# Relation of Determinant and trace of a matrix concerning derivative

If there is a function

$$F(t) = det(I_{n} + tA)$$

where

$$A$$ is an $$n \times n$$ matrix,

$$t$$ is an arbitrary real number,

and $$I_{n}$$ is $$n \times n$$ identity matrix,

is it true that the derivative of $$F(t)$$ at $$t = 0$$ is equal to the trace of $$A$$?

That is,

$$F'(0) = Tr(A)$$

I currently know that the trace is the sum of the diagonal entries of a matrix but I am not sure how I should go about differentiating the right hand side.

Is there a general formula for finding determinant that I could possibly differentiate?

It seems like there is something called 'big formula for determinant' but I am not sure how that can be used for this problem.

Any help would be appreciated.

• there is no general formula for the determinant, unless the size of the matrix is specified, which is why in my answer the differentiation uses the definition of the derivative Oct 10, 2019 at 3:34
• also, you cannot differentiate (and should not want to) the matrix $A$ either; it is like saying you want to differentiate a number, i.e. $A$ is a constant. You are differentiating with respect to $t$, and therefore, the whole expression $det(I+tA)$. The bit you wrote, meaning $I+tA$ are the first two terms of the taylos expension of the matrix exponential, if that can in any way be helpful. Oct 10, 2019 at 3:37
• @ElenKhachatryan When the problem specifies $I_n$ the size of the matrix is specified. But this is irrelevnat: there is a forumal for the determinat of ANY matrix, it uses the size as one of the parameters. Oct 10, 2019 at 3:42

The formula for the detrminant of a mattrix is $$\det(B)=\sum_{\sigma \in S_n} \epsilon(\sigma) b_{1 \sigma(1)} \cdot ... \cdot b_{n \sigma(n)}$$

Now, in your case $$b_{i,j}= \left\{ \begin{array}{lc} 1+ta_{i,i} & \mbox{if i=j} \\ ta_{i,j} &\mbox{ if } i \neq j \end{array} \right.$$

Now, split your formula in 2 terms $$\det(F(t))=b_{11}b_{22}\cdot...\cdot b_{nn}+\sum_{\sigma \in S_n, \sigma \neq e} \epsilon(\sigma) b_{1 \sigma(1)} \cdot ... \cdot b_{n \sigma(n)}$$

Now, each term in $$Q(t):=\sum_{\sigma \in S_n, \sigma \neq e} \epsilon(\sigma) b_{1 \sigma(1)} \cdot ... \cdot b_{n \sigma(n)}$$ contains two non-diagonal entries. Since $$Q(t)$$ is a polynomial, and each non-dagonal entry has a factor of $$t$$, it follows that $$t^2$$ is a factor of $$Q(t)$$ and hence $$Q'(0)=0$$

Next, let $$P(t):=b_{11}b_{22}\cdot...\cdot b_{nn}=(1+ta_{11})(1+ta_{22})...(1+ta_{nn})$$ By using the product rule you immediatelly get $$P'(0)=a_{11}+a_{22}+..+a_{nn}=tr(A)$$

Since $$\det(F(t))=P(t)+Q(t)$$ you get $$(\det(F(t)))'|_{t=0}=P'(0)+Q'(0)=tr(A)$$

• Although I have written an answer myself, I like your answer that much that I have to upvote it. :-) Oct 10, 2019 at 13:15

You have $$\det(\lambda I-A) = (\lambda-\lambda_1)\cdots(\lambda-\lambda_n)$$, where $$\lambda_1,\ldots,\lambda_n$$ are the (possibly complex) eigenvalues of $$A$$. Hence, \begin{align} \det(I+tA) &= \det((-t)(-t^{-1}I-A)) = (-t)^n\det(-t^{-1}I-A)\\ &= (-t)^n(-t^{-1}-\lambda_1)\cdots(-t^{-1}-\lambda_n)\\ &= (1+t\lambda_1)\cdots(1+t\lambda_n)\\ &= 1+(\lambda_1+\ldots+\lambda_n)t+a_2t^2+\ldots+a_nt^n. \end{align} Now, deriving with respect to $$t$$ and setting $$t=0$$ yields the trace of $$A$$.

• "deriving with respect to $t$"?
– JDZ
Oct 10, 2019 at 4:41
• @JDZ What's your problem with that expression? Oct 10, 2019 at 13:17
• math.stackexchange.com/questions/863148/derive-or-differentiate
– JDZ
Oct 10, 2019 at 17:38
• You could have made this discussion shorter by just saying that the correct English verb is to differentiate and pointing to that site. However, everybody understands what I mean by "derive with respect to $t$". So I leave it as is. Oct 10, 2019 at 17:43

Let $$B$$ be a curve of isomorphisms in a vector space $$V$$, with $$\dim V=n$$. Then if $$\omega$$ is a volume form on $$V$$, and $$v_1,...,v_n\in V$$ we have that \begin{align}(\det (B))^\cdot& \omega(v_1,...,v_n)=\big(\det( B)\omega(v_1,...,v_n)\big)^\cdot\\ &= (\omega(Bv_1,...,Bv_n))^\cdot \\ &=\sum_{i=1}^n \omega(Bv_1,...,\dot{B}v_i,...,Bv_n)\\ &=\sum_{i=1}^n \omega(Bv_1,...,BB^{-1}\dot{B}v_i,...,Bv_n)\\ &= \det (B )\sum_{i=1}^n \omega(v_1,...,B^{-1}\dot{B}v_i,...,v_n) \\ &= \det (B ){\rm tr}(B^{-1}\dot{B})\omega(v_1,...,v_n).\end{align}By arbitrariety of the vectors and of the volume form chosen, it follows that $$(\det (B))^\cdot = \det (B) {\rm tr}(B^{-1}\dot{B})$$. If you don't like the dot notation and want to make the dependence on a parameter $$t$$ explicit, this means that $$\frac{{\rm d}}{{\rm d}t} \det B(t) = \det (B(t)){\rm tr}(B(t)^{-1}B'(t)).$$ Now consider $$F(t)=\det({\rm Id}+tA)$$. For small enough $$t$$, the matrix $${\rm Id}+tA$$ is non-singular, so we may use the above to compute the derivative at $$t=0$$ as $$F'(0) = \det({\rm Id}) {\rm tr}({\rm Id}^{-1}A)={\rm tr}(A),$$as wanted.

• Beautiful answer computing the first derivative of the determinant! First time I've seen it done without resorting to something nasty like the cofactor expansion. Nov 12, 2022 at 1:24

Here is a general rule for the differential of the determinant of $$X$$. $$d\big(\det(X)\big) = \det(X)\;{\rm Tr}(X^{-1}dX)$$ In the current problem we have \eqalign{ &X = (I + At) \;\implies\; dX = A\,dt \\ &d\big(\det(X)\big) = \det(X)\;{\rm Tr}\big(X^{-1}A\,dt\big) \\ &\frac{d\big(\det(X)\big)}{dt} = \det(X)\;{\rm Tr}\big(X^{-1}A\big) \\ } Then as $$t\to 0$$ \eqalign{ X &\to I \\ \det(X) &\to \det(I) = {\tt 1} \\ {\rm Tr}(X^{-1}A) &\to {\rm Tr}(A) \\ \\ }