# Derivate of the cofactor and the determinant


Let $A(t)$ be a smooth path in $M_d(\mathbb{R})$, $A(0)=A,\dot A(0)=B$.

By differentiation the identity $(\Cof A)^T \circ A=\Det A \cdot \id$, one gets

$$(*) \, \, \big(d(\Cof)_A(B)\big)^T \circ A + (\Cof A)^T \circ B = \tr (\Cof A)^T B) \cdot \id= \langle \Cof A , B\rangle \cdot \id$$

(The derivative of the determinant is known as Jacobi's formula).

From this, at least in the case when $A$ is invertible we can deduce that

$$\big(d(\Cof)_A(B)\big)^T = \big(\langle \Cof A , B\rangle \cdot \id -(\Cof A)^T \circ B \big)A^{-1},$$ hence

$$d(\Cof)_A(B) = (A^{T})^{-1}\big(\langle \Cof A , B\rangle \cdot \id - B^T \circ \Cof A \big)$$

Questions:

1. Does equation $(*)$ uniquely determine $d(\Cof)_A(B)$ even when $A$ is singular? Is there a closed formula for $d(\Cof)_A(B)$ in this case?

2. Is there a more "direct way" to calculate $d(\Cof)_A(B)$? (without relying on Jacobi's formula)

Remark: As a corollary from $(*)$ we get $\tr \bigg( \big(d(\Cof)_A(B)\big)^T \circ A \bigg) = (d-1) \tr (\Cof A)^T B)$

This is interesting since the cofactor essentially measures the action of the linear map $A$ on $d-1$ dimensional parallelepiped, so maybe there is a "geometric" way to see this immediately.

Let $Cof_{ij}(A)=(-1)^{i+j}\det(\Delta_{ij}(A))$, where $\Delta_{ij}(A)$ is the submatrix obtained from $A$ by deleting row $i$ and column $j$. Notice that $\Delta_{ij}(\cdot)$ is a linear transformation. Let $Adj(D)$ be the adjugate matrix of $D$.

Let $\{e_1,\ldots,e_d\}$ be the canonical basis of $\mathbb{R}^d$.

Now, $d(Cof)_A(e_me_n^t)=\lim_{h\rightarrow 0}\dfrac{Cof(A+h e_me_n^t)-Cof(A)}{h}$.

$Cof_{ij}(A+h e_me_n^t)=(-1)^{i+j}\det(\Delta_{ij}(A+h e_me_n^t))=$

$(-1)^{i+j}\det(\Delta_{ij}(A)+h \Delta_{ij}(e_me_n^t))= (-1)^{i+j}\det(\Delta_{ij}(A)+h\ uv^t)=$

By the matrix determinant lemma,

$=(-1)^{i+j}\det(\Delta_{ij}(A))+(-1)^{i+j}h\ (v^tAdj(\Delta_{ij}(A))u)$

$=(-1)^{i+j}\det(\Delta_{ij}(A))+(-1)^{i+j}h\ tr(Adj(\Delta_{ij}(A))uv^t)$

$=(-1)^{i+j}\det(\Delta_{ij}(A))+(-1)^{i+j}h\ tr(Adj(\Delta_{ij}(A))\Delta_{ij}(e_me_n^t))$

$=Cof_{ij}(A)+h\ (-1)^{i+j}\ tr(Adj(\Delta_{ij}(A))\Delta_{ij}(e_me_n^t))$.

Thus, $d(Cof)_A(e_me_n^t)=((-1)^{i+j}\ tr(Adj(\Delta_{ij}(A))\Delta_{ij}(e_me_n^t)))_{i,j}$.

Since $B=\sum_{m,n=1}^db_{m,n}e_me_n^t$ then $d(Cof)_A(B)=((-1)^{i+j}\ tr(Adj(\Delta_{ij}(A))\Delta_{ij}(B)))_{i,j}$.

• Thanks. I do not understand some details, though: What are $u,v$ (perhaps $e_i,e_j$)? Also, do you know if my equation $(*)$ determines $d(\Cof)_A(B)$ when $A$ is singular? – Asaf Shachar Feb 13 '17 at 16:40
• $\Delta_{ij}(e_{m}e_n^t)$ is a rank one matrix. I called it $uv^t$. Let me think about your second question. – Daniel Feb 13 '17 at 16:42
• If $rank(A)\leq d-2$ then $Cof(A)=0$ and your formula becomes $\big(d(\Cof)_A(B)\big)^t \circ A =0$. I cannot see a way to determine $\big(d(\Cof)_A(B)\big)^t$ from the equality $XA=0$, there are infinitely many solutions. – Daniel Feb 13 '17 at 16:54
• You are right. Thanks again! – Asaf Shachar Feb 13 '17 at 17:00
• You are welcome. – Daniel Feb 13 '17 at 17:03

As mentioned by Daniel, when $$\operatorname{rank}(A) \le d-2$$, $$\operatorname{Cof}(A)=0$$, hence equation $$(*)$$ becomes

$$\big(d(\operatorname{Cof})_A(B)\big)^T A=0$$

However, the equation $$XA=0$$ has infinitely many solutions, as can be seen as follows:

Look at the linear map $$T:M_d \to M_d$$, defined by $$T(X)=X^TA$$. $$T$$ is not surjective, since every matrix in its image has rank smaller than $$d-1$$. Thus, $$T$$ is not injective.