# A_n is a matrix in size n xx n. Diagonal entry of A are 0 and the other entry are -1. How to determine eigenvalue of A_n ?

Apr 13, 2017

From your question, we have the matrix

${A}_{n} = \left[\begin{matrix}0 & - 1 & \cdots & \cdots & - 1 \\ - 1 & 0 & - 1 & \cdots & \vdots \\ \vdots & - 1 & \ddots & \text{ " & vdots \\ vdots & " " & " } & \ddots & - 1 \\ - 1 & \cdots & \cdots & - 1 & 0\end{matrix}\right]$

Using the determinant definition of:

$| \lambda {I}_{n} - {A}_{n} | = 0$,

we have:

$| \lambda {I}_{n} - {A}_{n} | = | \left(\lambda , 1 , \cdots , \cdots , 1\right) , \left(1 , \lambda , 1 , \cdots , \vdots\right) , \left(\vdots , 1 , \ddots , \text{ ", vdots), (vdots, " "," } , \ddots , 1\right) , \left(1 , \cdots , \cdots , 1 , \lambda\right) | = 0$

For simplicity, let us consider a $3 \times 3$ matrix

${A}_{3} = \left[\begin{matrix}0 & - 1 & - 1 \\ - 1 & 0 & - 1 \\ - 1 & - 1 & 0\end{matrix}\right]$

with the determinant...

$| \left(\lambda , 1 , 1\right) , \left(1 , \lambda , 1\right) , \left(1 , 1 , \lambda\right) | = 0$

If you diagonalize this matrix, you should get $\left[\begin{matrix}- 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right]$.

If you evaluate this as a determinant, you should easily get $\lambda = - 2 , 1 , 1$.

The $2 \times 2$ equivalent is $| \left(\lambda , 1\right) , \left(1 , \lambda\right) |$, which evaluates as $\lambda = \pm 1$.

If you notice, the eigenvalues all add up to zero, while the trace of the matrix is also zero.

Thus, the trace of the matrix is the sum of the eigenvalues.

This is true because diagonalizing the matrix preserves the eigenvalues, and the eigenvalues of a diagonal matrix can be obtained straight from treating each diagonal entry as its own matrix block set equal to $0$.

It does not, however, mean that all the eigenvalues are zero. It just means that whatever they are, they add up to zero. As Cesareo has mentioned, a general formula for this is that

${\left(\lambda - 1\right)}^{n - 1} \left(\lambda + n - 1\right) = 0$

is the characteristic equation for ${A}_{n}$.

Apr 15, 2017

Alternative approach

$A = \left(\begin{matrix}0 & - 1 & - 1 & \cdots \\ - 1 & 0 & - 1 & \cdots \\ - 1 & - 1 & 0 & \cdots \\ \vdots & \vdots & \vdots & \ddots\end{matrix}\right) = A ' + I$ where ${\left(A '\right)}_{i j} = - 1$

For possible eigenvalues $\left({\lambda}_{1. . . n}\right)$ of $A$ that satisfy $D = \det \left(A - \lambda I\right) = 0$, we can say that:

$D = \det \left(A ' - \left(\lambda - 1\right) I\right)$

$= \det \left(A ' - \lambda ' I\right) = 0$

We are now looking for the eigenvalues for:

$A ' = - \left(\begin{matrix}1 & 1 & 1 & \cdots \\ 1 & 1 & 1 & \cdots \\ 1 & 1 & 1 & \cdots \\ \vdots & \vdots & \vdots & \ddots\end{matrix}\right) = - \left(\begin{matrix}1 \\ 1 \\ 1 \\ \ldots\end{matrix}\right) \left(1 , 1 , 1 , \ldots\right)$

A' is singular of rank 1 and has $n - 1$ zero eigenvalues.

Because $A = m a t h b f u m a t h b f {v}^{T}$, it's other non-zero eigenvalue is $m a t h b f {v}^{T} m a t h b f u$:

$\lambda ' = - \left(1 , 1 , 1 , \cdots\right) \left(\begin{matrix}1 \\ 1 \\ 1 \\ \vdots\end{matrix}\right) = - n$

The corresponding eigenvalues for $A$, ie from $\lambda = \lambda ' + 1$, are therefore:

$\lambda = 1 , 1 - n$