I have a matrix of size $n \times n$, where diagonal entries are $n-1$ and non-diagonal entries are $2n-2$. I have solved for $2 \times 2$ and $3 \times 3$ matrices, but not getting for $n \times n$ matrix. Now I want to find its characteristic equation and it's polynomial. Please, any suggestions.
For the matrix $n \times n$, where diagonal entries zero and non-diagonal entries are $2n-2$, I have got the value $4(n-1)^2$.
On
Since this may be a homework question, I will work it out for the specific case $n=4$; the general case follows as an exercise. The following solution uses the fact that determinants are not affected by adding one row to another.
The characteristic polynomial is
$$p(x)=\det{\begin{pmatrix}3-x&6&6&6\\6&3-x&6&6\\6&6&3-x&6\\6&6&6&3-x\end{pmatrix}}$$
Subtract the last row from each row:
\begin{align*}p(x)&=\det{\begin{pmatrix}-3-x&0&0&3+x\\0&-3-x&0&3+x\\0&0&-3-x&3+x\\6&6&6&3-x\end{pmatrix}}\\ &=(3+x)^3\det{\begin{pmatrix}-1&0&0&1\\0&-1&0&1\\0&0&-1&1\\6&6&6&3-x\end{pmatrix}}\\ &=(3+x)^3\det{\begin{pmatrix}-1&0&0&1\\0&-1&0&1\\0&0&-1&1\\0&0&0&21-x\end{pmatrix}}\end{align*} In the last determinant, 6 times each row is added to the last row. Hence
$$p(x)=(x+3)^3(x-21)$$
For an $n\times n$ matrix, the polynomial should work out to $(-1)^n(x+n-1)^{n-1}(x-(n-1)(2n-1))$.
Let's try a simpler looking problem:
Consider the matrix $$A = \begin{pmatrix} 2 & 1 & 1 & 1 \\ 1 & 2 & 1 & 1 \\ 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 2 \end{pmatrix}.$$
We know that if $\lambda_1, \dots, \lambda_m$ are the eigenvalues of $A$ then its characteristic polynomial looks something like $f(x) = (x - \lambda_1)^{k_1} \cdots (x-\lambda_m)^{k_m}$.
Some suggestions: 1. Find what the eigenvalues are.
Find what dimensions that the eigenspaces are (i.e. how many linearly independent eigenvectors you have for a given eigenvalue).
Hopefully the eigenspaces give you enough information so you can try to construct the characteristic polynomial from that.