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I need to prove that characteristic polynomial of a square matrix $A$ is $$\det(A - rI) = \prod_{i = 1}^n(r - a_{i,i}) + q(r)$$

Where $q(r)$ is a polynomial of degree less than or equal to $n - 2$ and $a_{i,j}$ are the entries of matrix $A$.

Let $B = A - rI$ with entries $b_{i,j}$, then

$$\det B = \sum_{\sigma \in \operatorname{Perm}(n)}\operatorname{sign}(\sigma) \prod_{i = 1}^n b_{\sigma(i), i} $$

For $\tau \in \operatorname{Perm}(n)$ such that $\tau(i) = i$, we have

$$\det B = \sum_{\sigma \in \operatorname{Perm}(n) - \{\tau\}}\operatorname{sign}(\sigma) \prod_{i = 1}^n b_{\sigma(i), i} + \prod_{i=1}^n b_{\tau(i), i}$$

I know $b_{i,i} = a_{i,i} - r$ so $\prod_{i=1}^n b_{\tau(i), i} = \prod_{i = 1}^n(r - a_{i,i})$.

Now I just need to prove that $ \sum_{\sigma \in \operatorname{Perm}(n) - \{\tau\}}\operatorname{sign}(\sigma) \prod_{i = 1}^n b_{\sigma(i), i}$ is a $n-2$ degree polynomial in $r$.

I can see why this statement is true; If we have $\sigma \ne \tau$, we should at least have $\sigma(k) = j$ and $\sigma (j) = k$ for $j,k \le n$. So we lose two diagonal entries in the product $\prod_{i = 1}^n b_{\sigma(i), i}$, namely $b_{j,j}$ and $b_{k,k}$. Hence the degree of $\sum ...$ is $n-2$ atmost.

However this argument is hardly rigorous, how can I make this argument rigorous ?

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You are very close to a full proof, and you've stumbled upon all the right ideas. The result you want to make your argument rigorous is the following lemma.

Lemma: Let $S_n$ denote the symmetric group on $n$ letters. Let $e$ denote the identity permutation. Then any $\sigma \in S_n\backslash\{e\}$ has at most $n-2$ fixed points.

Proof: Let $\sigma \in S_n\backslash\{e\}$. Let $f_\sigma$ denote the number of fixed points of $\sigma$.

Since $\sigma$ is not the identity, it cannot fix every element, so we must have $f_\sigma < n$.

Now suppose that $f_\sigma = n-1$. Then $f_\sigma$ must fix all but one element, say $k$. Let $\sigma(k) = j$, where we have by assumption $k\neq j$. Since $k$ is the only non-fixed point, this means that $j$ must be a fixed point of $\sigma$. But this means that we have $$j=\sigma(j)= \sigma(k),$$ which contradicts the fact that $\sigma$ is a permutation, and hence a bijective map. It follows that $f_\sigma \neq n-1$ and hence we must have $f_\sigma \le n-2$. $\square$

Having this lemma should allow you to finish your proof.

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  • $\begingroup$ Amazing ! Thank you very much. $\endgroup$ – user8277998 Nov 18 '17 at 12:23

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