$$\det(A + B)$$

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.

  • 7
    $\begingroup$ 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... $\endgroup$
    – gt6989b
    Feb 12 '14 at 16:11
  • 1
    $\begingroup$ There are special cases, like en.wikipedia.org/wiki/Matrix_determinant_lemma $\endgroup$ Feb 12 '14 at 16:13
  • 3
    $\begingroup$ Also, in light of math.stackexchange.com/questions/298454/… , I think such a formula will always depend on more than just $\det A, \det B$ $\endgroup$ Feb 12 '14 at 16:14
  • 1
    $\begingroup$ @ABlumenthal I'm having a hard time comprehending your link although it seems to answer my question. Can you explain it to me? $\endgroup$ Feb 14 '14 at 5:54
  • $\begingroup$ 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). $\endgroup$
    – kon psych
    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}

  • 8
    $\begingroup$ 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) $\endgroup$ Sep 22 '18 at 22:09
  • 2
    $\begingroup$ Anything for $n=3$? $\endgroup$ Sep 22 '18 at 22:17
  • $\begingroup$ @Frpzzd Unfortunately, I did not find any interesting result for $n=3$. $\endgroup$ Oct 8 '18 at 1:24
  • $\begingroup$ @Frpzzd Please check my new answer. $\endgroup$ Mar 9 '19 at 11:50
  • 2
    $\begingroup$ @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). $\endgroup$
    – Charlie
    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 $\alpha$.


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