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I am to prove or find a counterexample for the following problem: Let $A, B, C, D, F \in GL(n, \mathbb{R})$. If $D^{-1}(A+B+C)D=F$ is correct then $A=F-B-C$ is correct.

I have not found a counterexample though there might be one - I don't know. This is how far I get:

Let $E=A+B+C$. Then the problem can be expressed as "If $D^{-1}ED=F$ is correct then $E=F$ is correct." Also, $D^{-1}ED=F \Leftrightarrow ED=DF$

But I have no clue how I could show $ED=DF \Rightarrow E=F$.

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    Please, try to make the title of your question more informative. E.g., *Why does $a imply $a+c?* is much more useful for other users than *A question about inequality.* From [How can I ask a good question?](http://meta.math.stackexchange.com/a/589/): *Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader.*2012-11-17
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    You cannot prove $ED=DF \Rightarrow E=F$ because in general it is not true. Check out matrix similarity.2012-11-17
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    @tst Do you have a counterexample for me?2012-11-17
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    It is easy to cook up counter examples once you know what's going on. Do read about matrix similarity (or matrix diagonalization). It will make this very clear.2012-11-17

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Hint: $E=\begin{pmatrix}0&1\\1&0\end{pmatrix}$ and $F=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ should work as a counterexample.

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    This would only work for $D=0$ which is not allowed because it is not part of the general linear group, or am I mistaken?2012-11-17
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    You are mistaken.2012-11-17
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    To clarify: You are right, $D=0$ is not part of the general linear group, but there is another $D$ such that $D^{-1}ED=F$.2012-11-17
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    Ah, I got it now. Thanks.2012-11-17
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    You're welcome.2012-11-17
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Hint: Try simple $2\times 2$ counterexamples. There is also counterexample where $B=C=0$.

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    $B$ and $C$ are non-zero, since they're from $GL(n, \mathbb{R})$.2012-11-17
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    Then try $B=-C=I_n$.2012-11-17