Are there at least 17 different bases in which $(1,2,3)$ can be expressed as $(1,1,1)$?

Question

The exercise in the title comes from a Linear algebra test for Physics students from my university. I tried to solve it like follows, but I'm sure there's a lot of more general statements that can prove this.

The fact that the vector (which we shall call $\vec{v}$) can be expressed as $(1,2,3)$ in the canonical base of $\mathbb{R}^3$ means that $\vec{v}=1(1,0,0)+2(0,1,0)+3(0,0,1)$. If we wanted $\vec{v}=(1,1,1)_{B'}$, a possibility would be a certain base $B'$ in which $\vec{v}=(1,0,0)+(0,2,0)+(0,0,3)=(1,2,3)$. But we can change that base in a particular way so that $v$ can still be expressed as $(1,1,1)_{B'}$. It can be done like this:

$$B'=\{(1,0,n),(0,2,-n),(0,0,3)\}, n\in\mathbb{R}$$

Since there's an infinite number of real numbers, we can assure that there are at least 17 bases $B'$ in which $(1,2,3)$ can be expressed as $(1,1,1)_{B'}$.

How would you solve it? Can you think of a more general way?

Note: this exercise comes from a freshman course of Linear algebra and is worth 7.5% of the final grade of a midterm which is supposed to take 2.5 hours to complete, so please, as we say in my country, don't kill flies with cannon fire. Use, if possible, the tools a student would have in that situation.

Answer 1

Your approach looks exactly right. If you wanted to put it a more formal way, you might say:

Let $c$ be any number, and consider the vectors:

$$b_1 = \begin{bmatrix}1\\0\\c\end{bmatrix}\quad b_2=\begin{bmatrix}0\\2\\-c\end{bmatrix}\quad b_3 = \begin{bmatrix}0\\0\\3\end{bmatrix}$$

1. These vectors span $\mathbb{R}^3$, and therefore constitute a basis. To prove this, it's enough to show how to make the three standard basis vectors out of the $b_i$. And indeed, $$e_1 = b_1 - c/3\,b_3$$ $$e_2 = (b_2 + c/3 b_3)/2 $$ $$e_3 = b_3/3$$

2. The vector $\vec{v} \equiv 1e_1 + 2e_2 + 3e_3$ can be expressed as $1 b_1 + 1b_2 + 1b_3$ in this basis.

   Indeed, using the standard basis to represent these in matrix form, we have that: $$b_1 + b_2 + b_3 = \begin{bmatrix}1\\0\\c\end{bmatrix}+ \begin{bmatrix}0\\2\\-c\end{bmatrix}+\begin{bmatrix}0\\0\\3\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$$ by cancellation, as required.

3. Because each choice of $c$ yields a different basis set, it is clear that there is an infinite number of possible bases with this property, and in particular at least 17.