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Well considering two $n \times n$ matrices does the following hold true: $$\det(A+B) = \det(A) + \det(B)$$ Can there be said anything about $\det(A+B)$?

If $A/B$ are symmetric (or maybe even of the form $\lambda I$) - can then things be said?

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    $\begingroup$ While this is a natural question to ask once you encounter the determinant, did you try any examples at all? $\endgroup$ Commented Aug 12, 2013 at 18:43
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    $\begingroup$ It would be a good exercise to determine for which matrices, the identity $\det(A+b)=\det(A)+\det(B)$ holds. I think that it would work for rather few of them. $\endgroup$
    – Prism
    Commented Aug 12, 2013 at 18:46
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    $\begingroup$ But you are expected to think a bit about things before asking questions. Sure, part of the question would still make sense to ask, but you would probably have known that the equality does not hold in general, and people might be focusing more on the more tricky part of the question. $\endgroup$ Commented Aug 12, 2013 at 18:58
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    $\begingroup$ @paul23 Working out some small, concrete examples is often the first step in figuring out how to prove something. $\endgroup$
    – littleO
    Commented Aug 12, 2013 at 19:47
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    $\begingroup$ In dimension $2$, you have the identity $\mathrm{det}(A+B) +\mathrm{tr}(AB) = \mathrm{det}(A) +\mathrm{det}(B) + \mathrm{tr}(A) \mathrm{tr}(B)$, but even this is false when the dimension is $3$ or higher. Have a look here. $\endgroup$
    – Mike F
    Commented Oct 7, 2015 at 22:18

5 Answers 5

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This does not hold true in general. For even $n$, let $A=-B$ and $\det(A) > 0$, so $\det(A+B)=0 < \det(A)+\det(B)$. Now, consider $A=B$. We have $\det(A+B)=\det(2A)=2^n \det(A) > 2 \det(A)=\det(A)+\det(B)$ for $n>1$ and $\det(A)>0$. Thus, either inequality can hold.

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In general, you can't expect a formula for $\det(A+B)$. But sometimes, when you're lucky, you can use the Matrix Determinant Lemma, which says the following:

$$\det(A+uv^T)=(1+v^TA^{-1}u)\det(A),$$

where $A$ is an invertible matrix and $v^TA^{-1}u$ is interpreted as a scalar. Therefore, if $A$ is invertible, and you can write $B$ as $uv^T$ for two vectors $u,v$, then now you have a formula. One could also note that $\det(uv^T)$ is always 0.


If you're looking for a matrix operation which is well-behaved with respect to matrix addition, look for the trace.

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No, there is no such law. Playing with simple matrices gives counterexamples, e.g.

$$\det\left(\begin{bmatrix} 1 & 0\\0&0\end{bmatrix}+\begin{bmatrix} 0 & 0\\0&1\end{bmatrix}\right)=\det\begin{bmatrix} 1 & 0\\0&1\end{bmatrix}=1,$$ but $$\det\begin{bmatrix} 1 & 0\\0&0\end{bmatrix}+ \det\begin{bmatrix} 0 & 0\\0&1\end{bmatrix}=0+0=0.$$

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Take $A=I_n=B$. Then $\det(A+B)=\det(2I_n)=2^n\det(I_n)=2^n$ and $\det(A)+\det(B)=1+1 = 2$ so for $n>1$, your equality does not hold, at least for these matrices. And for $n=1$, since the determinant is the only element of the matrix, we do have your equality. So $$\boxed{\Big[\forall A,B \in M_n\left(\Bbb R\right), \det(A+B)=\det(A)+\det(B)\Big]\iff n = 1\ }$$

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    $\begingroup$ Your boxed in statement isn't completely correct. You mean $n=1 \Rightarrow [\dots]$. It's not if and only if. $\endgroup$
    – abnry
    Commented Aug 12, 2013 at 19:06
  • $\begingroup$ @nayrb : I added brackets but no, I mean $\iff$. If you have the equality for all matrices, you have it for both matrix equal to $I_n$ so $n=1$ and if $n=1$ then you get the equality for all matrices. Or have I made a mistake somewhere? $\endgroup$
    – xavierm02
    Commented Aug 12, 2013 at 20:04
  • $\begingroup$ I'm sorry, I must have confused the definition of $M_n(R)$. Is $M_n(R) = \{\lambda I_n \in R^{n \times n}\;|\lambda \in R\}$? $\endgroup$
    – abnry
    Commented Aug 12, 2013 at 21:41
  • $\begingroup$ Okay, take this example. If $A = [[1, 0],[0, 0]]$ and $B = I_2 - B$, then $\det(A+B)=\det(I)=1$ but $\det(A)=\det(B) = 0$. So the space $M_n$ cannot be all real matrices, or even non singular ones. $\endgroup$
    – abnry
    Commented Aug 12, 2013 at 21:46
  • $\begingroup$ The second equation should read $B=I_2-A$. $\endgroup$
    – abnry
    Commented Aug 12, 2013 at 21:52
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Although the determinant function is not linear in general, I have a way to construct matrices $A$ and $B$ such that $\det(A + B) = \det(A) + \det(B)$, where neither $A$ nor $B$ contains a zero entry and all three determinants are nonzero:

Suppose $A = [a_{ij}]$ and $B = [b_{ij}]$ are 2 x 2 real matrices. Then $\det(A + B) = (a_{11} + b_{11})(a_{22} + b_{22}) - (a_{12} + b_{12})(a_{21} + b_{21})$ and $\det(A) + \det(B) = (a_{11} a_{22} - a_{12} a_{21}) + (b_{11} b_{22} - b_{12} b_{21})$.

These two determinant expressions are equal if and only if

$a_{11} b_{22} + b_{11} a_{22} - a_{12} b_{21} - b_{12} a_{21} = $ $\det \left[ \begin{array}{cc} a_{11} & a_{12}\\ b_{21} & b_{22} \end{array} \right]$ + $\det \left[ \begin{array}{cc} b_{11} & b_{12}\\ a_{21} & a_{22} \end{array} \right]$ = 0.

Therefore, if we choose any nonsingular 2 x 2 matrix $ A = [a_{ij}]$ with nonzero entries and then create $B = [b_{ij}]$ such that $b_{11} = - a_{21}, b_{12} = - a_{22}, b_{21} = a_{11},$ and $b_{22} = a_{12}$, we have solved our problem. For example, if we take $$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \quad \text{and}\quad B = \begin{bmatrix} -3 & -4 \\ 1 & 2\end{bmatrix} ,$$

then $\det(A) = -2, \det(B) = -2, $ and $\det(A + B) = -4$, as required.

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    $\begingroup$ how do you generalize this to large matrix? $\endgroup$
    – Turbo
    Commented Oct 24, 2016 at 22:43
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    $\begingroup$ Right now, I don't know how to generalize this. $\endgroup$
    – PolyaPal
    Commented Dec 4, 2016 at 4:16
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    $\begingroup$ what you have is a cool idea I think you should work on to generalize $\endgroup$
    – Turbo
    Commented Dec 5, 2016 at 4:53

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