Can $\det(A + B)$ be expressed in terms of $\det(A)$, $\det(B)$, $n$?

Question

Let $A$ and $B$ be $n \times n$ matrices. Can $\det(A + B)$ be expressed in terms of $\det(A)$, $\det(B)$, $n$?

Answer 1

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

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

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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{\det A}{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}

Answer 2

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.

Answer 3

From the MAA (mathematics association of america) there is a general formula here. https://web.archive.org/web/20240324155831/https://maa.org/sites/default/files/0746834207570.di020741.02p0009e.pdf

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 $\alpha$.

Answer 4

To add to Drake Marquis' nice answer, yet another formula for the $n = 2$ case is

$\det(A + B) = \det(A) + \det(B) + \text{tr}(A)\text{tr}(B) - \text{tr}(AB).$

The proof is given as follows:

$\det(A + B) = (A_{11} + B_{11})(A_{22} + B_{22}) - (A_{12} + B_{12})(A_{21} + B_{21})$

which expands into

$(A_{11}A_{22} - A_{12}A_{21}) + (B_{11}B_{22} - B_{12}B_{21}) + A_{11}B_{22} + B_{11}A_{22} - A_{12}B_{21} - B_{12}A_{21}.$

This can be written

$\det(A) + \det(B) + A_{11}B_{22} + B_{11}A_{22} - A_{12}B_{21} - B_{12}A_{21}.$

We now just need to verify the cross-terms. Now

$\text{tr}(A)\text{tr}(B) = (A_{11} + A_{22})(B_{11} + B_{22})$

which expands to

$A_{11}B_{11} + A_{11}B_{22} + A_{22}B_{11} + A_{22}B_{22}$

and

$\text{tr}(AB) = A_{11}B_{11} + A_{12}B_{21} + A_{21}B_{12} + A_{22}B_{22}.$

Therefore the difference $\text{tr}(A)\text{tr}(B) - \text{tr}(AB)$ is the cross term, completing the proof.

Answer 5

Assume $$X=[x_1,x_2], Y=[y_1,y_2]$$ $$|X+Y|=|x_1,x_2|+|y_1,x_2|+|x_1,y_2|+|y_1,y_2|$$

By inspection, select columns from one or the other exclusively form a determinant and sum the determinants.

We need an index and a column selection matrix.

Assume $k$ runs from $1$ to $2^N$ $N=\text {rank}[X+Y]$.

Also, k=Σbkj2^(j-1), bkj the bits of k.

A diagonal matrix is formed with bkj. It is the Bkd below.

Define also its complement Bkcd=I-Bkd

Then |X+Y|=Σk|Χ * Βkd+Y * Bkcd|

It is not perfect but you can get these matrices and do transformations e.g. where matrix pencils are involved (not just characteristic polynomials in one variable, or nonlinear eigenvalues etc.)

Also inversion of pencils based on these matrices.

It is not a simple sum, it involves 2^N terms from which some by chance may be zero.

Answer 6

For $n=2$, the identity follows from $$\det(A+\lambda B)=\det(A)+\operatorname{Tr}(A \operatorname{adj}(B))\lambda + (\det B) \lambda^2.$$