I have been working on this problem for a couple hours and am completely stuck. Any help would be greatly appreciated.

Let $A$ be the $n \times n$ matrix which has zeros on the main diagonal and ones everywhere else. Find the eigenvalues and eigenvectors of $A$.

  • $\begingroup$ what have you done? what does $A-\lambda I$ look like? $\endgroup$ – symplectomorphic May 23 '14 at 3:22
  • $\begingroup$ I have found out that λ1 = n-1 and λ2 through λn = -1, but I don't know how to prove that. $\endgroup$ – John May 23 '14 at 3:31
  • $\begingroup$ Hint $A+I$ is a matrix with rank $1$. This means the eigenspace of $-1$ has dimension $n-1$. $\endgroup$ – Pedro Tamaroff May 23 '14 at 3:33

Note that if $\lambda=-1$, then $$ A-\lambda I = \begin{bmatrix} 1 & \dotsc & 1 \\ \vdots & \ddots & \vdots \\ 1 & \dotsc & 1 \end{bmatrix} $$ But $A-\lambda I$ has rank $1$, so $\lambda_1=-1$ is an eigenvalue of $A$ with geometric multiplicity $n-1$. The eigenspace $E_{\lambda_1}$ is the nullspace of this matrix. Can you find it? Hint: it has dimension $n-1$.

Now, note that $\operatorname{trace}A=0$ so $\lambda_2=n-1$ is the other eigenvalue of $A$. The eigenspace $E_{\lambda_2}$ is the nullspace of $A-\lambda_2 I$. Can you find it too?

  • $\begingroup$ "${\rm rank}(A)=0$" should be ${\rm trace}(A)=0$, yes? Only a zero matrix has rank $0$. $\endgroup$ – David May 23 '14 at 3:39
  • $\begingroup$ whoops trace yes $\endgroup$ – Brian Fitzpatrick May 23 '14 at 3:39
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    $\begingroup$ Your $\lambda_2$ is clearly wrong unless $n=2$. $\endgroup$ – Ted Shifrin May 23 '14 at 3:42
  • $\begingroup$ Ok I think all the typos have been corrected, correct? $\endgroup$ – Brian Fitzpatrick May 23 '14 at 3:44

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