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When computing $\det(A+B)$ we notice that there is no relation between $\det A + \det B$.

However does the $\det(A+B)$ have any relation to the matrices $A+B$ as they stand?

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Yes. The former is the determinant of the latter. That probably isn't what you're actually wanting to know, though. Can you clarify? –  Cameron Buie Mar 14 '13 at 3:37
you cant define a relation between $det(A+B)$ and matrix $A+B$, since the former is just a number defined for a SQUARE matrix where the latter addition between two matrix are defined for same dimension matrix( can be square or rectangular) –  Learner Mar 14 '13 at 3:39
In case you would imagine that $\det(A+B)=\det A+\det B$ for all $B$, see this thread. –  1015 Mar 14 '13 at 3:43
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4 Answers 4

There are formulas for $\det(A+B)$ in terms of $A$ and $B$, but they are not nicely self-contained, as one needs to refer to compositions that can't be expressed in terms of basic matrix operations. For example, see Theorem 4.1 in U. Prells, M. I. Friswell and S. D. Garvey, “Use of geometric algebra: compound matrices and the determinant of the sum of two matrices”, Proc. R. Soc. Lond. A (2003) 459, pp. 273–285. Here's a PDF link.

To get a sense of what I mean by "compositions" here is a nice formula that works for $2\times 2$ matrices: $$\det(A+B) = \det(A) + \det(B) + \text{tr}(A^{\dagger} B),$$ where $A^{\dagger}$ is the adjugate of $A$ (essentially $\det(A) A^{-1}$ but still meaningful when $A$ is singular). The series acquires more terms as the size of the matrices increase.

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Nice formula for the 2-by-2 case. +1 –  user1551 Mar 14 '13 at 4:40
For the $3 \times 3$ case it's $\det(A) + \det(B) + \text{tr}(A^\dagger B) + \text{tr}(A B^\dagger)$. –  Robert Israel Mar 14 '13 at 6:51
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Here is something that can be said in $M_n(\mathbb{C})$, in the special case where the matrices commute.

If $AB=BA$, then it can be shown that there exists an invertible matrix $P$ such that $PAP^{-1}$ and $PBP^{-1}$ are both upper-triangular.

Then the diagonal of $PAP^{-1}$ is $\{\lambda_1,\ldots,\lambda_n\}$ where the $\lambda_j$', are the eigenvalues of $A$ with multiplicities, and $PBP^{-1}$ has diagonal $\{\mu_1,\ldots,\mu_n\}$ whre the $\mu_j$'s are the eigenvalues of $B$ with multiplicities.

Now observe that $P(A+B)P^{-1}$ is upper triangular with diagonal $\{\lambda_1+\mu_1,\ldots,\lambda_n+\mu_n\}$.

Note that $\det PCP^{-1}=\det P\det C \det P^{-1}=\det C$ and that the determinant of an upper triangular matrix is the product of its diagonal terms.

It follows that $$ \det (A+B)=(\lambda_1+\mu_1)\cdots(\lambda_n+\mu_n). $$ This yields $\det A +\det B$, plus $2^n-2$ mixed terms of the form $\lambda_{i_1}\cdots\lambda_{i_k}\cdot \mu_{i_{k+1}}\cdots\mu_{i_{n}}$ which could give anything.

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There is a formula of sorts, which I have actually had occasion to use. Assume and $A$ and $B$ are $n\times n$ and let $M$ denote the set of $2^n$ matrices we get by replacing, in turn, each subset of columns of $A$ with the corresponding columns of $M$. Then $$ \det(A+B) = \sum_{C\in M} \det(C). $$

If you take $B=xI$, you can derive Laplace's formula for the coefficients of the characteristic polynomial of $A$.

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Assuming everything is defined well and stuff. ($A$, $B$ are square matrices of same size)

The determinant of $A+B$ is simply the determinant of the sum of the matrices $A$ and $B$. However, it is NOT the sum of determinants.

In general,


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