How do i prove that $\det(tI-A)$ is a polynomial? In wikipedia, it's said "$\det(tI-A)$ can be explicitly evaluated using exterior algebra", but i have not learned exterior algebra yet and i just want to know whether it is polynomial, not how it looks like.
How do i prove that $\det(tI-A)$ is a polynomial in $\mathbb{F}[t]$ where $\mathbb{F}$ is a field and $A$ is an $n\times n$ matrix?
 A: To prove that $\det (tI-A)$ is a polynomial you must know some definition, or some properties, of the determinant. The most straightforward and least mystical approach is to use Laplace's formula:  http://en.wikipedia.org/wiki/Laplace_expansion
This would give a rather quick way of proving that $\det (tI-A)$ is a polynomial, and for almost the same amount of work, that it is a monic polynomial of degree $n$ (if $A$ is an $n\times n$ matrix). 
A: Difficulties arise if you try to use induction to prove that $\det(tI - A)$ is a monic polynomial of degree $n$.
However, the following result is amenable to induction:
Theorem: If $A$ and $B$ are $n$-by-$n$ matrices, then $\det(tB - A) = \det(B) t^n + p(t)$, where $p$ is a polynomial of degree at most $n-1.$
Proof. Proceed by induction on $n$; the result is clear when $n=1$, so, for the induction-step, assume that the result holds for all $k$-by-$k$ matrices, where $k \geq 1$.
Recall that if $M$ is a matrix (square, or otherwise), then $M_{ij}$ denotes the submatrix of $M$ formed by deleting the $i$th-row and $j$-th column of $M$; moreover, if $M$ and $N$ are $m$-by-$n$ matrices, and $\alpha$ and $\beta$ are scalars, then $\left[ \alpha M + \beta N \right]_{ij} = \alpha M_{ij} + \beta N_{ij}$.
If $A$ and $B$ are $(k+1)$-by-$(k+1)$ matrices, then, following a Laplace-expansion along row $i$,
\begin{align}
\det(tB - A) 
&= \sum_{j=1}^{k+1} (-1)^{i+j} (tb_{ij} - a_{ij}) \det([tB - A]_{ij}) \\
&= \sum_{j=1}^{k+1} (-1)^{i+j} (tb_{ij} - a_{ij}) \det(tB_{ij} - A_{ij}) \tag{from above} \\
&= \sum_{j=1}^{k+1} (-1)^{i+j} (tb_{ij} - a_{ij}) \left[\det(B_{ij})t^{k} + p_{ij} (t) \right] \\
&= \sum_{j=1}^{k+1} (-1)^{i+j}  b_{ij} \det(B_{ij}) t^{k+1}+\sum_{j=1}^{k+1} (-1)^{i+j} \left[b_{ij} tp_{ij} (t) - a_{ij}\det(B_{ij})t^{k} - a_{ij}p_{ij}(t) \right] \\
&= \det(B)t^{k+1} +q_{i}(t),
\end{align}
where $$q_i (t) := \sum_{j=1}^{k+1} (-1)^{i+j}\left[b_{ij} tp_{ij} (t) - a_{ij}\det(B_{ij})t^{k} - a_{ij}p_{ij}(t) \right].$$
Since $\deg(p_{ij}) \leq k-1$, it follows that $\deg(q_i) \leq k$.
The result now follows by the principle of mathematical induction.
Corollary: If $A$ is an $n$-by-$n$ matrix, then $\det(tI - A) = t^n + q(t)$, where $\deg(q) \leq n-1$.
Remark: The constant term of $p_{A,B} (t):= \det(tB - A)$ is $(-1)^n \det(A)$ since $p_{A,B} (0):= \det(0B - A) = \det(-A) = (-1)^n \det(A)$.
