_4$ is orthogonal to every vector in V).

Thanks in advance.

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    $\begingroup$ Do you have any thoughts on how to do this, or does your "thanks in advance" mean you expect a solution? $\endgroup$
    – user217285
    Sep 10, 2016 at 7:16
  • $\begingroup$ Just looking for anything that can help me understand how to do the question, sorry - wasn't clear. I don't really have any idea on how to start. $\endgroup$ Sep 10, 2016 at 7:17
  • $\begingroup$ Begin by finding any basis for $V$. This can be done by noting $x_1 = 2x_2 - 3x_3 - x_4$. Plugging in $(x_2,x_3,x_4) = e_1,e_2,e_3$ should do the trick. Then set up an equation for the coordinates of $v_4$ that will guarantee orthogonality to the $v_i$. $\endgroup$
    – user217285
    Sep 10, 2016 at 7:22
  • $\begingroup$ Hint: the equation for elements of $V$ can be written as $(1,-2,3,1)\cdot(x_1,x_2,x_3,x_4)=0$. $\endgroup$
    – amd
    Sep 10, 2016 at 7:25
  • $\begingroup$ Would I be using the gram-schmidt process to determine the orthogonal vector? $\endgroup$ Sep 10, 2016 at 7:29

2 Answers 2

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Here is the approach I would take to solve this problem:

  1. Obtain three linearly independent vectors $v_1,v_2,v_3\in V$ by solving $x_1=2x_2-3x_3-x_4$ with exactly one of $x_2,x_3,$ and $x_4$ non-zero. Note that $\dim V=3$ because it is a hyperplane in $\mathbb R^4$.

  2. Find vectors $w_1,w_2,w_3\in\mathbb R^4$ with $w_i\perp v_i$ (this is especially simple because the $v_i$ have exactly two nonzero components).

  3. Set $v_4=w_1+w_2+w_3$ - by linearity, $v_4\in V^\perp$. Then $\{v_1,v_2,v_3,v_4\}$ is linearly independent and hence is a basis for $\mathbb R^4$.

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For this problem, you can read a solution from the equation $x_1 -2x_2 + 3x_3 +x_4=0$, which we can rewrite as $(1,-2,3,1)(x_1,x_2,x_3,x_4)^T=0$. Recognizing this expression as a dot product, this means that $v_4=(1,-2,3,1)^T$ is orthogonal to $V$.

If we consider the left multiplicand as the matrix of a linear map $L:\mathbb R^4\to\mathbb R$, the equation tells us that $V=\ker L$. This matrix is already in row-reduced echelon form, so we can use a standard technique to read a basis for $V$ from it without further ado: $v_1=(2,1,0,0)^T$, $v_2=(-3,0,1,0)^T$ and $v_3=(-1,0,0,1)^T$.

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