Throughout, $A$ and $B$ denote $n \times n$ matrices over $\mathbb{C}$. Everyone knows that the determinant is multiplicative, and the trace is additive (actually linear). \begin{align*} \det(AB) = \det(A)\det(B) && \mathrm{tr}(A+B)= \mathrm{tr}(A) + \mathrm{tr}(B). \end{align*} On the other hand, the opposite equations \begin{align} \det(A+B) = \det(A)+\det(B) && \mathrm{tr}(AB)= \mathrm{tr}(A)\mathrm{tr}(B) \tag{1}. \end{align} don't hold for all $A,B$ unless $n=1$. For instance, taking $A=B=I$ in (1), we get \begin{align*} \det(A+B) = 2^n && \det(A)+\det(B) = 2 && \mathrm{tr}(AB)= n && \mathrm{tr}(A)\mathrm{tr}(B) = n^2. \end{align*} When $n=2$ a very curious thing happens, which is that, even though the equations (1) are typically false, their sum is actually valid. That is, \begin{align} \det(A+B) + \mathrm{tr}(AB)= \det(A)+\det(B)+ \mathrm{tr}(A)\mathrm{tr}(B) \tag{2} \end{align} for all $A$ and $B$ when $n=2$. However, (2) does not hold when $n>2$. Indeed, specializing to $B=I$ in (2), we get \begin{align} \det(A+I) + \mathrm{tr}(A)= \det(A)+1+ n \cdot \mathrm{tr}(A) \end{align} or, equivalently, \begin{align} \det(A+I) = \det(A)+(n-1) \cdot \mathrm{tr}(A) +1 \tag{3}. \end{align} If $n \geq 2$ and $A$ is projection onto the first coordinate, we have \begin{align} \det(A+I) = 2 && \det(A)+(n-1)\cdot \mathrm{tr}(A) +1 = n, \end{align} so (3) is only an identity for $n=2$ (or, trivially, when $n=1$).

Question: Is there any special significance to the equation \begin{align} \det(A+B) + \mathrm{tr}(AB)= \det(A)+\det(B)+ \mathrm{tr}(A)\mathrm{tr}(B) , \end{align} which is valid for all $2 \times 2$ matrices? It seems very strange to me that the sum of two obviously false equations should turn out true. Are there any nice applications of this identity?

  • $\begingroup$ Woah! This is such a crazy observation. (In a good way.) $\endgroup$ Oct 20, 2015 at 22:46

3 Answers 3


In a quite pedestrian way, this is just saying that $\det$ is a quadratic form with the trace as its polar form. Namely, for any matrix $A$, let $A^\star = \mathrm{Tr} A - A$ be its conjugate (as in “the conjugate root of the characteristic polynomial”). Then your formula is equivalent to $$ \det (A+B) - \det(A) - \det(B) = \mathrm{Tr}(A^\star \cdot B).$$ The left-hand side is the polar form associated to the determinant form.

In a slightly more abstract way, for any field $k$, the matrix algebra $k^{2\times 2}$ is a (split) quaternion algebra. The formula above is just the polar relation for the norm form in this algebra (which is the determinant of a matrix).

  • $\begingroup$ Thanks for the reply! I'm not sure I'm following, however. So, if I define $b(A,B) = Tr(A^\star B)$, then I get a bilinear form $b$ on the vector space of $2 \times 2$ matrices. It's even symmetric: $b(A,B) = Tr(A)Tr(B) - Tr(AB) = Tr(B)Tr(A) - Tr(BA) = b(B,A)$. OK, so that means it should satisfy a polarization identity: $b(A,B) = \frac{1}{2}\left( q(A) + q(B) - q(A-B) \right)$ where $q(X) = b(X,X)$ is the associated quadratic form. You are claiming the associated quadratic form is basically the determinant. However, I get $b(A,A) = Tr(A)^2 - Tr(A^2)$? What am I missing? $\endgroup$
    – Mike F
    May 5, 2015 at 7:26
  • $\begingroup$ OK never mind, I see. $Tr(A)^2 - Tr(A^2) = 2 \det(A)$ for $A$ 2-by-2. Which is pretty bizzare in and of itself... Thanks! $\endgroup$
    – Mike F
    May 5, 2015 at 7:34
  • $\begingroup$ This last fact is a simple consequence of the characteristic polynomial $A^2 - t A + d = 0$ - taking traces gives $$\mathrm{Tr} (A^2) - t \mathrm{Tr} (A) + \mathrm{Tr} (d) = 0,$$ or $$\mathrm{Tr} (A^2) - t^2 + 2d = 0.$$ $\endgroup$ May 5, 2015 at 10:05
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    $\begingroup$ That's a neat trick. So, trying the same thing for the $n=3$ case, I can use that $A$ satisfies its own characteristic polynomial: $A^3 + s_2(A) A^2 + Tr(A) A - \det(A) I = 0$, where $s_2(A)$ is defined to be the quadratic coefficient of the characteristic polynomial, and then take the trace to get $Tr(A^3) + s_2(A) Tr(A^2) + Tr(A)^2 - 3\det(A) = 0$ which shows that $\det(A) = \frac{1}{3} ( Tr(A^3) + s_2(A) Tr(A^2) + Tr(A)^2)$ $\endgroup$
    – Mike F
    May 5, 2015 at 16:41
  • $\begingroup$ The $n=2$ formula $\det(A) = Tr(A^2) - Tr(A)^2$ is a bit more intriguing to me, though, since it seems to suggest $\det(A)$ is something like the "variance" of the matrix a la the formula $Var(X) = E(X^2) - E(X)^2$. Certainly it is reasonable to think of the trace as some kind of expectation functional, being as how $\frac{1}{n} Tr(A)$ is the average of the spectrum. $\endgroup$
    – Mike F
    May 5, 2015 at 16:44

This is the case $n=2$ of a 1980 theorem of Amitsur. It is described (in more abstract notation) here and this question gives the reference to Amitsur's paper.

  • 1
    $\begingroup$ Thanks, that's nice to know! So, for instance, when $n=3$ we have the formula $\det(A+B) = \det(A)+\det(B) + s_2(A) \mathrm{tr}(B) + \mathrm{tr}(A) s_2(B) $ $- (\mathrm{tr}(A) + \mathrm{tr}(B)) \mathrm{tr}(AB) + \mathrm{tr}(A^2B + BAB)$ where $s_2(X)$ denotes the 2nd coefficient of the characteristic polynomial of $X$. I guess that last term could also be written as $\mathrm{tr}(A^2B) + \mathrm{tr}(AB^2)$. $\endgroup$
    – Mike F
    May 5, 2015 at 7:06
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    $\begingroup$ It would be be nice if you could quote the actual formula. $\endgroup$
    – nbubis
    May 5, 2015 at 15:39
  • $\begingroup$ @MikeF - Why does your formula true?.Please guide me. $\endgroup$
    – H....
    Feb 6, 2016 at 0:04
  • $\begingroup$ See my separate answer where I confirm this formula. Note the sign of the $s_2$ terms above assumes the convention $|x\mathbf 1-A|$ for the characteristic polynomial. $\endgroup$ Dec 26, 2019 at 9:25

@MikeF gives a further formula for $n=3$, under the answer by @Flounderer. (I would have made this answer just a comment there, but it's a bit too long.) The formula comes from de Smit apparently, as linked in that answer. See de Smit's line starting "$s_3(x+y) = \cdots$". I checked the formula using brute force in Mathematica. Note my added minus sign with the $s_2$ terms, because of the convention $|A-\mathbf 1x|$ for the characteristic polynomial: A = {{a11, a12, a13}, {a21, a22, a23}, {a31, a32, a33}} B = {{b11, b12, b13}, {b21, b22, b23}, {b31, b32, b33}} s2A = Coefficient[CharacteristicPolynomial[A, x], x] s2B = Coefficient[CharacteristicPolynomial[B, x], x] Simplify[ Det[A + B] == Det[A] + Det[B] - (s2A Tr[B] + Tr[A] s2B) - (Tr[A] + Tr[B]) Tr[A.B] + Tr[A.A.B] + Tr[A.B.B] ]


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