1
$\begingroup$

Is it easy to cite some example of $3 \times 3 $ non-similar matrices of real entries with equal determinant and trace?

$\endgroup$
1
6
$\begingroup$

Similar matrices must have the same eigenvalues. It is clear that $$ \begin{pmatrix} 0 & 1 & 0 \\ 0&0&0 \\ 0&0&0 \end{pmatrix} \quad \text{and} \quad \begin{pmatrix} 1 & 0 & 0 \\ 0&0&0 \\ 0&0& -1 \end{pmatrix} $$ do not have the same eigenvalues (the first has only $0$, the second $0,\pm 1$), but they have the same trace and determinant (all are zero).

$\endgroup$
1
  • $\begingroup$ @HenningMakholm Fair point. Deleted my comment. $\endgroup$ – Simon S Jun 15 '15 at 16:58
4
$\begingroup$

Even simpler examples than Chappers' one would be $$ \begin{pmatrix} 0&0&0\\0&0&0\\0&0&0 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} 0&0&1\\0&0&0\\0&0&0 \end{pmatrix} $$ (determinant and trace both $0$, but the zero matrix is similar only to itself) or $$ \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} 1&1&0\\0&1&1\\0&0&1 \end{pmatrix} $$ (determinant $1$, trace $3$, but the identity matrix is similar only to itself).

$\endgroup$
1
  • $\begingroup$ Yep. It is known that they need to have the same Jordan block structure to be similar so just having the same eigenvalues are not enough for similarity, the generalized eigenspaces must coincide too. $\endgroup$ – mathreadler Jun 15 '15 at 16:59
0
$\begingroup$

Since determinant is product of eigenvalues and trace is sum, the degrees of freedom will be $N-2$, so $N=3$ is the lowest example of where you can definitely construct matrices with same trace and determinant but still have some freedom left to chose eigenvalues.

However as Henning mentioned above, that eigenvalues are the same is a necessary but not sufficient condition for similarity. For instance even the 2x2 matrices

$$\left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right) \, \text{and}\, \left(\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array}\right) $$ have the same eigenvalues : 1 and 1, but are still not similar.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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