Suppose n is a positive integer and $T\in \mathcal L \left({F^n}\right)$ is difined by
$$T(x_1,…x_n)=(x_1+…+x_n,x_1+…+x_n)$$
In other words, T is the operator whose matrix (with respect to the standard basis) consists of all 1’s. Find all eigenvalues and eigenvectors of $T.$ Find all eigenvalues and eigenvectors of $T$.
I have a solution but I am trying to understand all the steps:
My attempt:
My first question is, how would we know that the matrix contains all 1's if it didn't tell us?
My proof: Let $\lambda$ be an eigenvalue of $T$. Then for $T\in \mathcal L \left({F^n}\right)$, $\lambda$ is an eigenvalue of T such that there exists a $v \in V$ for $v \neq 0$ such that $Tv= \lambda v$. Now from the equation in the question, $Tx=\lambda x$ we get
(1) $\lambda x_1=\lambda x_2=…\lambda x_n$ [my second question, how do we get this?]
I will write the rest of the proof after my questions are answered.
Best Answer
The matrix has only $1$s because of the formula(e) for $T(x_1,x_2,\dots,x_n)$.
If $v\neq 0$ is an eigenvector, there exists $\lambda$ such that $T(v)=\lambda v$, which means explicitly, coordinate by coordinate, that \begin{cases} x_1+x_2+\dots+x_n=\lambda x_1\\ x_1+x_2+\dots+x_n=\lambda x_2 \\ \vdots\\ x_1+x_2+\dots+x_n=\lambda x_n \end{cases} Thus $\;\lambda x_1=\lambda x_2=\dots=\lambda x_n$.
Now, either $\lambda =0$, and the corresponding eigenspace is the kernel of $T$, defined by the single equation $\;x_1+x_2+\dots+x_n=0$. It has dimension $n-1$.
Or $\lambda\ne 0$, and the above equation implies $\;x_1=x_2=\dots=x_n$. As one $x_i$ has to be non-zero, by definition of an eigenvector, they're all non-zero, and equal. We may suppose $x_i=1$ for all $i$,and the first series of equations simply reduces to $$1+1+\dots+1=\lambda\quad(n \text{ terms}).$$