Jacobi's Derivative of the Determinant I've been given the following theorem for the derivative of the determinant of a matrix:
"Let $A\in \mathbb{R}^{n\times n}$ be a square matrix. Then the Fréchet derivative of det$: \mathbb{R}^{n\times n}\to \mathbb{R}$ at $A$ is given by: 
d det(A)B = tr(adj(A)B)."
What I don't understand is what $B$ actually is. Where does it come from? Why did the theorem not follow up with "where $B$ is..." 
Could anyone please explain this to me? 
Thanks in advance!
 A: The Frechet derivative $dF(x)$ of a function $F$ defined on a Banach space $X$ is defined as a (continous) linear functional on $X$, i.e. a mapping $dF(x):X\to\mathbb{R}$, or, more precisely $dF(x)\in X^*$ ($dF(x)$ is an element in the dual space of $X$).
In your case you have $\det:\mathbb{R}^{n\times n}\to \mathbb{R}$, i.e.  $X = \mathbb{R}^{n\times n}$. The dual space of $X$ is equal to $X$, and hence $d\det(A)$ is a linear mapping from $\mathbb{R}^{n\times n}$ to $\mathbb{R}$ and it is precisely what you've written: $d\det(A)B = \mathrm{tr}(A^TB)$.
Another way to put it: $B$ is the direction in which you take the derivative, i.e. the limit of
$$
\frac{\det(A + tB) - \det(A)}{t}
$$
for $t\to 0$.
A: Note that your function $$ \det: \mathbb{R}^{n \times n } \rightarrow \mathbb{R} $$
Thus its frechet derivative $D\det :  \mathbb{R}^{n \times n } \rightarrow \mathit{L} (\mathbb{R}^{n \times n },  \mathbb{R} )  $, where  $\mathit{L} (\mathbb{R}^{n \times n },  \mathbb{R} )  $ is the vector space of all linear maps(here it is forms) from $\mathbb{R}^{n \times n } $  to $ \mathbb{R}$.   Hence, for any $ A  \in \mathbb{R}^{n \times n }  $, we have  $ D\det(A) \in \mathit{L} (\mathbb{R}^{n \times n },  \mathbb{R} )   $, i.e.  $ D\det(A) $ is a linear map $$ D\det(A) : \mathbb{R}^{n \times n } \rightarrow \mathbb{R}$$
Thus for any $B \in  \mathbb{R}^{n \times n } $, $ D\det(A) B $ is well defined and is a real number ($\in \mathbb{R}$). Hope this clarify the idea for you, and any question  is welcomed.
A: The derivative may be defined on functions mapping one normed vector space to another.  The idea is always that the derivative of a function $f$ at a point $x$ in its domain provides a linear approximation to the behavior of the function.  That is, the derivative $f'(x)$ is the linear operator in the approximation formula
\[
f(x+h) = f(x) + f'(x)h + E(x;h),
\]
where $E$ is an error function of order $||h||^2$.
In one-dimensional calculus, $f$ maps the real line into itself, so
$f'(x)$ may be thought of as a number, the slope of a tangent line. It could also be regarded as a linear operator mapping the real line to itself, the operator of multiplication by the number.
In vector calculus, $f$ might map $\mathbb{R}^m$ to $\mathbb{R}^n$.  A linear operator mapping $\mathbb{R}^m$ to $\mathbb{R}^n$ may be represented by multiplication by an $n \times m$ matrix. (This is where $h$ might be considered an $m$-vector, a "direction".)
In your case, $f(A) = \det A$, a function that maps the vector space of real $n \times n$ matrices into $\mathbb{R}$.   Therefore, its derivative is a linear operator mapping the space of $n \times n$ matrices into $\mathbb{R}$.
The answer to your question is: the $B$ in the formula is a generic real $n \times n$ matrix.  (The approximation is good when $||B||$ is small enough.)
That raises the question of the representation of the derivative of the determinant.  It can't simply be represented as square matrix multiplication.  With the elements of $B$ given by $(b^\nu_\mu)$, and elements of the linear transformation $f'(x)$ given by $(t^\mu_\nu)$, something like  $f'(x)B = \sum_{\mu,\nu} t_\mu^\nu b^\mu_\nu$ does the job.
I should point out that, in some restricted cases, there are simpler representations of the derivative of the determinant. For example, in the space of $n\times n$ conformal matrices, that is, matrices $A$ for which $A^T A = \alpha^2 I$, where $\alpha$ is a real number.
