Let $M$ be a symmetric $n \times n$ matrix.

Is there any equality or inequality that relates the trace and determinant of $M$?

  • $\begingroup$ Also probably not what you're looking for but I feel should be here: $\det(M)=\mathop{\mathrm{tr}}(\bigwedge^n M)$. $M$ need not be symmetric. $\endgroup$
    – blargoner
    Commented Nov 23, 2022 at 14:32

7 Answers 7


Not exactly what you're looking for but I would be remiss not to mention that for any complex square matrix $A$ the following identity holds:

$$\det(e^A)=e^{\mbox{tr}(A)} $$

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    $\begingroup$ This is not really an answer to the question. $\endgroup$ Commented Jan 4, 2017 at 19:50
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    $\begingroup$ @MarianoSuárez-Álvarez I'm sure I'm missing something obvious, but why doesn’t this answer the question? (I up voted it because it was exactly what I was looking for, but should I?) $\endgroup$
    – Clumsy cat
    Commented Apr 14, 2017 at 11:20
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    $\begingroup$ @TheoreticalPerson Strictly speaking, this does not answer the question because the equation does not relate $\tr(A)$ with $\det(A)$. However, the equation is certainly important and elegant, and in some sense one can argue that it really answers "the spirit" of the question. $\endgroup$
    – leonbloy
    Commented Jul 28, 2017 at 19:50
  • $\begingroup$ how come it doesn't answer how trace and determinant isn't related? $\endgroup$
    – Rainb
    Commented Jan 24, 2020 at 9:45
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    $\begingroup$ The special linear Lie algebra $\mathfrak {sl}(n,\mathbb F)$ of $n \times n$ matrices of trace $0$ generates the special linear group (matrices with determinant $1$). $\endgroup$ Commented Nov 18, 2022 at 23:43

The determinant and the trace are two quite different beasts, little relation can be found among them.

If the matrix is not only symmetric (hermitic) but also positive semi-definite, then its eigenvalues are real and non-negative. Hence, given the properties ${\rm tr}(M)=\sum \lambda_i$ and ${\rm det}(M)=\prod \lambda_i$, and recalling the AM GM inequality, we get the following (probably not very useful) inequality:

$$\frac{{\rm tr}(M)}{n} \ge {\rm det}(M)^{1/n}$$

(equality holds iff $M = \lambda I$ for some $\lambda \ge 0$)

Much more interesting/insightful/useful are the answers by Owen Sizemore and Rodrigo de Azevedo.

  • $\begingroup$ n is the order of the matrix ($M_{n\times n}$) right? $\endgroup$
    – TPArrow
    Commented Jan 4, 2017 at 14:41
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    $\begingroup$ Yes, I'm following the question statement. $\endgroup$
    – leonbloy
    Commented Jan 4, 2017 at 14:42
  • $\begingroup$ What does positive definite mean here can anyone tell me? Matrix without a symmetric one can be positive definite ? $\endgroup$
    – Daman
    Commented Jan 23, 2018 at 6:05

The trace of $\bf M$ is the directional derivative of the determinant in the direction of $\bf M$ at ${\bf I}_n$, i.e.,

$$ \det \left( {\bf I}_n + h {\bf M} \right) = 1 + h \, \mbox{tr} ({\bf M}) + O \left( h^2 \right) $$

In Tao's words, "near the identity, the determinant behaves like the trace"$^\color{red}{\star}$. More generally,

$$\det( {\bf A} + h {\bf B} ) = \det({\bf A}) + h \, \mbox{tr} \left( \mbox{adj} ({\bf A}) \, {\bf B} \right) + O \left(h^2\right)$$

which is a variation of Jacobi's formula. Note that it is not required that $\bf M$ be symmetric.

$\color{red}{\star}$ Terence Tao, Matrix identities as derivatives of determinant identities, January 13, 2013.

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    $\begingroup$ While this is true, it has nothing to do with the question! $\endgroup$ Commented Jan 4, 2017 at 19:49
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    $\begingroup$ The OP asked for a relation between determinant and trace. I gave him one. Nowhere was it written that derivatives were not allowed. $\endgroup$ Commented Jan 4, 2017 at 19:55
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    $\begingroup$ He wants an inequality or an equality involving the trace and the determinant of a symmetric matrix. What you wrote does not provide that. $\endgroup$ Commented Jan 4, 2017 at 20:03
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    $\begingroup$ @RodrigodeAzevedo "the OP asked for relation between determinant and trace." Not. "the trace and determinant of M", the determinant and the trace of the same matrix (linear operator). $\endgroup$
    – vesszabo
    Commented Jan 11, 2017 at 14:27

No, there is not. Consider the matrix with parameter $n$ $$\begin{bmatrix} 1 & n \\ n &1 \\ \end{bmatrix}$$ The trace is 2, while the determinant is $1-n^2$. You can vary $n$ to violate any possible inequality between the trace and the determinant.

  • $\begingroup$ it is not a symmetric matrix $\endgroup$
    – TPArrow
    Commented Jan 4, 2017 at 14:21
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    $\begingroup$ Even when the matrix is diagonal: there is no relation between a sum of numbers and their product. $\endgroup$
    – user65203
    Commented Jan 4, 2017 at 14:26
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    $\begingroup$ $2 > 1-n^2$ always holds! In fact, in dimension 2, you can say something: see my answer $\endgroup$
    – user14972
    Commented Jan 4, 2017 at 18:07
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    $\begingroup$ It doesn't hold if $n = 2i$. $\endgroup$ Commented Jan 4, 2017 at 18:27
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    $\begingroup$ @Erik It is pretty rare to use the term "symmetric" (rather than "Hermitian" or "self-adjoint") when working with complex matrices. I doubt anyone looking for inequalities is referring to anything but the real case. $\endgroup$
    – Erick Wong
    Commented Jan 4, 2017 at 19:59

Due to OP's fairly general formulation there's diverse bunch of answers by now. In addition to these, I'd like to mention some concrete relations expressing the determinant in terms of traces. They hold without the symmetry hypothesis, just assume dealing with a general complex matrix.

Despite being "quite different beasts", both $\det(M)$ and $\operatorname{tr}(M)$ of an $n\times n$ matrix $M$ are given by some $n$-variable symmetric polynomial, evaluated at the eigenvalues of $M$.

For any fixed matrix size $n$ there's a polynomial in $n$ variables, of total degree $n$, which yields $\det(M)$ when evaluated at $\left(\operatorname{tr}(M^n),\operatorname{tr}(M^{n-1}),\ldots,\operatorname{tr}(M)\right)$: $$ \det M\; =\; \begin{cases} \operatorname{tr}M & n=1\\[1.67ex] \frac{1}{2}\big((\operatorname{tr}M)^2 - \operatorname{tr}\left(M^2\right)\big) & n=2\\[1.67ex] \frac{1}{6}\big((\operatorname{tr}M)^3 - 3\operatorname{tr}\left(M^2\right)(\operatorname{tr}M) + 2\operatorname{tr}\left(M^3\right)\big) & n= 3\\[1.67ex] \quad\ldots & n\ge4 \end{cases} $$

Case $\mathbf{n=1}$ is clear.

$\mathbf{n=2}$ is straightforward when applying Cayley-Hamilton to $M$ $$ M^2\:-\:(\operatorname{tr}M)\:M\:+\:(\det M)\pmatrix{1&0\\ 0&1}\;=\;\pmatrix{0&0\\0&0} $$ and taking the trace.
Referring to Paul's answer/example featuring $M=\left(\begin{smallmatrix}1&k\\ k&1\end{smallmatrix}\right)$, thus $M^2=\left(\begin{smallmatrix}1+k^2&2k\\ 2k&1+k^2\end{smallmatrix}\right)$,
one obtains $$\det M\;=\;\frac{1}{2}\big(4-2(1+k^2)\big) = 1-k^2$$ for the sake of illustration, presumably not the quickest way towards $\det(M)$
(the parameter has been renamed to $k$ since $n$ denotes the matrix size).

Cases $\mathbf{n\ge3}$ get more expensive $\ldots$ and are available:
A suitable entry point is the corresponding subsection in the Wikipedia entry on determinants.

If you clicked the preceding link then you may scroll down just a bit to get into a det-tr-inequality for a positive-definite matrix, also worthwhile as answer to the OP. Or skip directly down to it ;-)


Up to sign, the trace and determinant of an $n \times n$ matrix are coefficients of its characteristic polynomial (specifically, the coefficients in degrees $n-1$ and $0$ respectively).

The only constraint that the matrix being symmetric adds is that the characteristic polynomial is totally real — that is, all of its roots are real.

(note every totally real polynomial is a characteristic polynomial; e.g. of the diagonal matrix whose entries are the roots)

Thus, your problem is equivalent to

Let $f$ be a monic polynomial of degree $n$ whose roots are all real. Is there any relationship between the coefficient in degree $n-1$ and the constant coefficient?

I believe you can only say anything when $n \leq 2$. When $n=2$, the requirement that the roots be real implies that $\mathrm{tr}(M)^2 \geq 4 \det(M)$.

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    $\begingroup$ Just to elaborate on the last point for the OP's benefit, the coefficients of the characteristic polynomial for $n = 2$ are, up to sign, exactly $\operatorname{tr} M$ and $\det M$; looking at the discriminant gives the indicated inequality. For higher $n$, there are more terms in the minimal polynomial. $\endgroup$
    – anomaly
    Commented Jan 4, 2017 at 18:31
  • $\begingroup$ +1 I was looking through the answers to post exactly this if no one had already! $\endgroup$
    – user541686
    Commented Jan 6, 2017 at 11:30

The only relation I can think of is this one : Since the matrix $A$ is symetric, it is diagonalizable, thus it can be written $A = P^{-1}DP$ where $D$ is diagonal and $P$ is invertible. Therefore $$\det A = \prod_{i = 1} ^n \lambda_i$$ and $$tr A = \sum_{i = 1} ^n \lambda_i$$ where $\lambda_i$ is an eigenvalue.

Some inequalities can be found between the sum and the product.

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    $\begingroup$ Note that these equalities are true in general and do not require symmetry or diagonalizability; this can i.e. be seen by trigonalisation in C. $\endgroup$
    – Maxim
    Commented Jan 4, 2017 at 15:37
  • $\begingroup$ You're right, this comes from the characteristic polynomial. $\endgroup$
    – Blencer
    Commented Jan 4, 2017 at 15:41
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    $\begingroup$ Another way of expressing this relationship: $$\left .\frac {d}{d\lambda} \det(A + \lambda I)\right |_{\lambda = 0} = \operatorname{tr}(A)$$ $\endgroup$ Commented Jan 4, 2017 at 17:37
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    $\begingroup$ @PaulSinclair No, the derivative should be the trace of the adjugate of $A$ rather than the trace of $A$. $\endgroup$
    – user1551
    Commented Jan 6, 2017 at 16:13
  • $\begingroup$ @user1551 - Gah - you're right. What I was intending is $$\left .\frac{1}{(n-1)!}\frac {d^{n-1}}{d\lambda^{n-1}} \det(A + \lambda I)\right |_{\lambda = 0} = \operatorname{tr}(A)$$, but I went for the wrong coefficient of the characteristic polynomial. $\endgroup$ Commented Jan 6, 2017 at 17:37

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