Alternative proof that $\mbox{tr}(A)$ is an eigenvalue of a rank-$1$ matrix

Question

I had the an assignment to prove that $\mbox{tr}(A)$ is an eigenvalue of $A$ if $\mbox{rank}(A)=1$. I was able to prove it the following way:

Let $A \in M_n(\mathbb{R})$ be a matrix such that $rankA=1$, then there exists a vector $v \in V$ so that every column of $A$ equals $cv$ for some $c \in \mathbb{R}$. In that case, we assert that for every vector $u \in V$ the following occurs $Au = \sum_{i=1}^nu_ic_iv \in Sp\{v\}$. Therefore, $v$ is an eigenvector of $A$. Now all that's left is to find that $Av=tr(A)v$ by working the indices.

Here is the problem: I rather dislike the proof, as it relies on "charging ahead" mindlessly with eigenvectors. Are there any alternative ways to prove this?

Answer 1

If $rank(A)=1 \implies\dim(N(A))=n-1$ then we have $n-1$ zero eigenvalues $\lambda_1,\lambda_2,...,\lambda_{n-1}$ and $Tr(A)=\sum \lambda_i=\lambda_n$.

Answer 2

The rank is $1$ implies that $Im(f)=Vect(v)$, where $f$ is the linear endomorphism whose matrix is $A$ in the canonical basis. let $(v,f_1,...,f_{n-1})$ be a basis of $\mathbb{R}^n$, $A(v)=cv, A(f_i)=c_iv$, the only non zero term of the matrix of $f$ in the basis $(v,f_1,..,f_{n-1})$ is $c$ this implies that $tr(A)=c$ since the trace does not depend of the basis.