The problem is to find the second order term in the series expansion of the expression $\mathrm{det}( I + \epsilon A)$ as a power series in $\epsilon$ for a diagonalizable matrix $A$. Formally, we will write the series as follows

$$ \det( I + \epsilon A ) = f_0(A) + \epsilon f_1(A) + \epsilon^2 f_2(A) + \cdots +\epsilon^N f_N(A), $$

In particular, we are looking for an expression in terms of the trace of the matrix $A$.

  • 1
    $\begingroup$ I'm posting this question and answer because I needed this identity a couple of weeks ago for a calculation and couldn't find it on the site. $\endgroup$
    – Spencer
    Commented Mar 4, 2015 at 4:39

2 Answers 2


Note that the determinant of a matrix is just a polynomial in the components of the matrix. This means that the series in question is finite. This also means that the functions $f_1,f_2,\dots,f_N$ are polynomials.

We start by computing the term which is first order in $\epsilon$.

Let $A_1, A_2, \dots , A_N$ be the column vectors of the matrix $A$. Let $e_1,e_2,\dots, e_N$ be the standard basis; note that these basis vectors form the columns of the identity matrix $I$. Then we recall that the determinant is an alternating multi-linear map on the column space.

$$ \mathrm{det}( I + \epsilon A ) = \mathrm{det}( e_1 + \epsilon A_1, e_2 + \epsilon A_2, \dots , e_N + \epsilon A_N ) $$

$$ = \mathrm{det}( e_1, e_2, \dots , e_N ) + \epsilon \lbrace \mathrm{det}(A_1,e_2,\dots, e_N) + \mathrm{det}(e_1,A_2,\dots, e_N) + \cdots + \mathrm{det}(e_1,e_2,\dots, A_N)\rbrace + O(\epsilon^2)$$

The first term is just the determinant of the identity matrix which is $1$. The term proportional to $\epsilon$ is a sum of expressions like $\mathrm{det}(e_1,e_2,\dots, A_j, \dots, e_N)$ where the $j$'th column of the identity matrix is replaced with the $j$'th column of $A$. Expanding the determinant along the $j$'th row we see that $\mathrm{det}(e_1,e_2,\dots, A_j, \dots, e_N) = A_{jj}$.

$$ \mathrm{det}( I + \epsilon A ) = 1 + \epsilon \sum_{j=1}^N A_{jj}+ O(\epsilon^2) = 1 + \epsilon \mathrm{Tr}(A) + O(\epsilon^2)$$

$$\boxed{ f_1(A) = \mathrm{Tr}(A) } \qquad \textbf{(1)} $$

We have the first term in our series in a computationally simple form. Our goal is to obtain higher order terms in the series in a similar form. To do this we will have to abandon the current method of attack and consider the determinant of the exponential map applied to a matrix.

If $A$ is diagonalizable then we can define $\exp(A)$ in terms of its action on the eigenvectors of $A$. If $a_1,a_2,\dots , a_N$ are eigenvectors of $A$ with eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_N$ then $\exp(A)$ is the matrix which satisfies,

$$ \exp(A) a_j = e^{\lambda_j} a_j.$$

It is not hard to show that $\det( \exp(A) ) = \exp(\mathrm{Tr}(A))$. Since $A$ is linear operator on a finite dimensional vector space it has a finite norm. This means that we can safely evaluate the exponential map as an infinite series. The infinite series is consistent with our definition in terms of the eigenbasis. Consider the following,

$$\det( \exp(\epsilon A) ) = \exp(\epsilon \mathrm{Tr}(A))$$

$$\det( I + \epsilon A + \frac{\epsilon^2}{2} A^2 +\cdots ) = 1 + \epsilon \mathrm{Tr}(A) + \frac{\epsilon^2}{2} (\mathrm{Tr}(A))^2 + \cdots \qquad (*)$$

On the left hand side of $(*)$ we can factor an $\epsilon$ an write,

$$ \det( I + \epsilon \lbrace A + \frac{\epsilon}{2} A^2 +\cdots \rbrace ) = 1 + \epsilon \mathrm{Tr}(A + \frac{\epsilon}{2} A^2 + \cdots)+ \epsilon^2 f_2( A + \frac{\epsilon}{2} A^2 + \cdots ) + O(\epsilon^3)$$

$$ = 1 + \epsilon \mathrm{Tr}(A) + \epsilon^2 \lbrace \frac{1}{2}\mathrm{Tr}( A^2)+f_2( A + \frac{\epsilon}{2} A^2 + \cdots ) \rbrace + O(\epsilon^3)$$

Now we compare the second order terms in $\epsilon$ and obtain,

$$ \frac{1}{2}\mathrm{Tr}( A^2)+f_2( A + \frac{\epsilon}{2} A^2 + \cdots ) = \frac{1}{2}(\mathrm{Tr}(A))^2, $$

Now allow $\epsilon \rightarrow 0 $,

$$ \boxed{f_2( A ) = \frac{\mathrm{Tr^2}(A)-\mathrm{Tr}(A^2)}{2}} \qquad \textbf{(2)}. $$

The higher order terms can be obtained systematically using the same trick as above though the computations become more involved.

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    $\begingroup$ Notice that this approach assumes that the ground field has characteristic $0$ and that the matrix is diagonalizable. The first assumption cannot be lifted (as witnessed by the $2$ in the denominator of (2)), while the second can. I am not sure of a good source for these formulas, but they are well-known among representation theorists. The idea is that in characteristic $0$, you can write the elementary symmetric polynomials in $n$ variables as polynomials in the power sums (see ... $\endgroup$ Commented Mar 4, 2015 at 17:49
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    $\begingroup$ ... en.wikipedia.org/wiki/… , Section 2.2), and then the same polynomials give you the $f_i\left(A\right)$ in terms of the $\operatorname{Tr}\left(A^i\right)$. The shortest (though not the only) proof of this proceeds by assuming WLOG that the matrix $A$ is diagonalizable (since the diagonalizable matrices, at least after passing to an algebraic closure, form a Zariski-dense subset of the whole matrix ring) and using eigenvalues in this case. $\endgroup$ Commented Mar 4, 2015 at 17:50
  • $\begingroup$ Hey that looks pretty cool, thanks for sharing. $\endgroup$
    – Spencer
    Commented Mar 4, 2015 at 18:16
  • $\begingroup$ This is what I am looking for. Can anyone recommend me some good (more extensive) reference for this? $\endgroup$
    – venrey
    Commented Sep 7, 2018 at 3:34
  • $\begingroup$ @darijgrinberg the formulas you are referencing let us express det(A) for an nxn matrix in terms of tr(A^j), j=1,...,n. how do you get the taylor coefficients of det(1+A) out of this? $\endgroup$
    – peter
    Commented Nov 20, 2018 at 13:08

The full formula for the expansion of the determinant of an nxn matrix A in a polynomial of traces of powers of A is: $$\det(A)=\sum_{\pi\in\Pi(n)}(-1)^{|\pi|-n}\prod_{m=1}^n\frac{(Tr(A^m))^{\pi_m}}{m^{\pi_m}\pi_m!}$$ where symbols are as follows: $\Pi(n)$ is the set of partitions of $n$ in positive integers; its elements are sequences of non negative integer multipliers, but a finite number of them different from zero, such that $$1\cdot \pi_1+2\cdot\pi_2+...n\cdot \pi_n=n.$$ E.g. the set of partitions of 3 has 3 elements, namely $$\Pi(3)=\{(3,0,\ldots),(1,1,0,\ldots), (0,0,1,0,\ldots)\},$$ meaning that the number 3 can be written as a sum of distinct positive integers with multiplicities in these ways: $$3=1\cdot3, 3=1\cdot 1+2\cdot 1, 3=3\cdot 1.$$ $|\pi|=\sum_{m=1}^{\infty}\pi_m$ is the length of the partition $\pi$, (the total number of non negative integers to be summed) e.g.$|(3,0,\ldots)|=3$, $|(1,1,0,\ldots)|=2$, etc. The formula can be derived, without assuming that A is diagonalizable, but of course assuming characteristic 0, by expanding $\ln(\lambda+A)=\ln\lambda +\ln(1+\frac{A}{\lambda})$ in a Taylor series, which converges for $\lambda>||A||$, then applying $\det e^B=e^{Tr(B)}$, expanding again the double series which is actually a polynomial because $\det(\lambda+A)$ is, and collecting terms of order 0 in $\lambda$. I believe there are other ways (i.e. combinatorial, using Newton's identities or something of the kind) to derive it, but I have not seen an explicit proof. The formula given in previous answer for a 2x2 matrix follows using the partitions of 2 $$\Pi(2)=\{(2,0,\ldots),(0,1,0,\ldots)\}$$ to compute $$\det(A)=(-1)^{2-2}\frac{(Tr(A))^2}{1^2\cdot 2!}+(-1)^{1-2}\frac{Tr(A^2)}{2^1\cdot 1!}.$$

As another example, using the partitions of 3 given above, one would find for a 3x3 matrix: $$\det(A)=(-1)^{3-3}\frac{(Tr(A))^3}{1^3\cdot 3!}+(-1)^{2-3}\frac{Tr(A)\cdot Tr(A^2)}{1^1\cdot1!\cdot 2^1\cdot 1!}+(-1)^{1-3}\frac{Tr(A^3)}{3^1\cdot1!}$$ or $$\det(A)=\frac{1}{6}\left((Tr(A)^3-3Tr(A^2)\cdot Tr(A)+2Tr(A^3)\right).$$

For a 4x4 matrix one ends up with $$\det(A)=\frac{(Tr(A))^4}{24}-\frac{(Tr(A))^2\cdot Tr(A^2)}{4}+\frac{(Tr(A^2))^2}{8}+\frac{Tr(A)\cdot Tr(A^3)}{3}-\frac{Tr(A^4)}{4}$$ and so on.

EDIT I'm editing my post, because I had forgotten to mention, that by the same token one can obtain a general formula for all the coefficients of the characteristic polynomial of A, not just the determinant. For an nxn matrix A, by defining $$\Delta_A(\lambda)=\det(\lambda+A)=\sum_{r=0}^{n}P_r(A)\lambda^r$$ (notice the unusual sign though) and expanding as above, one finds $$P_r(A)=\sum_{\pi\in\Pi(n-r)}(-1)^{|\pi|-n+r}\prod_{m=1}^{n-r}\frac{(Tr(A^m))^{\pi_m}}{m^{\pi_m}\pi_m!}$$ of which $\det(A)=P_0(A)$ is a particular case.

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    $\begingroup$ This is known in the literature as Girard-Waring formula. $\endgroup$
    – user1551
    Commented May 13, 2018 at 18:15
  • $\begingroup$ Thank you! I knew it is known and I had read it someplace, but I couldn't remember the name. $\endgroup$
    – wolpert
    Commented May 13, 2018 at 18:55
  • $\begingroup$ Is the characteristic polynomial written right? or is just assumed that $$ \det(\lambda + A) = \det(\lambda I + A) $$ I am just confused.. $\endgroup$
    – venrey
    Commented Sep 7, 2018 at 4:01
  • $\begingroup$ Sorry for the late answer. You are correct, $\det(\lambda+A)$ was assumed to mean $\det(\lambda I+A)$, the identity matrix being implied. This is not the common definition of the characteristic polynomial: that would be $\det(-\lambda I+A)$. I used a different sign in front of $\lambda$, because it gave (or seemed to me, at the time, to give) rise to fewer minus signs to keep track of, in the derivation I had sketched. $\endgroup$
    – wolpert
    Commented Sep 25, 2018 at 22:25
  • $\begingroup$ actually what was calculated in a previous answer was the second taylor coefficient of det(1+A), but the formula for det(A) in case A is 2x2 is the same. is this coincidence? why would the formula for the n-th taylor coefficient be the same as the det in terms of tr formula for an nxn matrix? $\endgroup$
    – peter
    Commented Nov 20, 2018 at 13:06

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