# Expressing the determinant of a sum of two matrices?

Can

$$\det(A + B)$$

be expressed in terms of

$$\det(A), \det(B), n$$

where $A,B$ are $n\times n$ matrices?

## #

I made the edit to allow $n$ to be factored in.

• Not in general. Even if $A,B$ are $n \times n$ identity matrices, $\det(A+B) = 2^n$ while $\det(A) = \det(B) = 1$, so the connection will depend on $n$ as well... Feb 12 '14 at 16:11
• There are special cases, like en.wikipedia.org/wiki/Matrix_determinant_lemma Feb 12 '14 at 16:13
• Also, in light of math.stackexchange.com/questions/298454/… , I think such a formula will always depend on more than just $\det A, \det B$ Feb 12 '14 at 16:14
• @ABlumenthal I'm having a hard time comprehending your link although it seems to answer my question. Can you explain it to me? Feb 14 '14 at 5:54
• I hope I am not making any mistake but what the link says for this case is that determinant of sum, is sum of determinants of $2^n$ matrices which are constructed by choosing for each column i either ith column of A or ith column of B (all possible choices are $2^n$ if you think about it). Apr 12 '16 at 7:26

When $$n=2$$, and suppose $$A$$ has inverse, you can easily show that

$$\det(A+B)=\det A+\det B+\det A\,\cdot \mathrm{Tr}(A^{-1}B)$$.

Let me give a general method to find the determinant of the sum of two matrices $$A,B$$ with $$A$$ invertible and symmetric (The following result might also apply to the non-symmetric case. I might verify that later...). I am a physicist, so I will use the index notation, $$A_{ij}$$ and $$B_{ij}$$, with $$i,j=1,2,\cdots,n$$. Let $$A^{ij}$$ donate the inverse of $$A_{ij}$$ such that $$A^{il}A_{lj}=\delta^i_j=A_{jl}A^{li}$$. We can use $$A_{ij}$$ to lower the indices, and its inverse to raise. For example $$A^{il}B_{lj}=B^i{}_j$$. Here and in the following, the Einstein summation rule is assumed.

Let $$\epsilon^{i_1\cdots i_n}$$ be the totally antisymmetric tensor, with $$\epsilon^{1\cdots n}=1$$. Define a new tensor $$\tilde\epsilon^{i_1\cdots i_n}=\epsilon^{i_1\cdots i_n}/\sqrt{|\det A|}$$. We can use $$A_{ij}$$ to lower the indices of $$\tilde\epsilon^{i_1\cdots i_n}$$, and define $$\tilde\epsilon_{i_1\cdots i_n}=A_{i_1j_1}\cdots A_{i_nj_n}\tilde\epsilon^{j_1\cdots j_n}$$. Then there is a useful property: $$\tilde\epsilon_{i_1\cdots i_kl_{k+1}\cdots l_n}\tilde\epsilon^{j_1\cdots j_kl_{k+1}\cdots l_n}=(-1)^sl!(n-l)!\delta^{[j_1}_{i_1}\cdots\delta^{j_k]}_{i_k},$$ where the square brackets $$[]$$ imply the antisymmetrization of the indices enclosed by them. $$s$$ is the number of negative elements of $$A_{ij}$$ after it has been diagonalized.

So now the determinant of $$A+B$$ can be obtained in the following way $$\det(A+B)=\frac{1}{n!}\epsilon^{i_1\cdots i_n}\epsilon^{j_1\cdots j_n}(A+B)_{i_1j_1}\cdots(A+B)_{i_nj_n}$$ $$=\frac{(-1)^s\det A}{n!}\tilde\epsilon^{i_1\cdots i_n}\tilde\epsilon^{j_1\cdots j_n}\sum_{k=0}^n C_n^kA_{i_1j_1}\cdots A_{i_kj_k}B_{i_{k+1}j_{k+1}}\cdots B_{i_nj_n}$$ $$=\frac{(-1)^s\det A}{n!}\sum_{k=0}^nC_n^k\tilde\epsilon^{i_1\cdots i_ki_{k+1}\cdots i_n}\tilde\epsilon^{j_1\cdots j_k}{}_{i_{k+1}\cdots i_n}B_{i_{k+1}j_{k+1}}\cdots B_{i_nj_n}$$ $$=\frac{(-1)^s\det A}{n!}\sum_{k=0}^nC_n^k\tilde\epsilon^{i_1\cdots i_ki_{k+1}\cdots i_n}\tilde\epsilon_{j_1\cdots j_ki_{k+1}\cdots i_n}B_{i_{k+1}}{}^{j_{k+1}}\cdots B_{i_n}{}^{j_n}$$ $$=\frac{\det A}{n!}\sum_{k=0}^nC_n^kk!(n-k)!B_{i_{k+1}}{}^{[i_{k+1}}\cdots B_{i_n}{}^{i_n]}$$ $$=\det A\sum_{k=0}^nB_{i_{k+1}}{}^{[i_{k+1}}\cdots B_{i_n}{}^{i_n]}$$ $$=\det A+\det A\sum_{k=1}^{n-1}B_{i_{k+1}}{}^{[i_{k+1}}\cdots B_{i_n}{}^{i_n]}+\det B.$$

This reproduces the result for $$n=2$$. An interesting result for physicists is when $$n=4$$,

$$\begin{split} \det(A+B)=&\det A+\det A\cdot\text{Tr}(A^{-1}B)+\frac{\det A}{2}\{[\text{Tr}(A^{-1}B)]^2-\text{Tr}(BA^{-1}BA^{-1})\}\\ &+\frac{1}{6}\{[\text{Tr}(BA^{-1})]^3-3\text{Tr}(BA^{-1})\text{Tr}(BA^{-1}BA^{-1})+2\text{Tr}(BA^{-1}BA^{-1}BA^{-1})\}\\ &+\det B. \end{split}$$

• This, in my opinion, is the better answer, as it not only falsifies the question, but also salvages the false statement with an interesting and relevant identity. (+1) Sep 22 '18 at 22:09
• Anything for $n=3$? Sep 22 '18 at 22:17
• @Frpzzd Unfortunately, I did not find any interesting result for $n=3$. Oct 8 '18 at 1:24
• @Frpzzd Please check my new answer. Mar 9 '19 at 11:50
• @FranklinPezzutiDyer, I've found a fine equality for the determinant of the sum of 3x3 matrices: $\det{\left( A + B \right)} = \det{A} + \det{A} \cdot Tr\left( A^{-1} \cdot B \right) + \det{B} \cdot Tr\left( B^{-1} \cdot A \right) + \det{B}$. I don't have a strict symbolic proof, but I've checked it using SymPy and this works well: A = Matrix(3, 3, symbols('a:3:3')); B = Matrix(3, 3, symbols('b:3:3')); expand(cancel((A + B).det())) == expand(cancel( A.det() + A.det() * (A.inv() * B).trace() + B.det() * (B.inv() * A).trace() + B.det() )) (online editor won't handle this -- check it locally). Feb 3 '20 at 20:14

When $n\ge2$, the answer is no. To illustrate, consider $$A=I_n,\quad B_1=\pmatrix{1&1\\ 0&0}\oplus0,\quad B_2=\pmatrix{1&1\\ 1&1}\oplus0.$$ If $\det(A+B)=f\left(\det(A),\det(B),n\right)$ for some function $f$, you should get $\det(A+B_1)=f(1,0,n)=\det(A+B_2)$. But in fact, $\det(A+B_1)=2\ne3=\det(A+B_2)$ over any field.

From the MAA (mathematics association of america) there is a general formula here. https://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/determinants-of-sums

There is a proof in the article, but in general: $$\det(A + B) = \sum_r \sum_{\alpha, \beta} (-1)^{s(\alpha) + s(\beta)} \det(A[\alpha | \beta]) \det(B(\alpha | \beta))$$ where $$r$$ runs over the integers from $$0,\dots,n$$; then the inner sum runs over all strictly increasing sequences $$\alpha$$ and $$\beta$$ of length $$r$$ chosen from $$1,\dots,n$$.

$$A[\alpha|\beta]$$ is the $$r$$ by $$r$$ square submatrix of $$A$$ lying in rows $$\alpha$$ and columns $$\beta$$.

$$B(\alpha|\beta)$$ is the $$(n-r)$$-square submatrix of $$B$$ lying in rows complementary to $$\alpha$$ and columns complementary to $$\beta$$.

$$s(\alpha)$$ is the sum of the integers in $$\alpha$$.