Let $T: V \to V$ for a finite-dimensional vector space $V$ be a linear operator whose matrix relative to the standard basis consists of all $1's$. find all eigenvalues and eigenvectors of $T$ and their geometric and algebraic multiplicities.
I didn't have much trouble finding the eigenvectors and eigenvalues, but I am not totally certain on the multiplicities. Here is what I have.
Let $A$ denote the matrix of $T$ relative to the standard basis, so $A_{ij} = 1$ for all $i,j$. If $v = (v_1, \ldots, v_n)$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $$Av = \begin{pmatrix} \sum\limits_{i} v_i \\ \vdots \\ \sum\limits_{i} v_i \end{pmatrix} = \lambda v = \begin{pmatrix} \lambda v_1 \\ \vdots \\ \lambda v_n \end{pmatrix}.$$ So $$\sum\limits_{i} v_i = \lambda v_k $$ for all $k$, so $$\lambda v_1 = \lambda v_2 = \ldots = \ldots,$$ which is true if and only if $\lambda = 0$ or $v_1 = v_2 = \ldots = v_n$. If $\lambda = 0$, then eigenvectors must satisfy $\sum\limits_{i} v_i = 0$. If $\lambda \neq 0$, then eigenvectors must satisfy $v_1 = \ldots = v_n$. Let $t$ equal this common term, then we have $\sum_i t = tn = \lambda t$, so $\lambda = n$.
So the eigenvalues are $\lambda_1 = 0$, in which case the sum of the eigenvectors must be $0$, and $\lambda_2 = n$, in which case all components of the eigenvector must sum to $1$.
By definition, the algebraic multiplicity is the dimension of the nullspace of $(T - \lambda I)^{\dim V}$ and the geometric multiplicity is the dimension of the nullspace of $T - \lambda I$. I cannot figure out how to compute these from my above work, though.
