# Characteristic polynomial with determinant of A-$t*I_n$

Let A be nxn matrix whose jth column is $$\vec e_{j+1}$$ for $$j and whose last column has entries $$-c_0,-c_1,...,-c_{n-1}$$, as below:

$$\begin{matrix} 0 & 0 & ... & -c_0 \\ 1 & 0 & ... & -c_1 \\ 0 & 1 & ... & -c_2 \\ ... & ... &... \\ 0 &...&1& -c_{n-1} \\ \end{matrix}$$

Viewing t as a variable, show that $$det(A - t*I) = (-1)^n(c_0 + c_1t + ... + c_{n-1}t^{n-1} + t^n)$$.

Biscally, if I compute determinant, statring from first row, I will have $$(-1)^nc_0 + (-t)*det(A-tI)_{11}$$, where $$(A-tI)_{11}$$ is $$A-tI$$ with first row and column removed. Then if I continue to compute $$(-t)*det(A-tI)_{11}$$, I will have $$(-1)^{n-1}t C_1$$ + another determinant.

Keep repeating the same process, I will end up with the answer. But is there a more elegant way to solve this question instead of directing solving the determinant? Thanks in advance.

There is a more elegant way. You first show that the claim holds for $$2\times 2$$ matrices. Which is easy enough to see. So now we assume that the claim holds for all $$(n-1)\times(n-1)$$ matrices of the given form. Formalizing what we mean of course, that is, we assume that a matrix of the form $$\begin{bmatrix}0&0&0&\ldots&-c_1\\-1&0& 0&\ldots&-c_1\\0&-1&0&\ddots&\vdots\\0&\ldots&\ddots&&\vdots\\0&\ldots&0&-1&-c_{n-1}\end{bmatrix}$$ has a determinate of the form $$(-1)^{n-1}(c_1+c_2t+\ldots+c_{n-1}t^{n-2}+t^{n-1}).$$
So now when we start your argument, instead of saying we repeat our argument, or something to that effect, we can say that by assumption (sometimes people say by induction, or inductive step) $$\det(A-t*I)_{11}=(-1)^{n-1}(c_1+c_2t+\ldots+c_{n-1}t^{n-2}+t^{n-1}),$$ hence $$\det(A-t*I)=(-1)^{n}(c_0+c_1t+\ldots+c_{n-1}t^{n-1}+t^{n-1}).$$
This is a standard argument using the Principle of Mathematical Induction. There are many ways to formulate these type of arguments, but essentially it works because we show it works for a 'smallest' case, and then show that if it works for one case, then it works for the next largest. Since our claim works for $$n=2$$ our argument shows it works for $$n=3.$$ Since our claim works for $$n=3$$ it works for $$n=4,$$ and so on and so forth.
Some other ways to use 'Induction' or formalize the argument, would be to say that suppose our claim didn't work. Well then it must not work for some smallest $$n>2$$. Well then your argument would allow us to deduce that it must also work for $$n-1$$, a contradiction, hence it must hold for all $$n$$. This uses the fact that we are working with counting numbers, and every set of counting numbers has a smallest element. This is sometimes called the Well Ordering Principle.
Another variant is basically well ordering, but I like it more for purely subjective reasons (I think it's cool sounding). We assume that it doesn't work for some $$n>2$$. By our argument if the claim held for $$n-1,$$ then it would hold for $$n,$$ hence it doesn't hold for $$n-1$$. Well then $$n-1>2,$$ and thus $$n-2>2$$. This tells us that there are infinitely many numbers bigger than $$2$$ and less than $$n.$$ Impossible, hence our claim must hold for $$n.$$ This is know as Infinite Descent.