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 $B=[T]_\beta$.


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


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




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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.