Expressing the determinant in terms of the trace of a matrix and the trace of its square

Question

How can I prove that $$\det(A) = \frac{ 1 }{ 2 } \begin{vmatrix}\operatorname{tr}(A) & 1 \\ \operatorname{tr}(A^{2}) & \operatorname{tr}(A)\end{vmatrix}$$ where vertical bars mean the determinant?

This is what I have so far.

Let $A = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]$. Then $\det(A) = \frac{ 1 }{ ad-bc } \left[\begin{matrix} d & -b \\ -c & a\end{matrix}\right]$. As such, $\operatorname{tr}(A) = a+d$, $ad+bc=2$ and $d=a=a+d$.

The left hand side is a real number and the right hand side is a matrix. Do you mean the determinant of the right side, too? — Thomas Andrews, Sep 20 '14 at 22:33
Also, the formula you've given for $\det(A)$ is actually the formula for $A^{-1}$. — Thomas Andrews, Sep 20 '14 at 22:34
Then edit your question to indicate what you really mean. It is still impossible for $\det(A)$ to be equal to a $2\times 2$ matrix. — Thomas Andrews, Sep 20 '14 at 22:54
Hint: trace is independent of basis, and depends only on the eigenvalues. — Adam Hughes, Sep 20 '14 at 23:16
There's an answer (by myself) to this $2\times 2$ case in another context. — Hanno, Dec 25 '18 at 20:29

score 0 · Accepted Answer · answered Sep 20 '14 at 23:15

If I understand what you want to prove, for a 2x2 matrix we have:

$$A=\begin{pmatrix} a&b\\ c&d \end{pmatrix},\text{ } det(A)=ad-bc\text{ , } Tr(A)=a+d$$

$$A^2=AA=\begin{pmatrix} a&b\\ c&d \end{pmatrix} \begin{pmatrix} a&b\\ c&d \end{pmatrix}=\begin{pmatrix} a^2+bc&ab+bd\\ ac+cd&bc+d^2 \end{pmatrix}$$

$$tr(A^2)=a^2+2bc+d^2$$

Now we have that:

$$\frac{1}{2}det\begin{pmatrix} tr(A)&1\\ tr(A^2)&tr(A) \end{pmatrix}=\frac{1}{2}det\begin{pmatrix} a+d&1\\ a^2+2bc+d^2&a+d \end{pmatrix}=$$

$$=\frac{1}{2}[(a+d)^2-(a^2+2bc+d^2)]=\frac{1}{2}[a^2+d^2+2ad-a^2-2bc-d^2]=$$

$$=\frac{1}{2}[2ad-2bc]=ad-bc=det(A)$$

Expressing the determinant in terms of the trace of a matrix and the trace of its square

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