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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? $\endgroup$ – Tobias Kildetoft Aug 12 '13 at 18:43
  • $\begingroup$ 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 Aug 12 '13 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. $\endgroup$ – Tobias Kildetoft Aug 12 '13 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 Aug 12 '13 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 Oct 7 '15 at 22:18
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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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    $\begingroup$ You still have a problem with your definition of $B$ in your counterexample. If $n$ is odd, then $\det(A+B)=2^n$, not zero. $\endgroup$ – M Turgeon Aug 12 '13 at 18:51
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    $\begingroup$ Thanks- the dangers of making counterexamples late at night! $\endgroup$ – Devlin Mallory Aug 12 '13 at 19:02
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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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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 matrix. And for $n=1$, since the determinant is the only element of the matrix, we do have your equality. So $\boxed{\left[\forall A,B \in M_n\left(\Bbb R\right), \det(A+B)=\det(A)+\det(B)\right]\iff n = 1}$

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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 Aug 12 '13 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 Aug 12 '13 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 Aug 12 '13 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 Aug 12 '13 at 21:46
  • $\begingroup$ The second equation should read $B=I_2-A$. $\endgroup$ – abnry Aug 12 '13 at 21:52
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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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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? $\endgroup$ – Brout Oct 24 '16 at 22:43
  • $\begingroup$ Right now, I don't know how to generalize this. $\endgroup$ – PolyaPal Dec 4 '16 at 4:16
  • $\begingroup$ what you have is a cool idea I think you should work on to generalize $\endgroup$ – Brout Dec 5 '16 at 4:53

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