# Let T : $\Bbb{R}^n\to \Bbb{R}^n$ be the linear operator with $T(e_i) = (1,…,1)$ for all $i = 1,\ldots,n$. Find an eigenbasis for $T$.

I know that if the matrix is an $n\times n$ matrix, the eigenvalues will be $n$ with alg. multiplicity $1$ and $0$ with alg. multiplicity $n-1$. I am having a hard time generalizing the eigenbasis for $n$ because I can't generalize the pattern for $n=2,3,4$, etc. Thanks!

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An eigenvector for $n$ is obviously $\sum_i e_i$. For 0, you can take, for example, the vectors $\{e_1 - e_i \mid 2 \le i \le n \}$ as basis for $\ker T$.
It’s clear that $\langle 1,\dots,1\rangle$ is an eigenvector for the eigenvalue $n$. What about vectors in the eigenspace of the eigenvalue $0$? (Note that that’s just the kernel, or null space, of $T$.) Since
$$T(v)=\sum_{k=1}^nv_k\langle 1,\dots,1\rangle\;,$$
$T(v)=0$ if and only if $\sum_{k=1}^nv_k=0$. One simple way to get $n-1$ such vectors that are linearly independent of one another and of $\langle 1,\dots,1\rangle$ is to use the vectors $e_1-e_k$ for $k=2,\dots,n$.