derivative of determinant of $I + tA$ when $t=0$

Question

Let us say that $A$ is a real $n \times n$ matrix and $I$ is the identity matrix (also $n \times n$). $t$ is any old real number. I'm entirely aware that there is no simple formula for $\det(I + tA)$, but I hope I can find $\frac{d}{dt}(\det (I + tA))|_{t=0}$. I just can't see how that can be done? I need it in order to move on with this bigger problem I'm addressing. It's probably simple and I'm just not seeing it, could I please get some hints?

Answer 1

Note that if $A$ has eigenvalues $\lambda_i$, then $tA + I$ has exactly the eigenvalues $t\lambda_i + 1$. This is seen easily by the equivalence: $$ Av_i = \lambda_iv_i \iff tAv_i + v_i = (t\lambda_i + 1)v_i \iff (tA+I)v_i = (t\lambda_i+1)v_i $$

Now, the determinant of $(A+tI)$ is the product of the eigenvalues, so it is just $\prod_i(t\lambda_i + 1)$.

We have Vieta's formulas, that help us understand what are the coefficients of such a polynomial (in $t$). In particular, here we have: $$ \prod_{i=1}^n (1+t\lambda_i) = 1 + t\sum_{i=1}^n \lambda_i + O(t^2) $$

To see this in a simpler fashion, look at the expression $(1 + t\lambda_1)(1+t\lambda_2)...(1+t\lambda_n)$. To see which terms of the product contain the power $t$, you should pick exactly one of the brackets in the product above, and multiply the $t$ term in that, with all the constant terms in the other brackets. This way, you will get an expression with just one $t$. But then, the constants are just one, so for each bracket you pick yet the term $t\lambda_i$, and the sum of these is $t \sum \lambda_i$.

Now, as $t \to 0$, we know that the determinant, being a continuous function, goes to $1$. So the differential quotient looks like: $$ \frac{\prod_{i=1}^n (1+t\lambda_i) - 1}{t} = \sum_{i=1}^n \lambda_i + O(t) $$

Finally, the limit of the right hand exists as $t \to 0$, and equals the trace of $A$. Hence, the required derivative at zero, is nothing but the trace, the sum of either the eigenvalues, or the diagonal entries of $A$.

NOTE : this step is important in proving the Jacobi determinant formula, which gives the formula for the derivative of determinant of a differentiable function.

Answer 2

Remember there is a canonical identification $M_n(\mathbb R)\simeq \mathbb R^{n^2}$ defined as: $$(a_{ij})\longmapsto (a_{11}, \ldots, a_{1n}, a_{21}, \ldots, a_{2n}, \ldots, \ldots, a_{n1}, \ldots, a_{nn}).$$ Define $f_A: \mathbb R\longrightarrow \mathbb R^{n^2}$ setting $$f_A(t):=I+tA.$$ This is differentiable and $f_A^\prime(0)=A$. Also, $\det:\mathbb R^{n^2}\longrightarrow \mathbb R$ is differentiable with $$\frac{\partial \det}{\partial a_{ij}}(A)=\mathsf{adj}^t(A)_{ij},$$ hence, by chain rule:

$$\frac{d}{dt}\biggr|_{t=0} \det(I+tA)=(\det\circ f_A)^\prime(0)=\langle \nabla \det(f_A(0)), f^\prime_A(0)\rangle =\langle \nabla \det(I), A\rangle=\textrm{tr}(A),$$ where $\nabla \det(I)$ is the gradient of $\det$ at the point $I$ and $\langle \cdot, \cdot\rangle$ stands for the usual inner product in $\mathbb R^{n^2}$.

Remark. For proving $$\frac{\partial \det}{\partial a_{ij}}(A)=\mathsf{adj}^t(A)_{ij}$$ I suggest proving firstly that any multilinear map $$f:\mathbb R^{n_1}\times \ldots \mathbb R^{n_p}\longrightarrow \mathbb R^m$$ is differentiable with $$Df(a_1, \ldots, a_p)(v_1, \ldots, v_p)=\sum_{j=1}^n f(a_1, \ldots, v_j, \ldots, a_p),$$ and then apply this to $\det:\mathbb R^{n^2}\longrightarrow \mathbb R$.