Second derivative of $\det(\mathbb{1}+tA)$?

I am given a function $F : \mathbb{R} \to \mathbb{R}$ defined by

$$F(t)=\det(\mathbb{1}+tA)$$

where $A \in \mathbb{R}^{n \times n}$. As far as I know, the following is true.

$$\frac{d}{dt}\bigg|_{t=0} F(t) = \text{tr}~ A$$

However, how to find the second derivative?

$$\frac{d^2}{dt^2}\bigg|_{t=0} F(t)$$

• Why isn't the first derivative $\mathrm{det}(A)$? – rafa11111 Jun 25 '18 at 13:50
• @rafa11111 math.stackexchange.com/a/1526473/493774 – Simon Mueller Jun 25 '18 at 13:53
• Following @SimonMueller, the OP may try to use the formula and derive an expression, but I imagine it may not look very nice. – Gregory Jun 25 '18 at 14:01
• Notice that $\det(1+tA)$ is a polynomial. It's second derivative at $t=0$ equals twice what's the term by $t^2$ – Jakobian Jun 25 '18 at 14:11
• Closely related: Second derivative of a determinant. – mathcounterexamples.net Jun 25 '18 at 14:15

Edit: Here's a totally different argument, probably simpler, with more linear algebra and less calculus. Say $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $A$, listed according to (algebraic) multiplicity. Since $\det(I+tA)$ is the product of the eigenvalues of $I+tA$ it follows that $$F(t)=\prod_{j=1}^n(1+t\lambda_j).$$Multiplying that out in the imagination makes it clear that $$F'(0)=\sum\lambda_j=\text{tr}(A),$$and $$F''(0)=2\sum_{j< k}\lambda_j\lambda_k.$$Since the eigenvalues of $A^2$ are $\lambda_j^2$ it follows that $$F''(0)=\left(\sum\lambda_j\right)^2-\sum\lambda_j^2=(\text{tr}(A))^2-\text{tr}(A^2),$$as below.

Update: There's a slightly iffy spot above that nobody seemed to notice. It's easy to see that $\lambda$ is an eigenvalue of $A^2$ if and only if $\lambda=\omega^2$ where $\omega$ is an eigenvalue of $A$, but above we need more than that: We need to know that the algebraic multiplicities are the same. Maybe that's obvious? (Ok, it's clear from the Jordan form, probably clear just from the fact that the algebraic multiplicity is the dimension of the "generalized eigenspace", but it should be easier than that...)

Cheap trick: It's clear if the eigenvalues are distinct. And matricies with distinct eigenvalues are dense, hence the trace of $A^2$ is $\sum\lambda_j^2$ for every $A$.

Or, in better taste: Since $\det(\lambda I-A)=\prod(\lambda-\lambda_j)$, \begin{align}\det(\lambda^2I-A^2)&=\det(\lambda I-A)\det(\lambda I+A) \\&=(-1)^n\prod(\lambda-\lambda_j)\prod(-\lambda-\lambda_j) \\&=\prod(\lambda^2-\lambda_j^2).\end{align} Hence $\det(\lambda I-A^2)=\prod(\lambda-\lambda_j^2)$.

Thought of this because I was bothered by the fact that the expression for $F(t)$ derived below doesn't look like a polynomial:

Exercise Use the fact that $\text{tr}(A^k)=\sum_j\lambda_j^k$ to show that $\exp\left(t\,\text{tr}A-\frac{t^2}2\text{tr}(A^2)+\frac{t^3}3\text{tr}(A^3)+\dots\right)=\prod(1+t\lambda_j)$ (for small $t$).

Original: Yes, $F'(0)=\text{tr} A$.

Say $B(t)=I+tA$. Note that $B(t)$ is invertible if $t$ is small enough. So if $t$ and $h$ are both small then $$F(t+h)=\det(B(t)+hA) =\det(B(t))\det(I+hB(t)^{-1}A).$$Taking the derivative with respect to $h$ shows that \begin{align}F'(t)&=\det(B(t))\text{tr}(B(t)^{-1}A) \\&=F(t)\text{tr}((I-tA+t^2A^2-t^3A^3\dots)A) \\&=(1+t\,\text{tr}A+\dots)(\text{tr}A-t\,\text{tr}(A^2)+\dots).\end{align}Hence $$F''(0)=(\text{tr}A)^2-\text{tr}(A^2).$$

Bonus: There's a differential equation above, saying that $$F'/F=\text{tr}A-t\,\text{tr}(A^2)+\dots.$$With the initial condition $F(0)=1$ this shows that $$F(t)=\exp\left(t\,\text{tr}A-\frac{t^2}2\text{tr}(A^2)+\frac{t^3}3\text{tr}(A^3)+\dots\right),$$which should allow you to find as many derivatives as you want.

• Using the logarithmic derivative could simpify the initial differentiation here maybe.. – Pixel Jun 29 '18 at 5:49
• David, "$λ$ is an eigenvalue of $A$ if and only if $λ^2$ is an eigenvalue of $A^2$"; are you sure ? – user91684 Jun 30 '18 at 10:38
• @loupblanc Aargh. Of course there's a true fact there, but the way II phrased it is nonsense - thanks. – David C. Ullrich Jun 30 '18 at 13:27
• @DavidC.Ullrich: Nice solution. I have a different solution in terms of well known results on linear ODEs. It more closely related to you bonus section. – Oliver Diaz Apr 8 at 19:50

If in $A(t)=(a_{ij}(t))_{1\leq i,j \leq n}$ each of the components is a $k$ times differentiable real-valued function, then we have (see e.g. this link): \begin{align*} \frac{d^{k}}{dt^k} \begin{vmatrix} a_{11}(t)&a_{12}(t)&\cdots&a_{1n}(t)\\ a_{21}(t)&a_{22}(t)&\cdots&a_{2n}(t)\\ \vdots&\vdots&&\vdots\\ a_{n1}(t)&a_{n2}(t)&\cdots&a_{nn}(t)\\ \end{vmatrix} =\sum_{{k_1+k_2+\cdots+k_n=k}\atop{k_j\geq 0, 1\leq j\leq n}}\frac{k!}{k_1!k_2!\ldots k_n!} \begin{vmatrix} a_{11}^{(k_1)}(t)&a_{12}^{(k_1)}(t)&\cdots&a_{1n}^{(k_1)}(t)\\ a_{21}^{(k_2)}(t)&a_{22}^{(k_2)}(t)&\cdots&a_{2n}^{(k_2)}(t)\\ \vdots&\vdots&&\vdots\\ a_{n1}^{(k_n)}(t)&a_{n2}^{(k_n)}(t)&\cdots&a_{nn}^{(k_n)}(t)\\ \end{vmatrix} \end{align*}

Here we have the special case $n=2$ and $A(t)=1+tA$ with $A$ an $n\times n$ constant matrix. We obtain \begin{align*} &\color{blue}{\frac{d^{2}}{dt^2} \begin{vmatrix} 1+a_{11}t&a_{12}t&\cdots&a_{1n}t\\ a_{21}t&1+a_{22}t&\cdots&a_{2n}t\\ \vdots&\vdots&&\vdots\\ a_{n1}t&a_{n2}t&\cdots&1+a_{nn}t\\ \end{vmatrix}}\\ &\qquad\quad=\sum_{{k_1+k_2+\cdots+k_n=2}\atop{k_j\geq 0, 1\leq j\leq n}}\frac{2!}{k_1!k_2!\ldots k_n!} \begin{vmatrix} (1+a_{11}t)^{(k_1)}&(a_{12}t)^{(k_1)}&\cdots&(a_{1n}t)^{(k_1)}\\ (a_{21}t)^{(k_2)}&(1+a_{22}t)^{(k_2)}&\cdots&(a_{2n}t)^{(k_2)}\\ \vdots&\vdots&&\vdots\\ (a_{n1}t)^{(k_n)}&(a_{n2}t)^{(k_n)}&\cdots&(1+a_{nn}t)^{(k_n)}\\ \end{vmatrix}\\ &\qquad\quad\color{blue}{=2\sum_{{k_1+k_2+\cdots+k_n=2}\atop{0\leq k_j\leq 1, 1\leq j\leq n}}\frac{1}{k_1!k_2!\ldots k_n!} \begin{vmatrix} (1+a_{11}t)^{(k_1)}&(a_{12}t)^{(k_1)}&\cdots&(a_{1n}t)^{(k_1)}\\ (a_{21}t)^{(k_2)}&(1+ta_{22})^{(k_2)}&\cdots&(a_{2n}t)^{(k_2)}\\ \vdots&\vdots&&\vdots\\ (a_{n1}t)^{(k_n)}&(a_{n2}t)^{(k_n)}&\cdots&(1+a_{nn}t)^{(k_n)}\\ \end{vmatrix}} \end{align*} Observe the simplification in the last step in the index range, where we set $0\leq k_j\leq 1$. Since if there would be an index $k_j=2$ the corresponding second derivative produces zeros in the $j$-th row, so that the determinant evaluates to $0$.

Example: $n=2$

Let's do the calculation for small $n=2$. We get \begin{align*} \frac{d^{2}}{dt^2} \begin{vmatrix} 1+a_{11}t&a_{12}t\\ a_{21}t&1+a_{22}t\\ \end{vmatrix} &=2\sum_{{k_1+k_2=2}\atop{0\leq k_j\leq 1, 1\leq j\leq 2}}\frac{1}{k_1!k_2!} \begin{vmatrix} (1+a_{11}t)^{(k_1)}&(a_{12}t)^{(k_1)}\\ (a_{21}t)^{(k_2)}&(1+a_{22}t)^{(k_2)}\\ \end{vmatrix}\\ &=2 \begin{vmatrix} (1+a_{11}t)^{(1)}&(a_{12}t)^{(1)}\\ (a_{21}t)^{(1)}&(1+a_{22}t)^{(1)}\\ \end{vmatrix}\\ &=2 \begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\\ \end{vmatrix}\\ &=2(a_{11}a_{22}-a_{12}a_{21})\tag{1} \end{align*}

On the other hand we have \begin{align*} \frac{d^{2}}{dt^2} \begin{vmatrix} 1+a_{11}t&a_{12}t\\ a_{21}t&1+a_{22}t\\ \end{vmatrix} &=\frac{d^{2}}{dt^2}\left[(1+a_{11}t)(1+a_{22}t)-a_{12}a_{21}t^2\right]\\ &=\frac{d^{2}}{dt^2}\left[1+a_{11}t+a_{22}t+a_{11}a_{22}t^2-a_{12}a_{21}t^2\right]\\ &=2(a_{11}a_{22}-a_{12}a_{21}) \end{align*} in accordance with (1).

Note the answer of @DavidCUllrich coincides with the result above \begin{align*} \color{blue}{(\text{tr}A)^2-\text{tr}(A^2)} &=(a_{11}+a_{22})^2-\text{tr}\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}^2\\ &=(a_{11}+a_{22})^2-\text{tr}\begin{pmatrix}a_{11}^2+a_{12}a_{21}&\ast \\ \ast&a_{12}a_{21}+a_{22}^2\end{pmatrix}\\ &\,\,\color{blue}{=2(a_{11}a_{22}-a_{12}a_{21})} \end{align*}

• @DavidC.Ullrich: Answer corrected and the results now coincide. Thanks for the support. – Markus Scheuer Jun 25 '18 at 18:08

The following is inspired by David C. Ullrich's answer, namely its bonus section. Recall that $$\color{blue}{\det \big( \exp(\mathrm M) \big) = \exp \big( \mbox{tr} (\mathrm M) \big)}$$

\begin{aligned} f (t) := \det ( \mathrm I_n + t \mathrm A ) &= \det \big( \exp \big( \ln \big( \mathrm I_n + t \mathrm A \big) \big) \big)\\ &= \exp \big( \underbrace{\mbox{tr} \big( \ln \big( \mathrm I_n + t \mathrm A \big) \big)}_{=: g (t)} \big) = \exp \big( g (t) \big) \end{aligned}

From the Maclaurin expansion of $$\ln (1+x)$$, we obtain

\begin{aligned} g (t) &= \mbox{tr} \big( t \mathrm A - \frac{t^2}{2} \mathrm A^2 + \frac{t^3}{3} \mathrm A^3 - \cdots \big)\\ &= t \,\mbox{tr} (\mathrm A) - \frac{t^2}{2} \mbox{tr} (\mathrm A^2) + \frac{t^3}{3} \mbox{tr} (\mathrm A^3) - \cdots\end{aligned}

and

$$g ' (t) = \mbox{tr} (\mathrm A) - t \, \mbox{tr} (\mathrm A^2) + t^2 \, \mbox{tr} (\mathrm A^3) - \cdots$$

and

$$g '' (t) = - \mbox{tr} (\mathrm A^2) + 2t \, \mbox{tr} (\mathrm A^3) - \cdots$$

Differentiating $$f$$ twice,

$$f '' (t) = \big( g''(t) + \left( g'(t) \right)^2 \big) \, f(t)$$

and, thus,

$$f''(0) = \big ( - \mbox{tr} (\mathrm A^2) + \left( \mbox{tr} (\mathrm A) \right)^2 \big) \, \underbrace{f(0)}_{=1} = \color{blue}{\big( \mbox{tr} (\mathrm A) \big)^2 - \mbox{tr} \left( \mathrm A^2 \right)}$$

Here is another solution based on the known properties of the Wronskian of linear differential equations:

Consider the system \begin{align} \dot{\mathbf{x}}=B(t)\mathbf{x},\qquad \mathbf{x}(0)=\mathbf{x}_0\tag{1}\label{one} \end{align} where $$t\mapsto B(t)$$ is a maps from a neigborhood of $$0$$ into the space of matrices in $$\mathbb{R}^n$$. Let $$\phi(t;\mathbf{y})$$ denote the solution to $$\eqref{one}$$ with $$\phi(0;\mathbf{y})=\mathbf{y}$$. It is known that Wronskian $$W(t)=\operatorname{det}(\partial_{\mathbf{y}}\phi(t;\mathbf{y}))$$ satisfies \begin{align} \dot{W}=\operatorname{Trace}(B(t))W,\qquad W(0)=1\tag{2}\label{two} \end{align}

Define $$\phi(t;\mathbf{y)}=(I+tA)\mathbf{y}$$. Then $$\phi$$ satisfies the differential equations $$\dot{\mathbf{x}}=B(t)\mathbf{x},\qquad \mathbf{x}(0)=\mathbf{y}$$ in a small neginborhood around $$0$$ (small enough so that $$(I+tA)$$ remains invertible) where $$B(t)=A(I+tA)^{-1}$$.

It follows that $$W(t)=\operatorname{det}(I+tA)$$, and by equation \eqref{two}, \begin{align} \ddot{W}&=\operatorname{Trace}(\dot{B}(t))W + \operatorname{Trace}(B(t))\dot{W}\\ &=\operatorname{Trace}(\dot{B}(t))W + \Big(\operatorname{Trace}(B(t))\Big)^2W \end{align} By an application of the chain rule, $$\dot{B}(t)=-A\big(I+tA\big)^{-2}$$; hence $$\dot{B}(0)=-A^2$$. Putting things together, we obtain \begin{align} \ddot{W}(0)&=\operatorname{Trace}(\dot{B}(0))W + \Big(\operatorname{Trace}(B(0))\Big)^2W\\ &=- \operatorname{Trace}(A^2) +\Big(\operatorname{Trace}(A)\Big)^2 \end{align}