Finding eigenvalues. I'm working on the following problem:
Define $T \in L(F^n)$ (T an operator) by
$T(x_1,...,x_n) = (x_1+...+x_n,...,x_1+...+x_n)$
Find all eigenvalues and eigenvectors of $T$.
I've found that the eigenvalues of $T$ are $\lambda = 0$ and $\lambda = n$. Is there an easy way to prove that these are the only eigenvalues of $T$? Determining and solving the characteristic polynomial is messy for arbitrary $n$.
 A: Try this:  by direct computation,
$T^2 = nT, \tag{1}$
since every entry of $T^2$ is $n$.
So $m_T(x) = x^2 - nx$ is the minimal polynomial of $T$; every eigenvalue $\lambda$ of $T$ satisfies
$m_T(\lambda) = 0, \tag{2}$
so the only possibilities are $\lambda = 0$ and $\lambda = n$.
A: $|A|=0$
Product of eigen values $=0$
So atleast one eigen value is $0$
matrix is all one,So its rank is $1$(Echelon form)
As Casteels said,rank of a matrix is the number of non-zero eigenvalue
Also Sum of eigen value $ =n$
So other eigen value is $n$
A: The eigenvalues of $T$ are the same as the eigenvalues of any matrix representation of $T$. If you compute the matrix with respect to the standard basis of $F^n$, you find
$$
\begin{bmatrix}
1 & 1 & \dots & 1 \\
1 & 1 & \dots & 1 \\
\vdots & \vdots & \ddots & \vdots \\
1 & 1 & \dots & 1
\end{bmatrix}
$$
We easily see that this matrix has $0$ and $n$ as eigenvalues. There can be no more, because the geometric multiplicity of $0$ (dimension of the eigenspace relative to $0$, which in this case is the null space) is $n-1$, since the matrix has rank $1$.
