Find the eigenvalues and eigen vectors of $A=(a_{ij})$ where $a_{ii}=0$ and $a_{ij}=1;i\neq j$.

Question

Give the eigenvalues and eigen vectors of an $n\times n$ matrix $A=(a_{ij})$ where $a_{ii}=0$ and $a_{ij}=1;i\neq j$ . Is the matrix diagonalizable?

I've tested the matrix for $n=2$ and I get $x^2-1=0$ where $x_{1}=1$ and $x_{2}=-1$

for $n=3$ I get $-x^3+3x+2$ and the eigenvalues are $x_{1}=2$ and $x_{2}=-1$

and for $n=4$ I get $x^4-6x^2-8x-3$and the eigenvalues are $x_{1}=-1$ and $x_{2}=3$

I can't see any kind of pattern.

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

1

The matrix can be written as $J-I$ where $J$ is the all $1$ matrix.

The eigen values of $J$ are $n$ with multiplicity $1$ and $0$ with multiplicity $n-1$.(HOW ?)

(HOW ?)Note that $J$ has rank $1$ since every minor of order $i$ for $i\ge 2$ has $\det=0$. Hence $0$ is an eigen value of $J$ with multiplicity $n-1$ .

Now $\text{trace } J=n\implies $ the other non-zero eigen value of $J=n$

Hence the eigen values of $J-I$ are $n-1$ with multiplicity $1$ and $-1$ with multiplicity $n-1$.

NOTE: Every real symmetric matrix is diagonalizable

edited Jun 12 '20 at 10:38

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answered Nov 14 '17 at 17:01

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I still can't understand why are the eigenvalues of $J$ $n$? And how did you get this $1$ and $0$? – Zauberkerl Nov 14 '17 at 17:12
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@Zauberkerl; See if things are clear now – Learnmore Nov 14 '17 at 17:17
could you just explain a little bit more detailed the very last step? – Zauberkerl Nov 14 '17 at 17:19
@Zauberkerl; Note that $(J-I)^T=J-I$ – Learnmore Nov 14 '17 at 17:20

Find the eigenvalues and eigen vectors of $A=(a_{ij})$ where $a_{ii}=0$ and $a_{ij}=1;i\neq j$.

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