Find $\det(A^{2}+A^{T})$ when eigenvalues are $1,2,3$ We have to find $\det(A^{2}+A^{T})$. It is given that eigenvalues of $A$ are $1,2,3$.
My attempt: Since the question implicitly states that the answer would be same for all $A$ with eigenvalues $1,2,3$ we find that the answer is $(1+1)*(2+4)*(3+9)$ by taking $A$ as a diagonal matrix :P
I am unable to prove it for general $A$. Any help will be appreciated.
 A: I assume that matrix $A$ is $3$-by-$3$.
Consider the following matrices:
$$A(a,b,c)=\left[\begin{array}{ccc}1 & a & b\\ 0 & 2 & c \\ 0 & 0 & 3\end{array}\right]$$
It's clear that the eigenvalues of $A(a,b,c)$ are $1$, $2$ and $3$ for all values of $a$, $b$ and $c$. Anyway:
$$\det(A(a,b,c)^2+A(a,b,c)^T) = a^2c^2 - 36a^2 + 13abc - 24b^2 - 10c^2 + 144$$
Then, for different matrices that have the same eigenvalues, you can obtain very different determinant values of $A^2 + A^T$. In other words, you are not able to say nothing about this determinant for a generic matrix with eigenvalues $1$, $2$ and $3$.

Anyway, you can say something about the trace of $A^2 + A^T$.
Indeed, we know that $tr(A) = tr(A^T) = 1+2+3 = 6$ and $tr(A^2) = 1 + 4 + 9 = 14$. Also, we can separate the trace:
$$tr(A^2+A^T) = tr(A^2) + tr(A^T) = 14 + 6 = 20.$$
This is true for all matrices $A$ which eigenvalues are $1$, $2$ and $3$. 

If you were meant to find $\det(A^2+A)$ then things are easier. Indeed:
$$\det(A+A^2) = \det(A(I+A)) = \det(A)\det(I+A)$$
The eigenvalues of $(I+A)$ are equal to those of $A$ plus $1$. Then:
$$\det(A+A^2) = \det(A)\det(I+A) = (\lambda_1\cdot\lambda_2\cdot\lambda_3)((\lambda_1+1)\cdot(\lambda_2+1)\cdot(\lambda_3+1)) = (1\cdot 2 \cdot 3)(2 \cdot 3 \cdot 4) = 144$$
Finally, $$\det(A+A^2) = 144$$ 
A: If $A$ is symmetric then it is very easy.
Then $$\det(A^2+A^T)=\det(A^2+ A)=\det(A)\det(A+ I))$$ which is $144$ if $A$ is symmetric and. It is to see how we can calculate this quantity if $A$ is none of these.

If the case is the one as pointed out by @user1551 in the comments then it is not difficult to calculate the determinant and it will be then $$\det(A^2+A^{-1})=\det (A^{-1})\det(A^3+I)=\frac{1}{\det(A)}\det(A+I)\det(A+\omega)\det(A+\omega^2)\\=\frac{(2\times 3\times 4)(1+\omega)(2+\omega)(3+\omega)(1+\omega^2)(2+\omega^2)(3+\omega^2)}{(1\times 2\times 3)}=\frac{(1+1)(1+8)(1+27)}{6}\\=84$$
A: You can calculate the eigenvalues of $\bf A^2$ as the squares of the eigenvalues of $\bf A$ and it is easy to show that $\bf A^T$ has the same eigenvalues. However, the transpose operator may shift the eigensystem. If $\bf A = PDP^{-1}$ then $\bf A^T = P^{-T}DP^T$ and $\bf A^2 = PD^2P^{-1}$ In general we can't say anything about the relations between $\bf P, P^{-1}, P^{T}$ so the eigensystem can change when we do the addition. 
But we can see that if for instance $\bf P^T = P^{-1}$ (and if $\bf A$ is diagonalizable) then it will be as you think.
