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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).

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  • $\begingroup$ I may not understand the notation, but is it not the case that for each $\mathbf{x}\in \mathbb{R}^3$, $[\mathbf{x}]_B=\mathbf{x}$? That is to say, $[\cdot]_B$ is the identity function on $\mathbb{R}^3$. $\endgroup$ Commented Sep 2, 2019 at 12:39
  • $\begingroup$ @AlbertoTakase The way i read is, $x$ is a vector, and $[x]_B$ is a column of three numbers describing the coefficients when the vector $x$ is decomposed into a linear combination of the three canonical basis vectors. $\endgroup$
    – Arthur
    Commented Sep 2, 2019 at 12:42
  • $\begingroup$ Essentially, the column vectors themselves (like $(1,0,0)$) don't mean anything unless some context, i.e. the basis being used, is given. $(1,0,0)$ isn't the same vector regardless of basis. Think about how the mass of $1$ pound differs from $1$ kilogram even though both are assigned the same 'value' (of $1$). $\endgroup$ Commented Sep 2, 2019 at 12:43
  • $\begingroup$ I see now (thanks to the two comments above): Implicitly there is a three dimensional vector space in the background. $\endgroup$ Commented Sep 2, 2019 at 12:46
  • $\begingroup$ All unit vectors point to a position on the unit sphere. If two vectors are labeled $(1,0,0)$ then those two vectors are the same vector. Given a single mathematician, he fixes 3 perpendicular directions on his sheet of paper. He labels these directions $(1,0,0)$, etc. All other bases must point to a value on his unit sphere. All you did was give the same basis two different names: B and V, because you gave them the same unit sphere values $\endgroup$
    – DWade64
    Commented Sep 2, 2019 at 13:22

2 Answers 2

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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.

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


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

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


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.

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    $\begingroup$ I think the equal signs can be a source of confusion. The main point of confusion in the OP was the mistaken belief that if the representation of $e_1$ in basis $B$ is the same as the representation of $v_1$ in the basis $V$, then the vectors are equal. Basically the OP was using the transitive property of equality. For this reason, I’ve seen some books use “$:=$” to emphasize that it should be read “is represented as”. $\endgroup$
    – Joe
    Commented Sep 2, 2019 at 13:38
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    $\begingroup$ @Joe Usually, I have seen $:=$ to mean "is defined as", and $\simeq$ as "is represented by". But yes, and overloaded equality sign is at (or at least close to) the core of the issue here. $\endgroup$
    – Arthur
    Commented Sep 2, 2019 at 14:19
  • $\begingroup$ I made an edit to my answer addressing the fact that I highlighted equalities "red" when I should not have. $\endgroup$ Commented Sep 2, 2019 at 14:34

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