S is set of all vectors of form (x,y,z) such that x=y or x =z. Basis?

[Hall]

$S$ is a set of all vectors of form $(x,y,z)$ such that $x=y$ or $x=z$. Can $S$ have a basis?

S contains either $(x,x,z)$ type of elements or $(x,y,x)$ type of elements.

Case 1: $(x,x,z)= x(1,1,0)+z(0,0,1)$
Hencr, the basis for case 1 is $A = \{(1,1,0), (0,0,1)$\}

And similarly for case 2 the basis would be $A'= \{ (1,0,1), (0,1,0)\}$.

But how to find the basis for $S$? A union of A and A' would give us a set whose linear span would go beyond S and hence cannot be a basis for S. Can S have a basis? How do we find it?

 

[PeroK]

Perhaps you should check whether $S$ is a vector (sub-) space?

 

[Hall]

Perhaps you should check whether $S$ is a vector (sub-) space?

$(x,x,z)$, $(x,y,x)$ $\in S$

$(2x, x+y, x+z)$ is not in S. S is not a subspace.

Oh yes, thanks.

 

[Office_Shredder]

I think that's 95% of a proof, but if you want to be very technical, you should prove the thing you wrote down isn't actually of the form (x'x',z') or (x',y',x'). This is typically done by actually picking specific x,y,z and observing that it doesn't work.

Also, in case you didn't realize, you didn't actually pick generic elements of S, since you forced x to be the same in both of them. In some cases you could get unlucky and trick yourself into thinking it is a subspace doing this (obviously it works out ok here)

 

[Hall]

I think that's 95% of a proof, but if you want to be very technical, you should prove the thing you wrote down isn't actually of the form (x'x',z') or (x',y',x'). This is typically done by actually picking specific x,y,z and observing that it doesn't work.

Also, in case you didn't realize, you didn't actually pick generic elements of S, since you forced x to be the same in both of them. In some cases you could get unlucky and trick yourself into thinking it is a subspace doing this (obviously it works out ok here)

Well, the 2nd paragraph is really pedantic. Thanks.

 

[Mark44]

Well, the 2nd paragraph is really pedantic.

On the contrary, I think @Office_Shredder makes a cogent point.