# determinant of matrices

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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Does the question say that the eigenvalues correspond to the same eigenvectors? – Martin Sleziak Dec 16 '12 at 19:15
>@Martin I have not much more information about the problem.thank you sir.sir, is it possible that the sum of the eigen values of two matrices will be eigen values of resulting one. – p.haz Dec 17 '12 at 11:22

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,

$$(1,2,3) + (-1,-4,-2) = (0,-2,1)$$

but

$$(2,1,3) + (-1,-4,-2) = (1,-3,1).$$

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>@Ted Thanks for your valuable comments.one another things ,If orders are fixed then pls tell me why we can use this summation to find out the determinant i.e what is the theoretical background of this method?.I know one thing after finding out the eigen values we can find out the determinant by finding out the product of these. – p.haz Dec 17 '12 at 2:40
There's no such thing as "orders are fixed". The eigenvalues of a matrix have no inherent order. – Ted Dec 17 '12 at 3:21
>Thanks ,for your valuable comments. – p.haz Dec 17 '12 at 11:32

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.

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@dineshileep : How do we kow both matrices give the same set of eigenvectors ? – hong wai Dec 17 '12 at 4:41
@hongwai if they commute, then they are simultaneously diagonalizable. If $AB-BA$ is strictly upper triangular, then they are simultaneously triangularizable. – dineshdileep Dec 17 '12 at 5:15