Proving Identity for Derivative of Determinant For a square matrix $A$ and identity matrix $I$, how does one prove that $$\frac{d}{dt}\det(tI-A)=\sum_{i=1}^n\det(tI-A_i)$$ Where $A_i$ is the matrix $A$ with the $i^{th}$ row and $i^{th}$ column vectors removed?
 A: One way to prove this claim is to take the matrix $tI-A$ and replace the first $t$ on the main diagonal with $t_1$, the second one with $t_2$ etc. Let the resulting determinant be $p(t_1,t_2,\ldots,t_n)$. Then
$$
\det(tI-A)=p(t,t,\ldots,t),
$$
and by the chain rule
$$
\frac{d}{dt}\det(tI-A)=\frac{\partial}{\partial t_1}p(t,t,\ldots,t)+\ldots+\frac{\partial}{\partial t_n}p(t,t,\ldots,t).
$$
The partial derivative w.r.t $t_i$ can be simply calculated as $\det(tI-A_i)$ e.g. using the determinant expansion along the $i$-th column.
A: Here is one way to see this:
Note that the map $\phi(t_1,...,t_n) = \det ( \sum_k t_k e_k e_k^T -A)$ is smooth, and if $\tau(t) = (t,....,t)$ then $f(t)=\det (tI-A) = \phi(\tau(t))$.
In particular, $f'(t) = \sum_k {\partial \phi(t,....,t) \over \partial t_k}$.
If we adopt the notation $\det B = d(b_1,...,b_n)$, where $b_k$
is the $k$th column of $B$, we have
\begin{eqnarray}
\phi(t,...,t+\delta,...t) &=& d(te_1-a_1,..., \delta e_k + t_ke_k -a_k,...,te_n -a_n) \\
&=& \phi(t,...,t) + \delta d(te_1-a_1,..., e_k,...,te_n -a_n) \\
&=& \phi(t,...,t) + \delta \det (tI-A_k)
\end{eqnarray}
and so ${\partial \phi(t,....,t) \over \partial t_k} = \det (tI-A_k)$.
