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Question:

Find two $2 \times 2$ matrices $A$ and $B$ with the same rank, determinant, trace and characteristic polynomial, but they are not similar to each other.

My thought:

I come up with two matrices:

$A=\begin{pmatrix} 0 &1 \\ 0 &0 \end{pmatrix}$ and $B=\begin{pmatrix} 0 &0 \\ 1 &0 \end{pmatrix}.$

It is easy to check that they have same rank, determinant, trace and characteristic polynomial. However, my question is I do not know how to prove two matrices are similar or not.

I have learnt the converse in my textbook, i.e. If two matrices are similar, they have the same determinant, characteristic polynomial,etc.

I have also known that (but I do now know the proof), we can check by using Jordan form of two matrices. I do not know if this claim is correct:

"If two matrices have the same Jordan form, they are similar to each other."

Yet, go back to the question, is it a quick way to prove? Thank you in advance.

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Note that determinant and trace can be found from the characteristic polynomial, so you might as well ask just for the same characteristic polynomial and rank. –  Marc van Leeuwen Apr 12 at 5:16

3 Answers 3

up vote 6 down vote accepted

Try $A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$. They are both rank 2.

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Thank you for your counter-example. I rather concern how to prove that they are not similar. Would you please give me a hint how to disprove it? Thank you so much. –  nam Apr 12 at 4:50
    
Compute the eigenvalues of the two matrices. One of them has a one dimensional space of eigenvalues, the other has a two dimensional space of eigenvalues. –  Stephen Montgomery-Smith Apr 12 at 4:56
3  
Or even easier: $PBP^{-1} = B = 2I$ for any invertible $P$. –  Stephen Montgomery-Smith Apr 12 at 4:56
    
Thank you for providing methods to check the similarity. I get it now. –  nam Apr 12 at 5:02

Your matrices are similar, related by $\begin{pmatrix}0&1\\1&0\end{pmatrix}$.

By inspection, what each of your matrices do is kill one basis vector while turning the other into the first one. That's intuitively similar, so to transform one to the other all we have to do is swap the basis vectors which is what $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ does.

Hint. The rank (if full), determinant, trace, and characteristic polynomial of a triangular matrix depends only on the diagonal elements.

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Thank you for answering. Would you please tell me how do you find this matrix such that my $A$ and $B$ are similarly related? –  nam Apr 12 at 4:48
1  
What you say in the hint is not quite true for the rank: the rank of nilpotent triangular matrices depends on their off-diagonal entries. –  Marc van Leeuwen Apr 12 at 5:15
    
@Marc: good point. –  Henning Makholm Apr 12 at 12:00
    
@nam: A matter of experience and intuitive understanding of what the matrices do, I'm afraid. See updated answer. –  Henning Makholm Apr 12 at 12:05

You need two matrices with distinct Jordan forms yet the same characteristic polynomial; this is only possible if the characteristic polynomial has at least one multiple root. Here that it means is it of the form $(X-a)^2$. Distinct Jordan forms can only be obtained if one of the matrices does not have a single block of size$~2$, in which case the only remaining possibility is two (trivial) blocks of size$~1$, so one of the matrices must be a multiple $aI$ of the identity. Here you can see why you example fails: you took $a=0$ but neither of your matrices equals the zero matrix, so they have equal Jordan forms (in fact $A$ is the Jordan form of $B$). You can see rather easily that you just interchanged the roles of the first and second standard basis vector in constructing $A$ and $B$, so they are similar by the permutation matrix interchanging those vectors.

Once you've seen that you need to choose (say) $A=aI$, one can take $B$ to be any other matrix with characteristic polynomial $(X-a)^2$ (there are plenty of them; the size $2$ Jordan block with diagonal entries $a$ is the most obvious choice, but for instance the companion matrix of $(X-a)^2$ will do fine as well). This ensures everything that was required, except equal ranks: the rank is not always determined by the characteristic polynomial. However if $a\neq0$, both matrices will be invertible, and therefore of rank$~2$ (for $a=0$ the matrix $A=0$ will have a different rank than $B$, namely $0$ versus $1$, so one should not choose $a=0$). So for an example take $A=aI$ with $a\neq0$, and $B$ any matrix with characteristic polynomial $(X-a)^2$. Moreover all examples are obtained like this (possibly with $A,B$ interchanged).

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