Is this proof correct? If {u,v,w} is a basis for V then {u+v+w,v+w,w}is also a basis?

  • Thread starter aortizmena
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Let u,v,w[itex]\in[/itex] V a vector space over a field F such that u≠v≠w. If { u , v , w } is a basis for V. Prove that { u+v+w , v+w , w } is also a basis for V.

Proof

Let u,v,w[itex]\in[/itex] V a vector space over a field F such that u≠v≠w. Let { u , v , w } be a basis for V. Because { u , v , w } its a basis, then u,v,w are linearly independent and <{ u , v , w }>=V.

Let x[itex]\in[/itex]V be an arbitrary vector then x can be uniquely expressed as a linear combination of { u , v , w }. Lets suppose x=au+bv+cw for some a,b,c[itex]\in[/itex]F.

On the other hand ,lets consider { u+v+w , v+w , w }[itex]\subseteq[/itex]V.

Then <{ u+v+w , v+w , w }>={d(u+v+w) + e(v+w) + f(w) | d,e,f[itex]\in[/itex]F}={du + (d+e)v +(d+e+f)w | d,e,f[itex]\in[/itex]F}.

If x[itex]\in[/itex]V then x=du + (d+e)v +(d+e+f)w its another unique representation of x[itex]\in[/itex]V . Then for any arbitrary x[itex]\in[/itex]V we have d=a, d+e=b and d+e+f=c [itex]\in[/itex]F.

Because { u , v , w } its a basis fpr V then { u+v+w , v+w , w } must also be a basis for V.

Edit:
I tried to give an alternate proof instead of proving <{ u , v , w }>=V<{ u+v+w , v+w , w }>
 
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Answers and Replies

  • #2
Hurkyl
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There's a bit of imprecision here, but I think you've actually omitted something important:

Then for any arbitrary x[itex]\in[/itex]V we have d=a, d+e=b and d+e+f=c [itex]\in[/itex]F.

In particular, you haven't shown that there is a unique solution for (d,e,f) in that system of equations.



To help understand this is important, try repeating your argument for {u-v, u-w, v-w}. Would it conclude that this is also a basis for V?
 
  • #3


Thanks Hurkyl but isn't saying that for for any arbitrary x∈V=<{ u , v , w }> with d=a, d+e=b and d+e+f=c it implies the unique solution d=a, e=b-d=b-a , f=c-d-e=c-a-(b-a)=c-a-b+a=c-b ?
 
  • #4
Hurkyl
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Thanks Hurkyl but isn't saying that for for any arbitrary x∈V=<{ u , v , w }> with d=a, d+e=b and d+e+f=c it implies the unique solution d=a, e=b-d=b-a , f=c-d-e=c-a-(b-a)=c-a-b+a=c-b ?

Yes, but I think it needs saying when coming from a student of linear algebra -- the professor needs to know you actually thought about that. (did you think about that? Or did you not think about it until I pointed it out)
 
  • #5


I thought it was implicit, that was my intention
 
  • #6
chiro
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Hey aortizmena and welcome to the forums.

Do you know how to show something is a basis using matrices and the properties of matrices?

Do you know the properties of basis in the context of linear algebra? Perhaps you should list them and go through them one by one on this forum as a way to think out aloud and to also show us your thinking so that we can help you correct it if need be.
 
  • #7


Thanks chiro, yes i know alternatives of proving the statement, i was just trying to give a proof with this alternative aproach. I know i can prove it by matrix properties, by linear independence of the vectors, and by proving <{ u , v , w }>=V<{ u+v+w , v+w , w }>.

Thanks though.
 
  • #8
Deveno
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Thanks chiro, yes i know alternatives of proving the statement, i was just trying to give a proof with this alternative aproach. I know i can prove it by matrix properties, by linear independence of the vectors, and by proving <{ u , v , w }>=V<{ u+v+w , v+w , w }>.

Thanks though.

you've shown spanning (although it's not real easy to decipher), but you haven't shown linear independence (this is the "uniqueness" part).

what i recommend is this:

show {u+v+w,v+w,w} is linearly independent. spanning is the easy part.

(because w is in span({u+v+w,v+w,w}), and v = 1(v+w) + (-1)w, so v is in the span, and u = ......?)

you are "almost" at the point where you've shown you have a basis. to finish it, if you wish to make your particular argument, you need to invoke dimensionality in some way.
 
  • #9
Hurkyl
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I thought it was implicit, that was my intention
But I'm asserting that it is wrong to leave it implicit in this context. (which, I assume, is something akin to a homework problem or class exercise in a first introduction to linear algebra)

Recognizing that the step must be done (and being able to do it) is the sort of thing I would expect a fellow mathematician to know, but I would expect a new student is fairly likely to get wrong.

If I were writing a similar proof for a colleague, I would be fairly likely to omit that last derivation. But I would also be fairly likely to actually mention it too. (without explicitly writing out the steps)

If I were writing a similar proof for a student of linear algebra to see, I would almost certainly write out that step. (unless I decided to state what needed to be done and leave it to the student to see if he can carry it out)
 

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