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$\newcommand{\Cof}{\operatorname{Cof}}$ $\newcommand{\Det}{\operatorname{Det}}$ $\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\tr}{\operatorname{Tr}}$

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.

2 Answers 2

1

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

  • 0
    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? – 2017-02-13
  • 0
    $\Delta_{ij}(e_{m}e_n^t)$ is a rank one matrix. I called it $uv^t$. Let me think about your second question. – 2017-02-13
  • 0
    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. – 2017-02-13
  • 0
    You are right. Thanks again! – 2017-02-13
  • 0
    You are welcome. – 2017-02-13
1

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.