If $\{e_1,e_2,e_3\}$ and $\{v_1,v_2,v_3\}$ are two different basis, $[v]_B\neq [v]_V$. So what is wrong in my argument

Question

What's wrong here ? Let $B=\{e_1,e_2,e_3\}$ the canonical basis of $\mathbb R^3$ and let $V\{v_1,v_2,v_3\}$ an other basis of $\mathbb R^3$. I denote $$\left[\begin{pmatrix}x\\ y\\ z\end{pmatrix}\right]_B=xe_1+ye_2+ze_3.$$

Let $v=\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\in\mathbb R^3$.

I know that $$[e_1]_B=[v_1]_V=\begin{pmatrix}1\\0\\0\end{pmatrix},\quad [e_2]_B=[v_2]_V=\begin{pmatrix}0\\1\\0\end{pmatrix}\quad \text{and}\quad [e_3]_B=[v_3]_V=\begin{pmatrix}0\\0\\1\end{pmatrix}.$$

So $$[v]_B=\left[\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\right]_B=1\left[e_1\right]_B+2\left[e_2\right]_B+3\left[e_3\right]_B=1\begin{pmatrix}1\\0\\0\end{pmatrix}+2\begin{pmatrix}0\\1\\0\end{pmatrix} +3\begin{pmatrix}0\\0\\1\end{pmatrix}=1[v_1]_V+2[v_2]_V+3[v_3]_V=[v]_V.$$ What is wrong in my argument ? (I know it's wrong since the writing of $v$ in $B$ and in $V$ should be different).

Answer 1

In your middle step you have $$ 1\begin{pmatrix}1\\0\\0\end{pmatrix}+2\begin{pmatrix}0\\1\\0\end{pmatrix} +3\begin{pmatrix}0\\0\\1\end{pmatrix} $$ Those columns still represent three vectors, expressed in some basis. Are they $[e_i]_B$, or are they $[v_i]_V$? Just because the components of $[e_i]_B$ (for some $i$) are the same as the components of $[v_i]_V$, that doesn't mean they represent the same vectors. A vector represented by a column of numbers only makes sense when the basis used is known and fixed.

Answer 2

I see two problematic parts: the "red" and "blue" parts. I highlighted them from your argument. I hope you are not colorblind.

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$\color{red}{\text{Let }v=\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\in\mathbb R^3}$.

I know that $$[e_1]={\color{red}{[v_1]_V}}=\begin{pmatrix}1\\0\\0\end{pmatrix},\quad [e_2]_B={\color{red}{[v_2]_V}}=\begin{pmatrix}0\\1\\0\end{pmatrix}\quad \text{and}\quad [e_3]_B={\color{red}{[v_3]_V}}=\begin{pmatrix}0\\0\\1\end{pmatrix}.$$

So $$[v]_B=\left[\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\right]_B=1\left[e_1\right]_B+2\left[e_2\right]_B+3\left[e_3\right]_B=1\begin{pmatrix}1\\0\\0\end{pmatrix}+2\begin{pmatrix}0\\1\\0\end{pmatrix} +3\begin{pmatrix}0\\0\\1\end{pmatrix}=1{{[v_1]_V}}+2{{[v_2]_V}}+3{{[v_3]_V}}{\color{blue}{{}=[v]_V}}.$$

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The blue equality you claim may not necessarily be true.

$v_1$, $v_2$, and $v_3$ have conflicting meaning. They are introduced as a basis; $v_i$ are then vectors. Then when $v:=(1,2,3)$ one could interpret $v_i$ to be the $i$-th projection of $(1,2,3)$; $v_i$ are then scalars.