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I'm having a small problem proving the following lemma:

Let $V$ be a finite-dimensional vector space over $F$ with $dim \ V=n$. Let $L \in \mathcal{L}(V)$ and $A \in M_n(F)$ where $A=[L]_B$ is a matrix representation of $L$ with respect to a basis $B$ of $V$. Then,

$A'=[L]_{B'}$ is a matrix representation of $L$ with respect to a basis $B'$ of $V$ $\iff$ $A$ and $A'$ are similar matrices.

No problems with the forward direction, proving that $A$ and $A'$ are similar. However, when I suppose similarity I'm not sure how to proceed. I do think I need to (or can) use the change of basis matrix here again (I used it in the proof in the forward direction), but I'm not sure how to piece everything together.

Any help with completing the proof is appreciated!

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1 Answer 1

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Assume that $A$ and $A'$ are similar. Let $P$ be the similarity matrix, i.e. $A'=P^{-1}AP$. Let $B'$ be the basis of $V$ formed by the columns of $P$, (read the column as the coordinate column of a vector w.r.t a fixed basis $B$ of $V$). Now let $[L]_B=A$, i.e. $[L(v)]_B=A\cdot[v]_B$. Now what will be $[L]_{B'}$?

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Is it trivial to show that $B'$ as you've defined it is actually a basis? Sorry, maybe I'm just missing the point. Also, I'm a little unsure of how exactly you're defining $P$ here, could you elaborate, please? Thank you! –  Bachmaninoff Oct 8 '12 at 1:38
    
@Bachmaninoff: How do you define two matrices to be similar? About $B'$ - Since $P$ is invertible, it's columns are linearly independent. Any $n$ linearly independent vectors in a space of dimension $n$ form a basis. –  Dennis Gulko Oct 8 '12 at 9:55
    
Oh, right! Not sure why that didn't occur to me. Thank you very much, Dennis! –  Bachmaninoff Oct 8 '12 at 16:55

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