$A\in M_n(\mathbb{R})$ 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
On
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 $tI-A$).
We can get rid of the annoying powers of $(-1)$ by writing $P(A) = \det(tI - A)$.
Note that if $Q$ is any invertible matrix then $P(A) = P(QAQ^{-1})$. So we may assume that $A$ is in Jordan canonical form. Also, since the determinant of a block matrix is just the product of the determinants of the blocks we can first study the case that $A$ is just one single Jordan block and then later use the product rule for derivatives to get the general case.
Finally, since the determinant of a Jordan block with $n$ times the eigenvalue $\lambda - t$ on the diagonal and $1$s on the first diagonal above it is just $(\lambda - t)^n$ we can use our knowledge about polynomials of this type to solve the problem and forget all about the fact that we were originally dealing with matrices.
I leave all the hairy details to you.