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Find an orthonormal basis for the subspace of $\mathbb{R}^4$ that consists of vectors perpendicular to $u = (1, -1, -1, 1)$.

I know the components of the vector $u$ is $u_1 = 1, u_2 = -1, u_3 = -1, u_4 = 1$.

I managed to find a vector $v$ that is perpendicular to the $u$.

This is done by $u_1v_1+ u_2v_2+ u_3v_3 + u_4v_4 = 0$.

This means $v_1 = -1, v_2 = 1, v_3= 1, v_4 = -1$ because $(1)(-1) + (-1)(1) + (-1)(1) + (1)(-1) = 0$.

So I found a vector $v$ perpendicular to the given vector $u$ which is $(-1, 1, 1, -1)$.

So my question would I be able to call $(-1, 1, 1, -1)$ my basis and perform Gram Schmidt process on it?

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  • $\begingroup$ Hi John. If I find two more vectors w and x perpendicular to the given and the other one I found then? If that is done will those 4 vectors be my basis? $\endgroup$
    – user983246
    Commented Dec 11, 2013 at 9:14
  • $\begingroup$ You need to find an orthonormal basis for the subspace of that consists of vectors $\perp$ to $u$. If you can find 3 such vectors which are linearly independent, then you can start doing gram-schmidt process. $\endgroup$
    – user99914
    Commented Dec 11, 2013 at 9:17
  • $\begingroup$ Thanks John. Thanks for adding the extra tag too! $\endgroup$
    – user983246
    Commented Dec 11, 2013 at 9:20

2 Answers 2

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A vector $v=(v_1,v_2,v_3,v_4)$ is orthogonal to $u$ iff $v_1-v_2+v_3-v_4=0$, or in other words $v_4=v_1-v_2+v_3$.

Then

$$ v=\left(\begin{matrix} v_1 \\ v_2 \\ v_3 \\ v_1-v_2+v_3 \\ \end{matrix}\right)= v_1\left(\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \\ \end{matrix}\right)+ v_2\left(\begin{matrix} 0 \\ 1 \\ 0 \\ -1 \\ \end{matrix}\right)+ v_3\left(\begin{matrix} 0 \\ 0 \\ 1 \\ 1 \\ \end{matrix}\right) $$

So you can start Gram-Schmidt on those three vectors.

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  • $\begingroup$ I'm confused are those components in that vector part of an outline to help me or are they actual vectors that could work with the the problem. Because the bottom components of each vector (1)(-1)(1) would be -1 instead of 0. And I do not think those 3 vectors are perpendicular to the given vector. $\endgroup$
    – user983246
    Commented Dec 11, 2013 at 9:31
  • $\begingroup$ @user983246 Those three vectors are indeed orthogonal to $u$ (that’s what my post shows), but not orthogonal between themselves. That’s why you need to apply Gram-Schmidt. So it's just an outline, not a full solution. $\endgroup$
    – Ewan Delanoy
    Commented Dec 11, 2013 at 9:40
  • $\begingroup$ When you mean u are you talking about my vector u that I posted or your own u? $\endgroup$
    – user983246
    Commented Dec 11, 2013 at 9:45
  • $\begingroup$ @user983246 Of course I'm speaking about your own $u$ $\endgroup$
    – Ewan Delanoy
    Commented Dec 11, 2013 at 9:53
  • $\begingroup$ Thanks for clearing it up my confusion, Ewan. Here is your check! $\endgroup$
    – user983246
    Commented Dec 11, 2013 at 9:57
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Here's a way without Gram-Schmidt.

Form the matrix $A$ with row $u$. Find one vector $v$ in the nullspace of $A$.

Form the matrix $B$ with rows $u,v$. Find one vector $w$ in the nullspace of $B$.

Form the matrix $C$ with rows $u,v,w$. Find a vector $x$ in the nullspace of $C$.

Now $v,w,x$ form an orthogonal basis of the space of vectors orthogonal to $u$. To make it orthonormal, just divide each by its length.

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  • $\begingroup$ That's an another interesting recommendation! $\endgroup$
    – user983246
    Commented Dec 11, 2013 at 10:01

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