The relation between trace and determinant of a matrix

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

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

Solution 1:

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:

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

Solution 2:

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.

Solution 3:

The trace of $\mathrm M$ is the directional derivative of the determinant in the direction of $\mathrm M$ at $\mathrm I_n$, i.e.,

$$\det (\mathrm I_n + h \mathrm M) = 1 + h \, \mbox{tr} (\mathrm M) + O (h^2)$$

In Tao's words, "near the identity, the determinant behaves like the trace" [0]. More generally,

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

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

[0] Terence Tao, Matrix identities as derivatives of determinant identities, January 13, 2013.

Solution 4:

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.