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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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