# Finding same-vectors that have same coordinates in two different basis

I have two different vector basis:

• Default: $\{e_1,e_2,e_3\} = \{(1,0,0);(0,1,0);(0,0,1)\}$

• Special basis: $\{e'_1,e'_2,e'_3\} = \{(1,1,1);(1,0,1);(0,2,1)\}$

My question is: How do I find which vectors (if there are any) whom have the same coordinates whether they are written in default or special basis vector space coordinates.

Example: The null vector has the coordinates $(0,0,0)$ in both default and special basis.

Through Gaussian elimination I established that the coordinates $(x_1,x_2,x_3)$ in regular basis, becomes $$([x'_1+x'_2];\,[x'_1+2x'_3];\,[x'_1+x'_2+x'_3])$$ after transformation to special basis.

Likewise, the coordinates $(x'_1,x'_2,x'_3)$ in special basis, becomes $$([2x_1+x_2-2x_3];\,[-x_1-x_2+2x_3];\,[-x_1+x_3])$$ after transformation to default basis.

As such, I tried setting:

$$\begin{cases} x_1 = 2x_1+x_2-2x_3 \\ x_2=-x_1-x_2+2x_3 \\ x_3 = -x_1+x_3 \end{cases}$$

Which has the trivial answer $x_1=x_2=x_3=0$.

Can I now conclude there are no (except the null vector) same-vectors which have same coordinates in both basis?

• I'm afraid I don't know what eigenvalues are yet. :Z – B. Lee Jan 24 '15 at 23:21
• for your basis change write a matrix and check their eigenvalues. For example wolframalpha.com/input/… – janmarqz Jan 24 '15 at 23:24
• @janmarqz Watched a video from Khan Academy on Eigenvalues and Eigenvectors. From what I understood, $\lambda$ needs to equal 1 in order for a vector to have the same coordinates in both basis. Does this mean there are no such vectors in my case? (Going by your Wolfram link) – B. Lee Jan 24 '15 at 23:55
• Updates my question to include an attempt which I'm unsure if it is enough to draw a conclusion from. – B. Lee Jan 25 '15 at 0:03
• since there is no eigenvalue = 1, then there is no vector with the same components in both bases, yes – janmarqz Jan 25 '15 at 3:11

Your solution looks perfectly valid. A very direct approach would be to say that you're looking for triples satisfying: $$xe_1+ye_2+ze_3=xe'_1+ye'_2+ze'_3$$ which doesn't require every doing any change of basis. Taking each coordinate alone, it yields the equations $$x=x+y$$ $$y=x+2z$$ $$z=x+y+z$$ which, when solved, yields $x=y=z=0$ as the only solution.
A slightly more sophisticated solution could take the above equations and rearrnage each to read, by subtracting lefthand side from righthand side: $$0=y$$ $$0=x-y+2z$$ $$0=x+y$$ where each equation defines a plane (or more conveniently, a linear subspace of dimension $2$) in $\mathbb R^3$ - the normal vectors to these planes being $(0,1,0),\,(1,-1,2),\,(1,1,0)$. The set of solutions to this will form a linear subspace, and thus be either a plane, a line, or a point (the origin). Since, clearly, $(0,1,0)$ and $(1,-1,2)$ are not parallel, the intersection of the planes normal to them is a line parallel to their cross product, $(2,0,-1)$. However, this line is not on the plane normal to $(1,1,0)$ (since their dot-product is not zero), thus its intersection with that plane is just a point - which must be the origin. More generally, if the product $$((e_1-e'_1)\times (e_2-e'_2))\cdot (e_3-e'_3)\neq 0$$ then the origin is the only vector satisfying this.