Showing determinants using trace in a 2x2 matrix I am confused about this homework question.
It says "Show that :
$\det(A) = \frac 12 \begin{vmatrix}\operatorname{tr}(A)&1\\\operatorname{tr}(A^2)& \operatorname{tr}(A)\end{vmatrix}$
for every $2\times 2$ matrix."
I am not sure how to "show" this. I'm guessing it means a proof of sorts.
But I am also confused as to why the $\det(A) = 1/2$ times a matrix. I'm pretty sure determinants are just a number, not a matrix.
Any tips would be great.
Thanks in advance.
 A: The "|" to the left and the right of your matrix probably indicate that you should take the determinant of this matrix, too. It is not completly uncommon to write $|A|$ instead of $det(A)$.
To prove the identity just take an abstract $2 \times 2$-Matrix and compute the left-hand and the right-hand side of the equation.
A: You could use an eigenvalue formulation, since if $\lambda_1,\lambda_2$ are the eigenvalues of $A$ then $\det(A)=\lambda_1 \lambda_2$, $tr A=\lambda_1+\lambda_2$, $tr(A^2)=\lambda_1^2+\lambda_2^2$. Plug these in your equation and obvious equality holds.
A: Here is a (requested) generalization for larger dimensions: If A is an n×n matrix, then one has the following expression for the determinant of A in terms of the determinant of a matrix whose entries are traces of powers of A:
$$\newcommand{\tr}{\operatorname{tr}}
\newcommand{\det}{\operatorname{det}}
\det(A) = \frac{1}{n!}\left|\begin{array}{cccccccc}
\tr(A) & 1 & . & . & . & \dots & . & . \\
\tr(A^2) & \tr(A) & 2 & . & . & \dots & . & . \\
\tr(A^3) & \tr(A^2) & \tr(A) & 3 & . & \dots & . & . \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
\tr(A^{n-1}) & \tr(A^{n-2}) & \tr(A^{n-3}) & \tr(A^{n-4}) & \tr(A^{n-5}) & \dots & \tr(A) & n-1 \\
\tr(A^n) & \tr(A^{n-1}) & \tr(A^{n-2}) & \tr(A^{n-3}) & \tr(A^{n-4}) & \dots & \tr(A^2) & \tr(A)
\end{array}\right|$$
In particular, we have for $3\times3$ that:
$$\det(A) = \frac{1}{6}\left|\begin{array}{cccc}
 \tr(A) & 1 & 0 \\
 \tr(A^2) & \tr(A) & 2 \\
 \tr(A^3) & \tr(A^2) & \tr(A) \end{array}\right|$$
The matrix on the right is defined in general by: $$B_{ij} = \begin{cases}
 i & \text{ if } j = i + 1 \\
 \tr(A^{i-j+1}) & \text{ if } j \leq i \\
 0 & \text{ if } j > i+1 
\end{cases}$$
I suggest using this recursively to create even more complicated formulas (all in lower Hessenberg form!) to be entered into the big book of bad algorithms.
A: Assuming you are familiar with Cayley–Hamilton theorem, this may be an alternative way:
Any $2\times2$ matrix satisfies $A^2-\text{tr}(A)A+\det(A)I_2=0 $. Taking trace of both sides, $\begin{align*}&\text{tr}(A^2)-\text{tr}(A)\text{tr}(A)+2\det(A)=0 \Rightarrow\\ &\det(A)=\frac12[\text{tr}(A)\text{tr}(A)-\text{tr}(A^2)]=\frac12\begin{vmatrix}\text{tr}(A)&1\\\text{tr}(A^2)&\text{tr}(A)\end{vmatrix}\end{align*}$
A: The RHS is determinant, i.e., $1/2[(tr A)^2-tr(A^2)]=1/2[(a_{11}+a_{22})^2-(a_{11}^2+2a_{12}a_{21}+a_{22}^2)]=\cdots$
