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How to find eigenvalues of following matrices of order $n$?

$$A=\begin{bmatrix} 1 & 1 & 1 & 1 & \cdots & 1 \\ 1 & 0 & 1 & 1 & \cdots & 1 \\ 1 & 1 & 0 & 1 & \cdots & 1 \\ 1 & 1 & 1 & 0 & \cdots & 1 \\ \vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\ 1 & 1 & 1 & 1 & \cdots & 0 \\ \end{bmatrix}_n$$

$$B=\begin{bmatrix} -1 & 1 & 1 & 1 & \cdots & 1 \\ 1 & 0 & 1 & 1 & \cdots & 1 \\ 1 & 1 & 0 & 1 & \cdots & 1 \\ 1 & 1 & 1 & 0 & \cdots & 1 \\ \vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\ 1 & 1 & 1 & 1 & \cdots & 0 \\ \end{bmatrix}_n$$

I had apply matrix calculator to find eigenvalues and I found that exactly $n-2$ eigenvalues of both matrices are $-1$,but I was not able to find rest of the eigenvalues.

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Using row reduction, $$ \det A=\begin{vmatrix} 1 & 1 & 1 & 1 & \cdots & 1 \\ 1 & 0 & 1 & 1 & \cdots & 1 \\ 1 & 1 & 0 & 1 & \cdots & 1 \\ 1 & 1 & 1 & 0 & \cdots & 1 \\ \vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\ 1 & 1 & 1 & 1 & \cdots & 0 \\ \end{vmatrix}_n =\begin{vmatrix} 1 & 1 & 1 & 1 & \cdots & 1 \\ 0 & -1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & -1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & -1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots &\ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & -1 \\ \end{vmatrix}_n=(-1)^{n-1} $$ One can see explicitly that there are $n-2$ linearly independent eigenvectors for the eigenvalue $-1$. Indeed if we solve the system $(A+I)x=0$, we easily get $$ x_1=0; \ \ x_2+x_3+\cdots+x_n=0 $$ (so the dimension of the solution space, the eigenvectors of $-1$, is $n-2$). We know from $\det A=(-1)^{ n-1}$ that the product of the eigenvalues is $(-1)^{n-1}$. So, as $A $ is real and symmetric (thus selfadjoint) the geometric multiplicities agree with the algebraic ones; we then have $$ (-1)^{n-1}=\lambda_1\lambda_2\,(-1)^{n-2}. $$ It follows that $\lambda_1\lambda_2=-1$.

We also know that the trace of $A$ is $1$; so $$ 1=\lambda_1+\lambda_2+(n-2)(-1)=\lambda_1\lambda_2+2-n. $$ Thus $$ \lambda_1+\lambda_2=n-1,\ \ \lambda_1\lambda_2=-1. $$ This is a quadratic, and one finds that $$ \lambda_1=\frac{n-1+\sqrt{n^2-2n+5}}2, \ \ \ \ \ \lambda_2=\frac{n-1-\sqrt{n^2-2n+5}}2 $$

The case for $B$ is similar. We get the same $n-2$ dimensional subspace of eigenvectors of $-1$. For the other two eigenvalues, working with the trace and the determinant we get $$ \det B=(-1)^{n-1}\,(2n-3) $$ Thus $$ \lambda_1\lambda_2\,(-1)^{n-2}=(-1)^{n-1}\,(2n-3),\ \ \lambda_1+\lambda_2=-1. $$ This reduces to $$ \lambda_1\lambda_2=3-2n,\ \ \lambda_1+\lambda_2=-1. $$ One then obtains $$ \lambda_1=-\frac{1+\sqrt{8n-11}}2,\ \ \ \ \ \lambda_2=-\frac{1-\sqrt{8n-11}}2. $$

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  • $\begingroup$ I got everything but how can I write the proof to show that algebraic multiplicity of $-1$ is $n-2$? $\endgroup$ – kalpeshmpopat May 11 '16 at 8:39
  • $\begingroup$ Both matrices are real and symmetric, so selfadjoint; the algebraic multiplicity agrees with the geometric multiplicity. $\endgroup$ – Martin Argerami May 11 '16 at 12:31

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