Find an orthogonal basis of ${\rm I\!R}^3$ which contains the vector $v=\begin{bmatrix}1\\1\\1\end{bmatrix}$.

I want to solve this without the use of the cross-product or G-S process. Please look at my solution and let me know if I did it right.

Let $u$ be an arbitrary vector $u=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}$ that is orthogonal to $v$.


$v\ \bullet\ u = x_1 + x_2 + x_3 = 0$
$0= x_1 + x_2 + x_3$
$x_1= -x_2 -x_3$

Any vector of the form $\begin{bmatrix}-x_2 -x_3\\x_2\\x_3\end{bmatrix}$ will be orthogonal to $v$.

Let $x_2 = x_3 = 1$
So, $u=\begin{bmatrix}-2\\1\\1\end{bmatrix}$ is orthogonal to $v$.

We now have two orthogonal vectors $u$ and $v$.

Put $u$ and $v$ as rows of a matrix, called $A$. Find a basis for $A^\bot = null(A)^T$:

Digression: I have memorized that when looking for a basis of $A^\bot$, we put the orthogonal vectors as the rows of a matrix, but I do not know why we put them as the rows and not the columns. Anyone care to explain the intuition?

$A=\begin{bmatrix}1&1&1\\-2&1&1\end{bmatrix} \sim \begin{bmatrix}1&0&0\\0&1&1\end{bmatrix}$
$x_3 = x_3$
$x_2 = -x_3$
$x_1 = 0$

A basis for $null(A)$ or $A^\bot$ with $x_3$ = 1 is: $(0,-1,1)$. Call this $w$.

Therefore, $w$ is orthogonal to both $u$ and $v$ and is a basis which spans ${\rm I\!R}^3$.
By definition of orthogonal vectors, the set $[u,v,w]$ are all linearly independent.

Is this correct? If so, what is a more efficient way to do this? If not, how do you do this keeping in mind I can't use the cross product G-S process?

  • $\begingroup$ 1st: I think you mean (Col A)$^\perp$ instead of A$^\perp$. Anyway, to answer your digression, when you multiply Ax = b, note that the i-th coordinate of b is the dot product of the i-th row of A with x. Now suppose x$\in$ Nul(A). Then b = 0, and so every row is orthogonal to x. $\endgroup$ – Joel Pereira Nov 12 '18 at 4:13

If you use the same reasoning to get $w=(x_1,x_2,x_3)$ (that you did to get $v$), then $0=v\cdot w=-2x_1+x_2+x_3$. So in general, $(\frac{x_2+x_3}2,x_2,x_3)$ will be orthogonal to $v$.

Now we get $-x_2-x_3=\frac{x_2+x_3}2$ (since $w$ needs to be orthogonal to both $u$ and $v$).

So, $-2x_2-2x_3=x_2+x_3$. Then $x_2=-x_3$.

So, say $x_2=1,x_3=-1$. Then we get $w=(0,1,-1)$.

  • $\begingroup$ Can you clarfiy why $−x2−x3=\frac{x2+x3}{2}$ tells us that $w$ is orthogonal to both $u$ and $v$? I can't immediately see why. $\endgroup$ – udpcon Nov 12 '18 at 4:13
  • 1
    $\begingroup$ We need a vector which simultaneously fits the patterns gotten by setting the dot products equal to zero. I've set $(-x_2-x_3,x_2,x_3)=(\frac{x_2+x_3}2,x_2,x_3)$. $\endgroup$ – Chris Custer Nov 12 '18 at 4:29

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