# Find eigenspace dimension and eigenvector of this matrix

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.

• 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? – 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 $A.(1,1,1,\ldots,1)=\bigl((n-1)a+b\bigr)(1,1,1,\ldots,1)$.
• @Poujh I don't know what a “rref” is. $(1,1,1,\ldots,1)$ is simply a vector. And $A$ times this vector is $\bigl((n-1)a+b,(n-1)a+b,,\ldots,(n-1)a+b\bigr))=\bigl((n-1)a+b\bigr)(1,1,\ldots,1)$. – José Carlos Santos Jan 19 '18 at 21:22