I'm new to more formal mathematics and I was wondering if I could write my attempt at a small proof here and see what others thought and if you could point out any fatal flaws or just offer some tips? I would appreciate it.


Let u,v, and w be distinct vectors in a vector space V. Show that if {u,v,w} is a basis for V, then {u+v+w, v+w, w} is also a basis for V.


: Let B = {u,v,w}. As B is a basis for V, the vectors in B generate V and are linearly independant. Therefore, the equation

a1 * u + a2 * v + a3 * w = 0

Is only true for a1, a2, a3 all equal to 0.

Now consider the set B' = {u+v+w, v + w, w}

If we examine the equation

c1 ( u+v+w) + c2 (v+w) + c3 (w) = 0

c1(u) + (c1 + c2)v + (c1 + c2 + c3)w = 0

Then we say that c1 = -c2, and c3 = -c1-c2, but from the first term we see that c1 must equal zero, therefore all the other coefficients are 0 as well, and so the set of vectors is linearly independant.

My questions

1.) I'm not sure if the part of the proof where I show that the solution to the equation for the vectors of B' being equal to 0 is connected well enough to the bit about B being a basis for V. Am I allowed to say that C1 must be equal to 0 only because I've already assumed that u,v, and w are all linearly independent already?

2.) I know that if B generates V that B' which is just set formed from L.I. vectors into other L.I. vectors must also generate V, but I'm not sure exactly why I can say this even though I feel it strongly.

  • 1
    $\begingroup$ Your argument is perfect. $\endgroup$ – Ted Shifrin Sep 15 '14 at 1:20

With regards to point 1), what you say is correct. As you say, you use the linear independence of $B$ to conclude that when $c_1(u) + (c_1 + c_2)v + (c_1 + c_2 + c_3)w = 0$, it must be the case that $c_{1} = 0$, and so on.

With regards to 2), one way to show that $B'$ spans the space is to show that it generates $B$: any linear combination of vectors from $B$ can then be converted into a linear combination of vectors from $B'$ by replacing $u,v,w$ by their expressions in terms of the list $B'$. So in this case $(v+w) - w = v$ and $(u+v+w) - (v+w) = u$.

By the way, you only need to show either that $B'$ form a linear independent set, or that they span the vector space, since the list has length $3$. (This is the result that a linear independent list of vectors, whose length is equal to the dimension of the space forms a basis, and similarly for spanning sets).


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