Can we reconstruct an $n\times n$ matrix if all we know is the determinant and trace of that matrix (and its size: i.e. what $n$ is)? I would think not, because two scalars wouldn't be enough to tell us about all $n^2$ entries. But then this brings up three questions for me.

  1. What else do we need to specify a matrix exactly? Obviously, I'm not talking about something for which we can immediately get back the matrix, like for instance, the transpose. But what if we knew all of the eigenvectors and eigenvalues? Or what if we know its Jordan form?
  2. If all we know is the trace and the determinant, what else can we figure out about the matrix? Obviously we know whether it is invertible or not, but what about say the eigenvalues -- can we figure them out in the $n\ge 3$ case? Can we figure something else out about the matrix?
  3. Are the trace and the determinant the only invariants of a matrix under change of basis?

Basically, I'm just trying to find out exactly what information is encoded in the trace and determinant of a matrix.

I know that the trace is the sum of the eigenvalues, the determinant is the product of the eigenvalues, and that the determinant of a matrix is the factor by which areas change under the linear transformation $x \mapsto Ax$, where $A$ is the matrix. Is there anything thing else that knowing both of these invariants tell us?


Specifying the determinant and trace doesn't restrict the set of all matrices as much as you'd like it to. In lower dimensions (namely $n=2$), things are a little better, but for $n\geq 3$ the situation isn't all that great.

One thing we do know is that the determinant and trace both show up in the characteristic polynomial in the matrix. For example, in the $2\times 2$ case, if the characteristic polynomial of $A$ factors into $(t-a)(t-b)$, then $$ (t-a)(t-b) = t^2 - (a+b)t + ab = t^2 - \text{trace}~A\cdot t + \det A. $$ At least in this case, having the determinant and trace completely determines the characteristic polynomial, and conversely. Sadly, this is known not to be enough to completely determine a matrix up to similarity. In particular, if the eigenvalues are repeated in the $2\times 2$ case then there are two possible forms of the Jordan matrices. On the other hand, if the eigenvalues are distinct then the matrix is diagonalizable, so in this specific case among $2\times 2$ matrices we can determine $A$ up to similarity explicitly.

In higher dimensions, we can show the same thing happens, but we get additional terms whose coefficients are traces of what are known as exterior powers of $A$. But if the matrix is unkonwn a priori then you won't be able to determine the characteristic polynomial with just the determinant and trace.

Of course, if one knows the Jordan form then one knows essentially everything one could possibly want about the matrix.


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