Why does $D \det(I)B=\frac{d}{d t}\det(I+t B)|_{t=0}$ hold? The following problem is from Taylor's PDE I. I do not get why $D \det(I)B=\frac{d}{d t}\det(I+t B)|_{t=0}$ hold. Since $\det(I)=1$ and $D$ is $n\times n$ matrix, the left side seems to be a matrix, while the right side is clearly a scalar. Could anyone tell me how I should correctly interpret $\det(I)$ in this case?

 A: Hoping I'm not completely off-base here!  I think that mjqxxx had it right in his comment, i.e. that $D \det(I)$ refers to the linear map from $M_n(\Bbb C) \to \Bbb C$ which is the derivative of $\det: M_n(\Bbb C) \to \Bbb C$ at the point $I \in M_n(\Bbb C)$; then $D\det(I) B$ makes sense as the directional derivative of $\det$ in the $B$ direction at the point $I$, so that
$D\det(I)B = \dfrac{d}{dt}(\det(I + tB) \mid_{t = 0} \tag{0}$
is in fact a legitimate notation.
Having said these things,  and clearly understanding the symbolism, I must say that I became quite intrigued by the "other question" presented in the body of this post, so I went ahead and answered it as well.
Soooooooo, in case that anyone is interested: 
Let $\lambda_i \in \Bbb C$, $1 \le i \le n$, be the $n$ eigenvalues of $B$; note if need be we allow multiple occurances of the same complex number in the list $\lambda_1$, $\lambda_2$, $\ldots$, $\lambda_n$ to accommodate repeated eigenvalues of $B$.  Bearing this convention in mind, we recall that the determinant of a complex matrix is the product of its eigenvalues; we have
$\det(B) = \prod_1^n \lambda_i.  \tag{1}$
We next observe that, for any complex square matrix $C$ and any eigenvalue $\mu$ of $C$, with eigenvector $v \ne 0$,
$Cv = \mu v, \tag{2}$
for any complex numbers $a, b \in \Bbb C$ we have
$(aC + bI)v = aCv + bIv = a\mu v + bv = (a\mu + b)v; \tag{3}$
that is,  the eigenvalues of $C$ themselves are transformed in accord with the linear mapping $C \mapsto aC + bI$,
whilst the eigenvectors are unaffected; furthermore, if $v$ is a generalized eigenvector, that is
$Cv - \mu v = w, \tag{4}$
where $w \ne 0$ is also a generalized eigenvector, then
$(aC + bI)v - (a\mu + b)v = aCv + bv - a\mu v - bv$
$= aCv - a\mu v  = a(Cv - \mu v) = aw; \tag{5}$
for $a \ne 0$, (5) indicates that $v$ is a generalized eigenvector corresponding to eigenvalue $a\mu + b$ of $aC + bI$; these considerations lead to the conclusion that the entire invariant subspace structure of $C$ is preserved under $C \mapsto aC + bI$, provided that $a \ne 0$; thus the multiplicities of the transformed eigenvalues $a\mu + b$ are the same as those of the original $\mu$.
We apply the above remarks to the matrix $B$; it follows that the eigenvalues of $I + tB$ are precisely the numbers $1 + t\lambda_i$; thus
$\det(I + tB) = \prod_1^n (1 + t\lambda_i). \tag{6}$
The product on the right-hand side of this equation may be expanded in terms of the elementary symmetic polynomials $\sigma_k(t\lambda_1, t\lambda_2, \ldots, t\lambda_n)$ in the $n$ quantities $t\lambda_i$, yielding
$\det(I + tB) = \prod_1^n (I + t\lambda_i) = \sum_0^n \sigma_k(t\lambda_1, t\lambda_2, \ldots, t\lambda_n); \tag{7}$
a detailed proof of (7) may be found in my answer to this question.  Each $\sigma_k(t\lambda_1, t\lambda_2, \ldots, t\lambda_n)$ is in fact a homogeneous polynomial of degree $k$ in the $\lambda_i$; thus (7) becomes
$\det(I + tB) = \prod_1^n (I + t\lambda_i) = \sum_0^n t^k \sigma_k(\lambda_1, \lambda_2, \ldots, \lambda_n)$
$= 1 + t\sigma_1(\lambda_1, \lambda_2, \ldots, \lambda_n) + t^2\sigma_2(\lambda_1, \lambda_2, \ldots, \lambda_n) + \ldots + t^n\sigma_n(\lambda_1, \lambda_2, \ldots, \lambda_n); \tag{8}$
it follows immediately from (8) that
$\dfrac{d}{dt}(\det(I + tB) \mid_{t = 0} = \sigma_1(\lambda_1, \lambda_2, \ldots, \lambda_n); \tag{9}$
since
$\sigma_1(\lambda_1, \lambda_2, \ldots, \lambda_n) = \sum_1^n \lambda_i = \text{Tr}(B), \tag{10}$
we may combine (9) and (10) to arrive at
$\dfrac{d}{dt}(\det(I + tB) \mid_{t = 0} = \text{Tr}(B), \tag{101}$
the result desired (by Yours Truly in any event).  QED!
Note:  One reason I find this problem so engaging is that I think it the may very well lead to nice proofs of things like Liouville's formula and so forth; I have always found the usual demonstrations (see the wikipedia page I just linked) rather cumbersome. End of Note.
A: suppose the eigenvalues of $B$ are $\lambda_1, \lambda_2, \cdots, \lambda_n.$  then  the eigenvalues of $I + tB$ are $1 + t\lambda_1, 1 + t\lambda_2, \cdots, 1 + t\lambda_n.$  we will use the fact that the determinant of a matrix is the product of its eigenvalues.
now we have $$det(I + tB) = (1 + t\lambda_1) (1 + t\lambda_2) \cdots (1 + t\lambda_n) = det(I)+ t(\lambda_1 + \lambda_2 \cdots +\lambda_n) + \cdots  $$
therefore $$\frac{d}{dt}det(I + tB)\big|_{t = 0}  = trace(B).$$
