Relation between the characteristic polynomial $A_{n\times n}$ and that of $A_{n-1\times n-1}$? I was trying to find the minimal polynomial of the matrix $A \in M_{n\times n}(\mathbb{R})$ where $A$ has every element $A_{ij}=1$
and saw that the characteristic follows the pattern $t^2-2t, -(t^3-3t^2)...$ and assume it would be easy to prove the characteristic would be $(-1)^n(t^n-nt^{n-1})$ with mathematical induction on $n$, but I couldn't.
Is there any formula that connects $\det(A_{n\times n}-tI)$  and $\det(A_{n-1\times n-1}-tI)$ ? Because that would make the proof easier, if not any hints on how to show that?
 A: Hint:
The expansion of the characteristic polynomial of an $n\times n$ matrix $A$ is
$$\det (XI_n-A)=X^n+\sum_{i=1}^n(-1)^{i} f_i(A) X^{n-i},$$
where $f_i(A)$ is the sum of all principal minors of order $i$ of the matrix $A$.
Can you compute these minors in the present case?
A: You have correctly guessed what the characteristic polynomial is. When matrices are very symmetric like this, the follow idea usually works: let $A$ be your matrix of size $n\times n$ that has all $1$s, and let $A(t) = tI - A$
Note I am not introducing $-t$ to avoid the signs you see in your computation.
If you add all rows to the first, you will end up the following matrix:
$$\begin{pmatrix}
t-n & t-n & \cdots & t-n \\
-1 & t-1 & \cdots & -1 \\
\vdots & \vdots & \ddots & \vdots \\
-1 & -1 & \cdots & t-1
\end{pmatrix}
$$
so you can extract $t-n$ and be left with $(t-n)\cdot \det B(t)$ where  $B(t)$ is the following matrix:
$$\begin{pmatrix}
1 & 1 & \cdots & 1 \\
-1 & t-1 & \cdots & -1 \\
\vdots & \vdots & \ddots & \vdots \\
-1 & -1 & \cdots & t-1
\end{pmatrix}
$$
If you add this first row to the other $n-1$ rows, you will obtain a matrix that has is diagonal except for the first row:
$$\begin{pmatrix}
 1 & 1 & \cdots & 1\\
0 & t & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & t
\end{pmatrix}
$$
The determinant of this is $t^{n-1}$, and hence your determinant (without the alternating sign) is $t^{n-1}(t-n)$.
