0
$\begingroup$

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 $v$.

Thus,

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

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $w=(0,1,-1)$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.