Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question

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 $\ker T$.

share|cite|improve this answer

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 $k=2,\dots,n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.