Let's say I have a canonical coordinate system in $\mathbb{R}^2$ described by the basis $\{e_1=\begin{bmatrix}0\\1\end{bmatrix}, e_2=\begin{bmatrix}1\\0\end{bmatrix}\}$.

In it I have a vector $\vec{x}=\begin{bmatrix}2\\1\end{bmatrix}$. I also have two perpendicular vectors $\vec{v_1}=\begin{bmatrix}1\\2\end{bmatrix}$ and $\vec{v_2}=\begin{bmatrix}2\\-1\end{bmatrix}$ that could describe a local coordinate system. This coordinate's system basis would be $B=\{\begin{bmatrix}2\\-1\end{bmatrix}, \begin{bmatrix}1\\2\end{bmatrix}\}=\{v_1, v_2\}$.

I want to transform $\vec{x}$'s coordinates from the canonical basis to $B$. From watching this Khan s Academy video I understood that to transform from coordinates in $B$, e.g. $[\vec{x}]_B$ back to the canonical coordinate system I could calculate $[\vec{x}]_Bc$ where $c=\begin{bmatrix}2&1\\-1&2\end{bmatrix}$ and to do the opposite I could do $\vec{x}c^{-1}=[\vec{x}]_B$.

So I found $c^{-1}=\begin{bmatrix}2/5&-1/5\\1/5&2/5\end{bmatrix}$.

But $\vec{x}c^{-1}=\begin{bmatrix}2\\1\end{bmatrix}\begin{bmatrix}2/5&-1/5\\1/5&2/5\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix}=[\vec{x}]_B$. I don't understand this result. I thought I could now sketch a graph with an x-axis and a y-axis and put this resulting vector into it, and the axis would be representing my $\vec{v_1}, \vec{v_2}$ and the calculated coordinates would draw my vector $\vec{x}$ into this graph.

I thought this should look more like this.


1 Answer 1


Everything is fine except one mistake. This is where you made mistake

$\vec{x}c^{-1}=\begin{bmatrix}2\\1\end{bmatrix}\begin{bmatrix}2/5&-1/5\\1/5&2/5\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix}$

order of multiplication matters. such a multiplication is not defined. you should write it as

$\vec{x}c^{-1}=\begin{bmatrix}2/5&-1/5\\1/5&2/5\end{bmatrix} \begin{bmatrix}2\\1\end{bmatrix}= \begin{bmatrix}\frac{3}{5}\\\frac{4}{5}\end{bmatrix}$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .