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Let $X$ be a finite dimensional vector space over $K$ and define $T:X\rightarrow X$ to be a linear transformation on $X$. If $\alpha, \beta$ are two different basis for $X$ then we know that the matrix representation $A=[T]_\alpha$ is similar to $B=[T]_\beta$. Is the converse true? That is suppose $A$, $B$ are $n\times n$ ($n=\dim X$) matrix, then if $A=[T]_\alpha$ and $A$ is similar to $B$, is it true that there exists basis $\beta$ such that $B=[T]_\beta$.

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Yes. Let $A$ represent $T$ in the basis $e_1,e_2,\ldots,e_n$. Since $A$ and $B$ are similar, there is an invertible matrix $P$ such that $A=PBP^{-1}$. Then $P^{-1}(e_1),P^{-1}(e_2),\ldots,P^{-1}(e_n)$ is another basis since $P$ is invertible. $A$ is uniquely determined by the vectors $A(e_1),A(e_2),\ldots,A(e_n)$, and you can get the columns of $A$ from this. Namely, if $A=(a_{ij})$ then

$$A(e_j)=\sum_{i}{a_{ij}e_i}$$

Let $f_j=P^{-1}(e_j)$. Let $B=(b_{ij})$ be the matrix representation of $B$ in the basis $(f_i)$. Notice that

$$B(f_j)=\sum_{i}{b_{ij}f_i}$$

and

$$A(e_j)=P(B(f_j))=\sum_{i}{P(b_{ij}f_i)}=\sum_{i}{b_{ij}P(f_i)}=\sum_i{b_{ij}e_i}$$

so $a_{ij}=b_{ij}$ for all $i,j$.

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  • $\begingroup$ Thanks so much for your answer, I will look at it later. $\endgroup$ – amathnerd Nov 2 '14 at 2:06
  • $\begingroup$ Ok I think you made your matrix commute in the last line $P(B(f_j))=PBP^{-1}e_i$. In this proof, where did you use the fact $A$ is a representation of linear transformation $T$. How do you know $f_i$ represents $T$ as $B$? $\endgroup$ – amathnerd Nov 2 '14 at 2:14
  • $\begingroup$ See my edit. P can commute past the real numbers $b_{ij}$ since they are real numbers. $A$ is assumed to represent $T$ in the basis I chose, and the entries $b_{ij}$ represent $B$ in the transformed basis, and are the same as the $a_{ij}$. By possibly changing $P$, you may assume $B$ to be the matrix representation in the basis you desire. $\endgroup$ – Matt Samuel Nov 2 '14 at 2:26
  • $\begingroup$ I understand your argument. But I dont think $a_{ij}$ and $b_{ij}$ are equal because $\sum_{i}a_{ij}e_i$ is not the same as $\sum_{i}b_{ij}e_i$. If $a_{ij}=b_{ij}$, then you just proved similar matrix are the same ($A~B$ implies $A=B$), which is not true $\endgroup$ – amathnerd Nov 2 '14 at 2:42
  • $\begingroup$ Notice that $b_{ij}$ is the matrix representation of $B$ in a different basis than $e_1,e_2,\ldots,e_n$, namely $f_1,f_2,\ldots,f_n$. In other words, $A(e_j)=\sum_i{a_{ij}e_i}$ and $B(f_j)=\sum_i{a_{ij}f_i}$, but $B(e_j)\neq \sum_i{a_{ij}e_i}$. $\endgroup$ – Matt Samuel Nov 2 '14 at 2:45

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