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There is a theorem in the course notes but the lecturer didn't given the proof:

If $A,B\in M_n(K)$, where $K$ is a field. Then $A\sim B$ if and only if they represent the same linear transformation $T$ of an $n$-dimensional vector space $V$ over $K$ with respect to bases $\mathcal{B}$ and $\mathcal{C}$. i.e., $$ A\sim B\iff A=[T]_\mathcal{B},\ B=[T]_{\mathcal{C}}. $$

I know how to prove the '$\Leftarrow$' but I dont think I can prove the other direction '$\Rightarrow$'.

Could please somebody help me?

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  • $\begingroup$ Try with $T=$ multiplication by $A$ in the standard basis of $K^n$. $\endgroup$ – Quimey May 29 '14 at 14:41
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$A\sim B$ means that there exists an invertible matrix $P$ such that $B=P^{-1}AP$. An invertible matrix $P$ can be considered as the change of basis matrix $P=[I]^E_C$ from the standard basis $E$ of $K^n$ to the basis $C$ of consisting of the columns of the matrix $P$. Thus if you define the linear transformation $$T:K^n\to K^n$$ by $$T(v)=B\cdot v$$ you will get that $[T]_E=B$ and $[T]_C=[I]^E_C\cdot B\cdot [I]^C_E=PBP^{-1}=A$.

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  • $\begingroup$ his question is from an abstract vector space. Not $T: \mathbb{K}^n \rightarrow \mathbb{K}^n$. $\endgroup$ – James S. Cook May 29 '14 at 14:49
  • $\begingroup$ I don't see your point @JamesS.Cook. The OP wants to show that THERE EXISTS an $n$-dimensional vector space $V$ over $K$ such that... What is the problem with taking the $n$-dimensional vector space $V=K^n$? $\endgroup$ – DKal May 29 '14 at 14:53
  • $\begingroup$ Ah, yes, I retract my stupid comment +1. I suppose I'm interested in the slightly more obtuse question involving an abstract vector space. $\endgroup$ – James S. Cook May 29 '14 at 15:02
  • $\begingroup$ So why do A and B represent the same linear transformation? $\endgroup$ – JUNG WON CHO Nov 23 '18 at 23:57
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If $A \sim B$ then there exists an invertible matrix $P$ for which $B=P^{-1}AP$. Pick a basis $\beta = \{ v_1, \dots, v_n \}$ for $V$ over $\mathbb{K}$ and let $\Phi_{\beta}: V \rightarrow \mathbb{K}^n$ form the coordinate map defined by linearly extending $\Phi_{\beta}(v_i)=e_i$ where $(e_i)_j = \delta_{ij}$ for all $1 \leq i,j \leq n$. Let $Pe_i = \bar{e}_i$ and define $\bar{v}_i=\Phi_{\beta}^{-1}(\bar{e}_i)$. Suppose $T: V \rightarrow V$ is the linear transformation for which $[T]_{\beta,\beta}=A$. In other words, let $T$ be the linear transformation which is represented by $A$ in the $\beta$ basis. In particular, $A = [ \Phi_{\beta} \circ T \circ \Phi_{\beta}^{-1} ]$ where $[..]$ denotes the standard matrix.

I claim that $B = [T]_{\bar{\beta},\bar{\beta}}$.

Proof: Essentially, by definition, $$ [T]_{\bar{\beta},\bar{\beta}} = [ \Phi_{\bar{\beta}} \circ T \circ \Phi_{\bar{\beta}}^{-1} ]$$ But, the coordinate maps are isomorphisms and we can insert the $\beta$-maps as follows: \begin{align} [T]_{\bar{\beta},\bar{\beta}} &= [ \Phi_{\bar{\beta}} \circ \Phi_{\beta}^{-1} \circ \Phi_{\beta} \circ T \circ \Phi_{\beta}^{-1} \circ \Phi_{\beta}^{-1}\Phi_{\bar{\beta}} ] \\ &= [ \Phi_{\bar{\beta}} \circ \Phi_{\beta}^{-1}][\Phi_{\beta} \circ T \circ \Phi_{\beta}^{-1}][\Phi_{\beta} \circ \Phi_{\bar{\beta}}^{-1}] \end{align} Ok, but what is $[\Phi_{\beta} \circ \Phi_{\bar{\beta}}^{-1}]$? Recall $Pe_i = \bar{e}_i$ and $\bar{v}_i=\Phi_{\beta}^{-1}(\bar{e}_i)$. Thus $\bar{e}_i = P^{-1}e_i$ and $\Phi_{\beta}(\bar{v}_i)=\bar{e}_i$ The standard matrix is easily calculated as follows: \begin{align} [ \Phi_{\beta} \circ \Phi_{\bar{\beta}}^{-1}] &= [ \Phi_{\beta} \circ \Phi_{\bar{\beta}}^{-1}(e_1) | \cdots | \Phi_{\beta} \circ \Phi_{\bar{\beta}}^{-1}(e_n)] \\ &= [ \Phi_{\bar{\beta}}(\bar{v}_1) | \cdots | \Phi_{\bar{\beta}}(\bar{v}_n)] \\ &= [ \bar{e}_1 | \cdots | \bar{e}_n] \\ &= [ Pe_1 | \cdots | Pe_n] \\ &= P[ e_1 | \cdots | e_n] \\ &= P. \end{align} Finally, it is easy to see $[ \Phi_{\bar{\beta}} \circ \Phi_{\beta}^{-1}]^{-1}=P^{-1}$ and my claim follows.

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