Given is the following nxn matrix for $n >1$:

$$ \begin{pmatrix} b & a & .. & a \\ a & b & ... & : \\ : & ... & ... & a \\ a & .. & a & b \end{pmatrix} $$

a) Show that $b-a$ is an eigenvalue.
b) Determine the dimension of the eigenspace for $b-a$.
c) Find an eigenvector for an eigenvalue different from $b-a$

I have already shown a). For b), I think that the dimension is $n-1$, given that there are $n-1$ "free variables". Is that correct ? And for c), I don't really see how to proceed. Thanks for your help.

  • $\begingroup$ Hint: the matrix consisting of all ones is of rank 1 and has eigenvalues $0$, corresponding to the eigenspace $span\{(1,-1,0,0,\dots),(1,0,-1,0,\dots),(1,0,0,-1,\dots),\dots\}$ and $n$ corresponding to eigenspace $span\{(1,1,1,1,\dots)\}$. How does your matrix relate to some multiple of the matrix of all ones and the identity matrix? $\endgroup$ – JMoravitz Jan 19 '18 at 20:57

b) The dimension of the eigenspace associated with $b-a$ is $n-1$ because it contains $n-1$ linearly independente vectors:$$(1,-1,0,0,\ldots,0),(1,0,-1,0,\ldots,0),\ldots,(1,0,0,0,\ldots,-1).$$The dimension cant-t be $n$ because otherwise the matrix would be a diagonal one.

c) If $A$ is your matrix, then $A.(1,1,1,\ldots,1)=\bigl((n-1)a+b\bigr)(1,1,1,\ldots,1)$.

  • $\begingroup$ Thanks a lot ! But I don't really understand the answer for c). The (1,1,...,1) refer to the rref where we only have "a" on the first row, is that right ? If so, why do we multiply with ((n-1)a+b) ? n-1 refers to the number of dimension of the eigenspace for b), is that right ? But how does that help us for c) ? $\endgroup$ – Poujh Jan 19 '18 at 21:11
  • $\begingroup$ @Poujh I don't know what a “rref” is. $(1,1,1,\ldots,1)$ is simply a vector. And $A$ times this vector is $\bigl((n-1)a+b,(n-1)a+b,,\ldots,(n-1)a+b\bigr))=\bigl((n-1)a+b\bigr)(1,1,\ldots,1)$. $\endgroup$ – José Carlos Santos Jan 19 '18 at 21:22
  • $\begingroup$ Thanks a lot ! (By rref, I meant reduced row echelon form, sorry if it wasn’t clear) $\endgroup$ – Poujh Jan 19 '18 at 21:29

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