0

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 $\lambda=\frac{n}{2}$)

Rodrigo de Azevedo
  • 1
Hypernova
  • 742

1 Answers1

1

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.

user1551
  • 139,064