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I know that if two matrices $A$ and $B$ are similar implies that they have the same rank.

However, if they have the same rank are they similar?

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  • $\begingroup$ please can someone help me? $\endgroup$ – Mohamez Jul 7 '13 at 20:42
  • $\begingroup$ Are you looking for a proof? Or will a counterexample suffice? $\endgroup$ – Ataraxia Jul 7 '13 at 20:42
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    $\begingroup$ $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $\begin{pmatrix}2&0\\0&2\end{pmatrix}$ are not similar. $\endgroup$ – Hagen von Eitzen Jul 7 '13 at 20:45
  • $\begingroup$ @HagenvonEitzen darn it, I was just typing that haha $\endgroup$ – Ataraxia Jul 7 '13 at 20:45
  • $\begingroup$ No, but if they are of the same size, they are congruent by Gaussian elimination. Not that bad. $\endgroup$ – Julien Jul 7 '13 at 20:55
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As mentioned by the others, the answer is negative. Actually we can say something more: if $n\ge2$, then for any $n\times n$ nonzero matrix $A$, there is always a dissimilar matrix $B$ of the same rank; if $n=1$, the statement also holds when the characteristic of the field is not $2$.

Proof. The case $n=1$ is trivial. Suppose $n\ge2$. Let $k=\operatorname{rank}(A)$. If $A$ is not diagonalisable, let $B$ be a diagonal matrix of rank $k$. If $A$ is diagonalisable, let $B$ be the direct sum of a $k\times k$ Jordan block for eigenvalue $1$ and a zero block.

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    $\begingroup$ "when the characteristic of the field is not $2$": It would still be true if "characteristic" were replaced with "cardinality". $\endgroup$ – Jonas Meyer Jul 8 '13 at 1:53
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No not necessarily. To find a counterexample, just take any set of matrices with distinct eigenvalues, but have the same number of non-zero eigenvalues.

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