# Prove that $\alpha+\beta+\gamma=0$

Let $\mathbb{R}^2=Span\{\alpha,\beta,\gamma\}$. Does anybody know how to prove that if there is a linear tranformation $T: \mathbb{R}^2\to \mathbb{R}^2$ such that $T(\alpha)=\beta, T(\beta)=\gamma, T(\gamma)=\alpha$, it is $\alpha+\beta+\gamma=0$?

• Think about what $\alpha + \beta + \gamma = 0$ means geometrically. It means that after the transformation is applied, if you connect each vector head-to-tail, you'll end up back where you started. Dec 12 '17 at 18:38

## 4 Answers

$\{\alpha,\beta,\gamma\}$ is linearly dependent (otherwise $\dim(\mathbb{R}^2)=3$, a contradiction) so $c_1\alpha+c_2\beta+c_3\gamma=0$ where $c_1,c_2,c_3$ are not all zero.

Then $0=T(c_1\alpha+c_2\beta+c_3\gamma)=c_1T(\alpha)+c_2T(\beta)+c_3T(\gamma)=c_1\beta+c_2\gamma+c_3\alpha$

Similarly, we can get $0=c_1\gamma+c_2\alpha+c_3\beta$

Adding these up we get $(c_1+c_2+c_3)(\alpha+\beta+\gamma)=0$

Do you think you can show that $c_1+c_2+c_3\ne 0$? I encourage you to figure this out for yourself, so I will show this in a spoiler.

Since $\mathrm{span}\{\alpha,\beta,\gamma\}=\mathbb{R^2}$ we have that the column space of the matrix $A=\begin{bmatrix}\alpha&\beta&\gamma \end{bmatrix}$ has dimension $2$. So $\mathrm{nullity}(A)=1$ by the Rank-Nullity Theorem. Therefore, the nullspace of $A$ is spanned by one vector. But we know that $[c_1,c_2,c_3]^T$, $[c_3,c_1,c_2]^T$, and $[c_2,c_3,c_1]^T$ are nonzero vectors in the nullspace of $A$, so they all must be parallel. From this, we easily find that $c_1=c_2=c_3$ and they are not all zero, so $c_1+c_2+c_3\ne 0$

• I don't understand. Suppose $\gamma = \frac12(\alpha+\beta)$ to start with. Then won't you in fact have $c_1+c_2+c_3=0$? Dec 12 '17 at 19:05
• @TedShifrin Yes, but then there either exists no such linear transformation, or $\{\alpha,\beta,\gamma\}$ does not span $\mathbb{R}^2$ Dec 12 '17 at 19:08

First, $T^3=I$. So the minimal polynomial is $p\in\mathbb R[t]$, with $\deg(p)\le2$, and such that $p$ divides $$t^3-1=(t-1)(t^2+t+1).$$ Since $t^2+t+1$ is irreducible in $\mathbb R[t]$, the minimal polynomial is $t-1$ or $t^2+t+1$. But $T\ne I$. So the minimal polynomial is $t^2+t+1$.

Hence $$\alpha+\beta+\gamma=(I+T+T^2)\alpha=0.$$

• very nice proof. Could you please give your advice on mine, do you think it can work? Thanks
– user
Dec 12 '17 at 20:04

Since in $\mathbb{R}^2$ you can't have more of two vectors linearly independent, thus if $$Span\{\alpha,\beta,\gamma\}=\mathbb{R}^2$$ one vector must be a linear combination of two.

Let's exclude the trivial cases with 2 vectors collinear, assume wlog:

$$\gamma = a\alpha + b\beta$$

with $a,b\neq0$

By definition of T:

$$\gamma=T(\beta)=T(\frac1b \gamma-\frac{a}{b}\alpha)=\frac1b \alpha-\frac{a}{b}\beta$$

Since the representation of $\gamma$ is unique:

$$a=\frac1b$$

$$b=-\frac{a}{b}$$

$$\implies a=b=-1 \implies \alpha+\beta+\gamma=0 \quad \square$$

• That doesn't show $\alpha+\beta++\gamma=0$. Dec 12 '17 at 18:38
• This doesn't answer the question; the OP wants to show a specific linear combination of $\{\alpha, \beta, \gamma\}$ is $0$,, not just that one combination exists. Dec 12 '17 at 18:38
• You are right sorry, I complete teh answer!
– user
Dec 12 '17 at 18:39
• I’ve completed the answer!
– user
Dec 12 '17 at 19:46
• You didn't explain why $a=\frac1b$. You need to explain why there can't be 2 different linear combinations of $\alpha$ and $\beta$ to give you $\gamma$. This is what the spoiler in my answer explains. Dec 12 '17 at 19:46

Let's assume wlog $\alpha$ and $\beta$ as basis and $\gamma=a\alpha+b\beta$, thus T is represented by:

$$A=\begin{bmatrix}0 & a \\ 1 & b\end{bmatrix}\implies T(\gamma)=\begin{bmatrix}0 & a \\ 1 & b\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}ab \\ a+b^2\end{bmatrix}=\alpha=\begin{bmatrix}1 \\ 0\end{bmatrix}$$

That is

$$\begin{cases}ab=1\\a+b^2=0\end{cases}\implies\begin{cases}a=\frac1b\\b^3+1=0\end{cases}\implies a=b=-1\implies \alpha+\beta+\gamma=0 \quad \square$$

NOTE

Since ab=1, trivial cases with a=0 or b=0 are excluded.