linear algebra - What are the eigenvalues of matrix that have all elements equal 1?

As in subject: given a matrix $A$ of size $n$ with all elements equal exactly 1.

What are the eigenvalues of that matrix ?

Answer

Suppose $\,\begin{pmatrix}x_1\\x_2\\...\\x_n\end{pmatrix}\,$ is an eigenvector of such a matrix corresponding to an eigenvalue $\,\lambda\,$, then

$$\begin{pmatrix}1&1&...&1\\1&1&...&1\\...&...&...&...\\1&1&...&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\...\\x_n\end{pmatrix}=\begin{pmatrix}x_1+x_2+...+x_n\\x_1+x_2+...+x_n\\.................\\x_1+x_2+...+x_n\end{pmatrix}=\begin{pmatrix}\lambda x_1\\\lambda x_2\\..\\\lambda x_n\end{pmatrix}$$

One obvious solution to the above is

$$W:=\left\{\begin{pmatrix}x_1\\x_2\\..\\x_n\end{pmatrix}\;;\;x_1+...+x_n=0\right\}\,\,\,,\,\,\lambda=0$$

For sure, $\,\dim W=n-1\,$ (no need to be a wizard to "see" this solution since the matrix is singular and thus one of its eigenvalues must be zero)

Other solution, perhaps not as trivial as the above but also pretty simple, imo, is

$$U:=\left\{\begin{pmatrix}x_1\\x_2\\..\\x_n\end{pmatrix}\;;\;x_1=x_2=...=x_n\right\}\,\,\,,\,\,\lambda=n$$

Again, it's easy to check that $\,\dim U=1\,$ .

Now, just pay attention to the fact that $\,W\cap U=\{0\}\,$ unless the dimension of the vector space $\,V\,$ we're working on is divided by the definition field's characteristic (if you're used to real/complex vector spaces and you aren't sure about what the characteristic of a field is disregard the last comment)

Thus, assuming this is the case, we get $\,\dim(W+U)=n=\dim V\Longrightarrow V=W\oplus U\,$ and we've thus found all the possible eigenvalues there are.

BTW, as as side effect of the above, we get our matrix is diagonalizable.