Let A and B be any two 3 by 3 matrices with eigen values 1,2,3 and -1,-4,-2 respectively.Then my Question is to find out the determinant of A+B? Thanks in advance for valuable comments.
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It's not possible. If you take two diagonal matrices with those entries but in different orders, the sums will have different eigenvalues. So the eigenvalues of the sum are not determined by the eigenvalues of the individual matrices. Writing $(a,b,c)$ for the diagonal matrix with entries $a,b,c$ in that order,
$$(1,2,3) + (-1,-4,-2) = (0,-2,1)$$
$$(2,1,3) + (-1,-4,-2) = (1,-3,1).$$
In general, its a hopless situation. You can't make any comment on the determinant sum of the matrices. If you can simultaneously triangularize both the matrices, then you can make a statement about the determinant of their sum. Let $T_1$ denote the triangular matrix with its diagonal entries as eigenvalues of $A$ and $T_2$ be the same for $B$. Then, let $U$ be the set of orthonormal vectors that simultaneously triangularize them both. So $A=UT_1U^H$ and $B=UT_2U^H$, then $A+B=U(T_1+T_2)U^H$. So here, you can talk about the determinant. The order of arrangement of eigenvalues does matter in the sense, this will dictate the arrangement of columns in the matrix $U$. So there should be an arrangement of eigenvalues of $A$ and $B$, such that $U$ simultaneously triangularizes them.
In fact, simultaneously diagonalization for $A$ and $B$ also gives the same results.