# 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... – gt6989b Feb 12 '14 at 16:11
• There are special cases, like en.wikipedia.org/wiki/Matrix_determinant_lemma – A Blumenthal 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 $\det A, \det B$ – A Blumenthal 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? – frogeyedpeas 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). – kon psych Apr 12 '16 at 7:26

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.

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 $$n=3$$,

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

• 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) – Frpzzd Sep 22 '18 at 22:09
• Anything for $n=3$? – Frpzzd Sep 22 '18 at 22:17
• @Frpzzd Unfortunately, I did not find any interesting result for $n=3$. – Drake Marquis Oct 8 '18 at 1:24
• @Frpzzd Please check my new answer. – Drake Marquis Mar 9 at 11:50