# A limit with determinants

I'd be thankful if some could explain to me why the second equality is true... I just can't figure it out. Maybe it's something really simple I am missing?

$\displaystyle\lim_{\epsilon\to0}\frac{\det(Id+\epsilon H)-\det(Id)}{\epsilon}=\displaystyle\lim_{\epsilon\to0}\frac{1}{\epsilon}\left[\det \begin{pmatrix} 1+\epsilon h_{11} & \epsilon h_{12} &\cdots & \epsilon h_{1n} \\ \epsilon h_{21} & 1+\epsilon h_{22} &\cdots \\ \vdots & & \ddots \\ \epsilon h_{n1} & & &1+\epsilon h_{nn} \end{pmatrix}-1\right]$

$\qquad\qquad\qquad\qquad\qquad\qquad=\displaystyle\sum_{i=1}^nh_{ii}=\text{trace}(H)$

• @angryavian: A bit confusing but $Id$ is short for Identity. – Alex R. Jun 9 '16 at 18:23
• One possible approach: Group terms in the expansion of the determinant by powers of $\varepsilon$. You can show that the determinant is $$1+\varepsilon \sum_{i=1}^n h_{ii} + O(\varepsilon^2),$$ which is exactly what you need. – Nicholas Stull Jun 9 '16 at 18:25
• (specifically, use the Leibniz expansion). – Omnomnomnom Jun 9 '16 at 18:26
• Relevant post from Terrence Tao's blog – Omnomnomnom Jun 9 '16 at 18:28
• @Omnomnomnom, I went ahead and posted it. There are a couple of details missing, but I think it has enough included in order to expand it into a rigorous proof rather easily if needed. – Nicholas Stull Jun 9 '16 at 20:15

As I suggested in my comment, we proceed by expanding $$|Id + \varepsilon H| = |A| = \left| \begin{array}{cccc} 1+\varepsilon h_{11} & \varepsilon h_{12} & \cdots & \varepsilon h_{1n}\\ \varepsilon h_{21} & 1+\varepsilon h_{22} & \cdots & \ \\ \vdots & \ & \ddots & \ &\\ \varepsilon h_{n1} & \ & \ & 1+\varepsilon h_{nn} \end{array} \right|$$ in powers of $\varepsilon$. First, we have: $$|A| = (1+\varepsilon h_{11}) \left| \begin{array}{cccc} 1+\varepsilon h_{22} & \varepsilon h_{23} & \cdots & \varepsilon h_{2n}\\ \varepsilon h_{32} & 1+\varepsilon h_{33} & \cdots & \ \\ \vdots & \ & \ddots & \ &\\ \varepsilon h_{n2} & \ & \ & 1+\varepsilon h_{nn} \end{array} \right| + \varepsilon\sum_{j=2}^n (-1)^{1+j} h_{1j} \det(A_{1j})$$ where $\det(A_{1j}) = O(\varepsilon)$ (here, I am using the usual notation $A_{1j}$ to be the matrix obtained by deleting the first row and $j$th column from $A$).

Justification of this part: The minimal power of $\varepsilon$ would occur when the maximal number of diagonal terms is included. In this case (dealing with $(n-1)\times (n-1)$ minors, this would mean $n-2$ diagonal terms, since all terms of the cofactor expansion along the first row (with the exception of the first one, which I am separating from the rest of the computation) would exclude the $1+\varepsilon h_{11}$, as well as the diagonal entry that sits in the $j$th column. Finally, if there are $n-2$ diagonal terms multiplied together (in the minimal case), then there must be one off-diagonal term, introducing the (claimed) factor of $\varepsilon$.

We hence conclude that $$|A| = (1+\varepsilon h_{11}) \left| \begin{array}{cccc} 1+\varepsilon h_{22} & \varepsilon h_{23} & \cdots & \varepsilon h_{2n}\\ \varepsilon h_{32} & 1+\varepsilon h_{33} & \cdots & \ \\ \vdots & \ & \ddots & \ &\\ \varepsilon h_{n2} & \ & \ & 1+\varepsilon h_{nn} \end{array} \right| + O(\varepsilon^2)$$

Continuing (inductively) to expand the determinant in this manner, we see that \begin{align} |A| &= \prod_{j=1}^n (1+\varepsilon h_{jj}) + O(\varepsilon^2)\\ &= 1+\varepsilon \sum_{j=1}^n h_{jj} + O(\varepsilon^2) \end{align} which exactly yields the desired equality, since this immediately says $$|A|-1 = \varepsilon \sum_{j=1}^n h_{jj} + O(\varepsilon^2),$$ hence \begin{align} \lim_{\varepsilon \to 0} \frac{1}{\varepsilon}(|A|-1) &= \lim_{\varepsilon\to 0} \frac{1}{\varepsilon} \left( \varepsilon \sum_{j=1}^n h_{jj} + O(\varepsilon^2) \right)\\ &= \lim_{\varepsilon\to 0} \left( \sum_{j=1}^n h_{jj} + O(\varepsilon) \right)\\ &= \sum_{j=1}^n h_{jj} = \text{Trace}(H) \end{align}

• Thank you very much! Exacly what I was looking for – 35T41 Jun 9 '16 at 21:20

Here is a conceptual proof that avoids expanding a complicated determinant:

The determinant of a linear operator (or of a square matrix) is the product of the eigenvalues, counting multiplicity. The trace of a linear operator (or of a square matrix) is the sum of the eigenvalues, counting multiplicity.

Now suppose that $\lambda_1, \dots, \lambda_n$ are the eigenvalues of $H$, counting multiplicity. Then the eigenvalues of $I + \epsilon H$ are $$1 + \epsilon \lambda_1, \dots, 1 + \epsilon \lambda_n,$$ counting multiplicity. Thus $$\det(1 + \epsilon H) = (1 + \epsilon \lambda_1) \cdots (1 + \epsilon \lambda_n).$$ It is now clear that \begin{align*} \lim_{\epsilon\to 0} \frac{ \det(1 + \epsilon H) - 1}{\epsilon} &= \lambda_1 + \dots + \lambda_n\\ &= \text{trace } H, \end{align*} as desired.

• A very nice alternative approach to this question (and much easier to express rigorously than using the Leibniz Expansion in retrospect). – Nicholas Stull Jun 11 '16 at 5:41
• Thank you, this is indeed a nice approach, but H doesn't have to be a diagnolizable matrix – 35T41 Jun 11 '16 at 7:20
• My proof above does not require $H$ to be diagonalizable because each operator (or square matrix) on an $n$-dimensional complex vector space has exactly $n$ eigenvalues, counting multiplicity. – Sheldon Axler Jun 11 '16 at 15:22

Hint:

$$\left|\begin{matrix} 1+\epsilon h_{11}&\epsilon h_{12}\\ \epsilon h_{21}&1+\epsilon h_{22}\\ \end{matrix}\right|=1+\epsilon h_{11}+\epsilon h_{22}+\epsilon^2\left(h_{11}h_{22}-h_{12}h_{21}\right).$$

Only the main diagonal generates terms in $\epsilon$. This generalizes to higher order, for instance using the expansion by minors.

actually, you're trying to calculate the differential $\phi$ of the function $\det : M_n(\mathbb{R}) \rightarrow \mathbb{R}$ at $I_n$, which is defined as $$\forall H \in M_n(\mathbb{R}), \phi(H) = \lim\limits_{t \to 0} \dfrac{\det(I_n+tH)-\det(I_n)}{t}= \dfrac{d}{dt}_{t=0} \det(I_n+tH)$$

First, since $\det$ is a polynomial, it is differentiable at any point and $\phi$ is a linear form on $M_n(\mathbb{R})$. Therefore, it suffices to calculate its value on the canonical basis of $M_n(\mathbb{R})$.

• For $H=E_{i,i}$, $\det(I_n+tE_{i,i})=1+t$, so $\phi(E_{i,i})=1$.
• For $H=E_{i,j},$ where $i \neq j, \det(I_n+tE_{i,j})=1$, so $\phi(E_{i,j})=0$.

So, $\phi=\textrm{Trace}$, since these two linear forms coincide on a basis of $M_n(\mathbb{R})$.

• Maybe I'm misunderstanding but, how did you conclude $\phi$ is a linear form? The determinant is multilinear in columns. It just so happens that the answer is indeed linear. – Alex R. Jun 9 '16 at 18:50

2. One way to prove the problem statement using Laplace/cofactor expansion, but easier to understand than other similar approaches and without creating a mess, is to use total derivative. Let $$g(t_{11},t_{12},\ldots,t_{nn})=\det\pmatrix{ 1+t_{11}h_{11}&t_{12}h_{12}&\cdots&t_{1n}h_{1n}\\ t_{21}h_{21}&1+t_{22}h_{22}&\cdots&t_{2n}h_{2n}\\ \vdots&&\ddots&\vdots\\ t_{n1}h_{n1}&t_{n2}h_{n2}&\cdots&1+t_{nn}h_{nn}},$$ where each $t_{ij}=t_{ij}(\epsilon)=\epsilon$ (yes, here each $t_{ij}$s is actually identical to $\epsilon$, but we'd like to view it as a function of $\epsilon$). Let also $f(\epsilon)=g\left(t_{11}(\epsilon),t_{12}(\epsilon),\ldots,t_{nn}(\epsilon)\right)$. Then $$f'(\epsilon)=\sum_{i,j}\frac{\partial g}{\partial t_{ij}}\frac{\partial t_{ij}}{\partial \epsilon}=\sum_{i,j}\frac{\partial g}{\partial t_{ij}}$$ and in turn $$f'(0)=\sum_{i,j}\left.\frac{\partial g}{\partial t_{ij}}\right|_{t_{ij}=0}.$$ However, to evaluate $\left.\frac{\partial g}{\partial t_{ij}}\right|_{t_{ij}=0}$, all other $t_{rs}$ are treated as zero and we are effectively finding the derivative of $g$ with respect to a single variable $t_{ij}$. By Laplace expansion along the $i$-row, we see that $$\left.\frac{\partial g}{\partial t_{ij}}\right|_{t_{ij}=0}= \begin{cases} h_{ii}&\text{ if } i=j,\\ 0&\text{ otherwise}. \end{cases}$$ Therefore $f'(0)=\sum_i h_{ii}=\operatorname{trace}(H)$.