Let the entries of $n \times n$ matrix $A$, where $n \ge 3$, be $$a_{ij} = \frac{((-1)^{i+j}+1)}{2}$$ Find all eigenvalues and multiplicity of $A$.

By nullity I know that $\lambda=0$ has multiplicity $n-2$, but the two other eigenvalues are hard to calculate. I know that those two eigenvalues sum up with $n$ by the trace. How could I calculate the rest eigenvalues without just predicting the eigenvalue? (E.g. Plugging $\lambda=\frac{n}{2}$)

  • $\begingroup$ "By nullity, I know $\lambda=0$ has multiplicity $2$" How? What is the rank of $A$? $\endgroup$ Jun 2 '20 at 3:11
  • $\begingroup$ @above that's a typo. It's $n-2$ $\endgroup$
    – Hypernova
    Jun 2 '20 at 3:50
  • $\begingroup$ Ok thanks for the clarificaiton $\endgroup$ Jun 2 '20 at 4:12

Let $v=\left(-1,1,-1,1,\ldots,(-1)^n)\right)^T$ and $e=(1,1,\ldots,1)^T$. Then $A=\frac{vv^T+ee^T}{2}=\frac12\pmatrix{v&e}\pmatrix{v&e}^T$. Therefore $A$ has the same multi-set of nonzero eigenvalues as $$ B=\frac12\pmatrix{v&e}^T\pmatrix{v&e}=\frac12\pmatrix{n&\frac{(-1)^n-1}{2}\\ \frac{(-1)^n-1}{2}&n}, $$ i.e. its eigenvalues are $\frac12\left[n\pm\frac{(-1)^n-1}{2}\right]$ and $n-2$ copies of zeros.


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