Diagonalize the matrix
(All ones along both diagonals, and zeros elsewhere.)
If is even the Eigen basis is
With associated eigenvalues 0(n / 2 times ) and 2(n / 2times ).
Consider the matrix as:
We can write A as A = , where
Now consider the transpose of the matrix as represented below:
We can see that the matrix is symmetric since .
Now evaluate as represented below:
The matrix is orthogonal since .
In general, if a matrix is symmetric, it is orthogonally diagonalizable and all its eigenvalues are real.
If it is also orthogonal, its eigenvalues must be 1 or -1 .
For a matrix, if n is even, then both eigen-values will be of multiplicity and if is odd, then the eigenvalue 1 will be of multiplicity and eigen-value -1 will be of multiplicity .
The Eigen value of will be +1 or -1 and when added with Eigen value of which is +1, therefore the eigen values are 0 and 2 with multiplicities .
The Eigen basis for the Eigen value is:
role="math" localid="1660204574030" And for the Eigen value 0 is :
If n is even the Eigen basis is as represented below:
With associated eigenvalues 0( n/2 times) and 2(n/ 2times) .
Therefore, is even the Eigen basis is
With associated eigenvalues 0 (n/ 2times) and 2(n / 2times) .
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