Let u = (1, 2, 3, 4) and v = (4, 3, 2, 1) be two vectors in $R^4$. These vectors define the subset of $R^4$
$V = \{x \in R^4 | u \bullet x = 0$ and $v \bullet x = 0\}$
Here $u \bullet x$ denotes the dot product of the two vectors u and x, and $v \bullet x$ denotes the dot product of the two vectors v and x
Question: Find a basis of V and explain why the basis you have found is a basis
I have done many basis finding examples with linear algebra but I stuck with this dot product one, this is my first thought of the problem:
Because $u \bullet x = 0$ and $v \bullet x = 0$ then $u \bullet x = v \bullet x$
Then using the dot product property, we will have $(u-v)x=0$ then I substitute u and v in and use gaussian row reduce echelon to solve it
Can you guys help me if the way I did is a correct way or do you guys have any better alternative way that I can solve this problem? Appreciate a lot.
You are after a basis of the set of solutions of the system$$\left\{\begin{array}{l}x+2y+3z+4t=0\\4x+3y+2z+t=0.\end{array}\right.$$If you multiply the first equation by $-4$ and then add it to the second one, you will get$$\left\{\begin{array}{l}x+2y+3z+4t=0\\-5y-10z-15t=0.\end{array}\right.$$Now, dividing the second equation by $-5$, you get$$\left\{\begin{array}{l}x+2y+3z+4t=0\\y+2z+3t=0.\end{array}\right.$$And now, if you replace the first equation by the first equation minus twice the second one, you get$$\left\{\begin{array}{l}x-z-2t=0\\y+2z+3t=0.\end{array}\right.$$So,$$V=\{(z+2t,-2z-3t,z,t)\mid z,t\in\Bbb R\},$$a basis of which is $\{(1,-2,1,0),(2,-3,0,1)\}$.
Your approach does not work, because$$\{x\in\Bbb R^4\mid u\bullet x=v\bullet x=0\}\varsubsetneq\{x\in\Bbb R^4\mid(u-v)\bullet x=0\}.$$