TITLE: derivative of principal submatrix
QUESTION [0 upvotes]: $A\in M_n(\mathbb{R})$ and $A_i\in M_{n-1}(\mathbb{R})$ be the principal submatrix of A otained by deleting $i^{th}$ row and collumn of $A$.
I need to show ${d\over dt}P(A(t))=\sum_{i=1}^{n}P(A_i(t))$  where 
$P(A(t))=(-1)^n\det(A-tI)$ and $P(A_i(t))=(-1)^{n-1}\det(A_i-tI)$
I have no idea how to proceed. thankx

REPLY [1 votes]: The proof by Jordan blocks is fine, although it needs a bit of care as we are working over $\mathbb{R}$ and not $\mathbb{C}$.
But the result is far less deep than that.  
Look at the Laplace expansion of the determinant whose entries $a_{ij}(t)$ depend on $t$. When we differentiate this, each of the products is replaced by a sum of $n$ terms got by differentiating the various factors. Gathering up the terms sensibly it is clear that the derivative is the sum of $n$ determinants, each got by replacing a row of the original determinant by its derivative $a'_{ij}(t)$. 
Now all is easy; in the case at hand only the diagonal entry gives a non-zero derivative, and that derivative is $1$. (I use @Vincent's sensible suggestion that it is always better to look at $tI-A$).
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