Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A,B\in M_{n\times n}(\mathbb{R})$, prove that

$$(\frac{d}{dt}\det(A+tB))_{t=0} = Tr(cof(A)^{T}\cdot B), $$

where $cof(A)$ is cofactor matrix of $A$. I tried expanding it using row expansion, but it gets messed up. Any help?

share|improve this question
    
Maybe this could help: math.stackexchange.com/a/157072/19379 –  M Turgeon Sep 26 '12 at 1:19
add comment

2 Answers

up vote 2 down vote accepted

This identity can be computed directly from the definition of the determinant. It helps to regard $A$ as a matrix whose entries are differentiable functions with respect to the real parameter $t.$ Recall the definition of the determinant -

$\det A = \displaystyle\sum_{\sigma \in S_n} \textrm{ sgn } (\sigma ) \displaystyle\prod_{i=1}^n a_{i\sigma (i) }$

and that the cofactor matrix $A_{ij}$ is defined entrywise by the determinant of a the matrix formed by deleting row $i$ and column $j$; $\textrm{ Adj } A$ is the transpose of this matrix. Now, noting the product rule,

$\frac{d}{dt} \det (A(t) +tB) = \displaystyle\sum_{\sigma \in S_n } \textrm{sgn } (\sigma) \displaystyle\prod_{i=1}^n [a + tb]_{i\sigma (i)} = \displaystyle\sum_{\sigma \in S_n} \textrm{sgn }(\sigma ) \displaystyle\sum_{j=1}^n b_{j \sigma (j) } \displaystyle\prod_{j\ne i} [a + tb]_{i \sigma (i) }= \displaystyle\sum_{j=1}^n \displaystyle\sum_{\sigma \in S_n} b_{j\sigma (j)} \displaystyle\prod_{j\ne i} [a + tb]_{i \sigma (i) }$

Letting $t=0$ gives

$\displaystyle\sum_{j=1}^n \displaystyle\sum_{\sigma \in S_n} \textrm{sgn } (\sigma ) b_{j \sigma (j)} \displaystyle\prod_{j\ne i} a_{i \sigma (i) } = $

$\displaystyle\sum_{j=1}^n \displaystyle\sum_{k=1}^n b_{jk} \displaystyle\sum_{\sigma (j) = k } \textrm{sgn }(\sigma ) \displaystyle\prod_{j\ne i} a_{i \sigma (i) } = \displaystyle\sum_{j=1}^n \displaystyle\sum_{k=1}^n b_{jk} \det A_{jk} = \textrm{ tr }(\textrm{Adj }(A) B) $


What's sort of neat is that this identity can be used to prove Cramer's Rule. Note that as a special case, $\frac{d}{dt} (I+ tB)_{t=0} = \textrm{ Tr } (B)$ Hence

$\frac{d}{dt} \det (A +tB)_{t=0} =$ $\det (A) \frac{d}{dt} \det (I +tA^{-1}B)_{t=0} =$ $\det (A) \textrm{ Tr} (A^{-1} B)$

Note that the diagonal entries of $A^{-1} B$ are defined by $\displaystyle\sum_{k=1}^n a^{-1}_{ik} b_{ki}$ for some $1\le i \le n.$ If we take $B$ to be the matrix with $b_{ji}=1$ and zero entries elsewhere, we have $\textrm{ tr } A^{-1} B = a^{-1}_{ij}.$ Applying a similar remark to $\textrm{Adj }(A) B,$ we deduce that

$\det (A) A^{-1} = \textrm{ Adj } (A),$ as the matrices are equal entrywise.

share|improve this answer
    
How are diagonal entries of $A^{-1}B$ defined by $\sum_{k=1}^n{a_{ik}^{-1}b_{ki}}$? Don't you get something different? –  dmm Sep 26 '12 at 13:54
    
It's abusive notation, but $a_{ij}^{-1}$ here denotes the $ij%th entry of $A^{-1}.$ –  Tim Duff Sep 26 '12 at 17:13
add comment

In my answer here, I provide this definition of the determinant:

$$\begin{align*} \det(A) & = \sum_{\sigma\in S_n}(-1)^{\sigma}\prod_{i=1}^nA_{i,\sigma(i)} \end{align*}$$ (although I do not prove that this is equivalent to other standard definitions.) This definition is useful for things like this, but you have to be comfortable with the permutation group.

Now $$\begin{align*} \det(A+tB) & = \sum_{\sigma\in S_n}(-1)^{\sigma}\prod_{i=1}^n(A_{i,\sigma(i)}+tB_{i,\sigma(i)}) \end{align*}$$

Seeing as how we are going to evaluate at 0 after taking the derivative, the only terms that matter after expanding the product are the linear terms in $t$: $$\begin{align*} \partial_t\det(A+tB)|_{t=0} & = \partial_t\left.\left(\sum_{\sigma\in S_n}(-1)^{\sigma}\sum_{i=1}^n\left(tB_{i,\sigma(i)}\prod_{j\neq i}A_{i,\sigma(i)}\right)\right)\right|_{t=0}\\ & = \sum_{\sigma\in S_n}(-1)^{\sigma}\sum_{i=1}^n\left(B_{i,\sigma(i)}\prod_{j\neq i}A_{j,\sigma(j)}\right)\\ & = \sum_{i=1}^n\sum_{\sigma\in S_n}(-1)^{\sigma}\left(B_{i,\sigma(i)}\prod_{j\neq i}A_{j,\sigma(j)}\right)\\ \end{align*}$$

Meanwhile, $$ \begin{align*} \left(\operatorname{cof}(A)^T\cdot B\right)_{ii} & = \sum_{k=1}^n\operatorname{cof}(A)^T_{ik}B_{ki}\\ &=\sum_{k=1}^n\operatorname{cof}(A)_{ki}B_{ki}\\ &=\sum_{k=1}^n\sum_{\sigma\in S_n;\;\sigma(k)=i}(-1)^{\sigma+i+k}B_{ki}\prod_{j\neq k}A_{j,\sigma(j)}\\ &=\sum_{\sigma\in S_n}(-1)^{\sigma}B_{i,\sigma(i)}\prod_{j\neq k}A_{j,\sigma(j)}\\ \end{align*} $$

And so taking the trace of $\left(\operatorname{cof}(A)^T\cdot B\right)$ would be placing a $\sum_{i=1}^n$ in front of this and LHS=RHS.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.