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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.

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  • $\begingroup$ 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 $\endgroup$ – Elen Khachatryan Oct 10 '19 at 3:34
  • $\begingroup$ 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. $\endgroup$ – Elen Khachatryan Oct 10 '19 at 3:37
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    $\begingroup$ @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. $\endgroup$ – N. S. Oct 10 '19 at 3:42
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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 $$(\det(F(t)))'|_{t=0}=P'(0)+Q'(0)=tr(A)$$

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  • $\begingroup$ Although I have written an answer myself, I like your answer that much that I have to upvote it. :-) $\endgroup$ – amsmath Oct 10 '19 at 13:15
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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 $A$.

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  • $\begingroup$ "deriving with respect to $t$"? $\endgroup$ – JDZ Oct 10 '19 at 4:41
  • $\begingroup$ @JDZ What's your problem with that expression? $\endgroup$ – amsmath Oct 10 '19 at 13:17
  • $\begingroup$ math.stackexchange.com/questions/863148/derive-or-differentiate $\endgroup$ – JDZ Oct 10 '19 at 17:38
  • $\begingroup$ 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. $\endgroup$ – amsmath Oct 10 '19 at 17:43
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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.

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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 $t\to 0$ $$\eqalign{ X &\to I \\ \det(X) &\to \det(I) = {\tt 1} \\ {\rm Tr}(X^{-1}A) &\to {\rm Tr}(A) \\ \\ }$$

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