Prove that, if $\{u,v,w\}$ is a basis for a vector space $V$, then so is $\{u+v, v+w, u+v+w\}$. [duplicate]

I'm trying to prove the following statement:

In a vector space $V$ over a field $\mathbb{F}$, if $\{u,v,w\}$ is a basis for $V$, then $\{u+v, u+v+w, v+w\}$ is also a basis.

$$\underline{\text{My proof}}$$

Suppose that $\{u,v,w\}$ is a basis for $V$. We need to show that $\{u+v, u+v+w, v+w\}$ is linearly independent and spans $V$.

Now since $\{u,v,w\}$ is a basis, it is linearly independent, so the only solution to $\alpha u+\beta v+\gamma w=0$ is the trivial solution $\alpha=\beta=\gamma=0.$ In addition, $\{u,v,w\}$ spans $V$, so, for all $v_i \in V,$ we have $$v_i=au+bv+cw$$ for some $a,b,c \in \mathbb{F}.$

Proof that $\{u+v, u+v+w, v+w\}$ is linearly independent: Suppose that $\delta(u+v)+\varepsilon(u+v+w)+\zeta(v+w)=0.$ Then we have $$(\delta+\varepsilon)u+(\delta+\varepsilon+\zeta)v+(\varepsilon+\zeta)w=0.$$

But we've just proved that $\alpha u+\beta v+\gamma w=0 \implies \alpha=\beta=\gamma=0$, therefore, if we take $$\alpha=\delta+\varepsilon, \beta=\delta+\varepsilon+\zeta, \gamma=\varepsilon+\zeta,$$ then we have $$\begin{cases} \delta+\varepsilon=0 \\ \delta+\varepsilon+\zeta=0 \\ \varepsilon+\zeta=0\end{cases} \quad ,$$ the only solution to which is $\delta=\varepsilon=\zeta=0,$ hence $\{u+v, u+v+w, v+w\}$ is linearly independent.

Now we need to show that any $v_i$ can be written as some linear combination of $u+w, u+v+w, v+w.$

Let $a,b,c \in \mathbb{F}.$ If we can show that $$x_1(u+v)+x_2(u+v+w)+x_3(v+w)=au+bv+cw$$ has a solution (for $x_1, x_2, x_3 \in \mathbb{F}$, then it follows immediately that $\{u+v, u+v+w, v+w\}$ spans $V$.

Proof that $\{u+v, u+v+w, v+w\}$ spans $V$: Now, $$x_1(u+v)+x_2(u+v+w)+x_3(v+w)=au+bv+cw \\ \ \\\iff (x_1+x_2)u+(x_1+x_2+x_3)v+(x_2+x_3)w=au+bv+cw \\ \ \\ \iff \begin{cases} x_1+x_2=a \\ x_1+x_2+x_3=b \\ x_2+x_3=c\end{cases}$$

By substitution (a trivial exercise, whose steps I have left out), we get $$\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}=\begin{bmatrix} b-c \\ a-b+c \\ b-a\end{bmatrix} \quad ,$$ hence every $v_i \in V$ can be written in the form $x_1(u+v)+x_2(u+v+w)+ x_3(v+w)$, hence $\{u+v, u+v+w, v+w\}$ spans $V$.

Since $\{u+v, u+v+w, v+w\}$ is both linearly independent and spans $V$, it is a basis for $V$. $\square$ Note: you may be wondering: why are $b-c, \quad a-b+c, \quad b-a \in \mathbb{F}$? Answer: Since $\mathbb{F}$ is a field, it is closed under addition and subtraction, and $a,b,c$ were chosen to be in $\mathbb{F}.$

Question:

Is this proof correct? Can it be shortened?

Thanks!

Solution 1:

The determinant of $\{u+v, u+v+w, v+w\}$ relative to the given basis is

$$\det\begin{pmatrix}1&1&0\\1&1&1\\0&1&1\end{pmatrix}=-1\ne0$$ and the result follows.

Solution 2:

You don't need to do nearly so much work to show that you have a spanning set.

Note that $$u=(u+v+w)-(v+w)$$ is obvious. As is$$w=(u+v+w)-(u+v)$$

Then note that $$v=(u+v)-u=(u+v)-(u+v+w)+(v+w)$$

Since you can express $u,v,w$ in terms of the three vectors you are given, anything you can express in terms of $u,v,w$ can be expressed in terms of $(u+v), (u+v+w), (v+w)$. This proves the vectors span the space.