How to calculate the largest eigenvalue of a matrix where every entry is 1.

Question

I have an $n\times n$ matrix $A$ where every entry is equal to $1$. Using $\det(A-(\lambda)I)=0$ is the method I normally use to find eigenvalues, but I am unsure how that will help in this situation.

Answer 1

In this case it is easier to find out the eigenvalue directly. Note that if $v = (1,1, \cdots, 1)$, then $Av = n v$. If $w$ is orthogonal to $v$, then $Aw=0$. Thus $A$ is eigenvalue $n$ with eigenvector $v$, and eigenvalue $0$ with multplicity $n-1$ (The eigenspace is the orthogonal complement of $\langle v\rangle$.

Answer 2

John's answer is the most intuitive. Another approach:

With a bit of work, we can show inductively $$\det(A - \lambda I_n) = (-1)^n(\lambda^n - n\lambda^{n-1})$$ which has roots $\lambda = n$ and $\lambda = 0$ with multiplicity $n-1$.