# How write a matrix in terms of determinant and trace?

1. May 19, 2014

### 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...

2. May 19, 2014

### 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.

3. May 19, 2014

### Jhenrique

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

4. May 20, 2014

### Ben Niehoff

No.

4chars

5. May 20, 2014

### Jhenrique

what?

6. May 20, 2014

### 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.