Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
please can someone help me? – Mohamez Jul 7 '13 at 20:42
Are you looking for a proof? Or will a counterexample suffice? – Ataraxia Jul 7 '13 at 20:42
$\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $\begin{pmatrix}2&0\\0&2\end{pmatrix}$ are not similar. – Hagen von Eitzen Jul 7 '13 at 20:45
@HagenvonEitzen darn it, I was just typing that haha – Ataraxia Jul 7 '13 at 20:45
No, but if they are of the same size, they are congruent by Gaussian elimination. Not that bad. – 1015 Jul 7 '13 at 20:55
up vote 3 down vote accepted

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.

share|cite|improve this answer
"when the characteristic of the field is not $2$": It would still be true if "characteristic" were replaced with "cardinality". – Jonas Meyer Jul 8 '13 at 1:53

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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