How write a matrix in terms of determinant and trace?

[Jhenrique]

Given the following: $$\\ \begin{bmatrix}
A & 0\\
0 & B\\
\end{bmatrix}$$ the eigenvalues is exactaly A and B. So analogously, is possible to write a matrix with only two elements, T and D, such that the trace is T and the determinant is D?

I tried something: $$\\ \text{tr} \left(
\begin{bmatrix}
\frac{1}{2}T & 0\\
0 & \frac{1}{2}T\\
\end{bmatrix}
\right) = T $$
$$\\ \text{det} \left(
\begin{bmatrix}
\sqrt{\frac{D}{2}} & \sqrt{-\frac{D}{2}} \\
\sqrt{-\frac{D}{2}} & \sqrt{\frac{D}{2}} \\
\end{bmatrix}
\right) = D $$
But I can't join the 2 formulas...

 

[Ben Niehoff]

There are infinitely many ways. You could write

$$\begin{pmatrix}T & D \\ -1 & 0\end{pmatrix}$$
for example. Or you could try

$$\begin{pmatrix}\frac{T}{2} & \frac{T}{2} + \sqrt D \\ \frac{T}{2} - \sqrt D & \frac{T}{2}\end{pmatrix}$$
Since $T$ and $D$ are the only invariants, these matrices ought to be similar. Maybe you can work out the similarity transformation.

 

[Jhenrique]

There are infinitely many ways. You could write

$$\begin{pmatrix}T & D \\ -1 & 0\end{pmatrix}$$
for example. Or you could try

$$\begin{pmatrix}\frac{T}{2} & \frac{T}{2} + \sqrt D \\ \frac{T}{2} - \sqrt D & \frac{T}{2}\end{pmatrix}$$
Since $T$ and $D$ are the only invariants, these matrices ought to be similar. Maybe you can work out the similarity transformation.

Yeah, yeah! The ideia is express a matrix in terms of the invariants, exist some general formula for this?

 

[Ben Niehoff]

No.

4chars

 

[Jhenrique]

No.

4chars

what?

 

[Matterwave]

He's saying "no". But his reply needs to be at least 4-characters long (to prevent spam), hence the "4chars" at the end.